Lecture 13: Differentiation Derivatives of Trigonometric Functions

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Lecture 13: Differentiation Derivatives of Trigonometric Functions"

Transcription

1 Lecture 13: Differentiation Derivatives of Trigonometric Functions Derivatives of the Basic Trigonometric Functions Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions 1/25

2 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. 2/25

3 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. 2/25

4 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin 2/25

5 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) 2/25

6 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y 2/25

7 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is 2/25

8 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is cos (x ± y) 2/25

9 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is cos (x ± y) = cos x cos y sin x sin y 2/25

10 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have 3/25

11 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f (x) 3/25

12 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h 3/25

13 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h 3/25

14 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h 3/25

15 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h 3/25

16 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h Recall that using the Squeeze Theorem we prove that sin x lim x 0 x 3/25

17 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h Recall that using the Squeeze Theorem we prove that sin x lim = 1 x 0 x 3/25

18 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim h 0 h 4/25

19 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h 4/25

20 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = x sin x 4/25

21 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = sin x = sin x (0) + cos x (1) x 4/25

22 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = sin x = sin x (0) + cos x (1) = cos x x 4/25

23 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x 4/25

24 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x You use the sum formula for cos to prove the corresponing ifferentiation formula for cos x, which is x cos x 4/25

25 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x You use the sum formula for cos to prove the corresponing ifferentiation formula for cos x, which is cos x = sin x x 4/25

26 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are 5/25

27 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) 5/25

28 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x 5/25

29 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = 5/25

30 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x 5/25

31 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. 5/25

32 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x 5/25

33 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) 5/25

34 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) x ( π2 x ) 5/25

35 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) ( π2 x ) x = sin x( 1) 5/25

36 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) ( π2 x ) x = sin x( 1) = sin x 5/25

37 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x 6/25

38 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x 6/25

39 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x 6/25

40 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x 6/25

41 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 6/25

42 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) cos 2 x 6/25

43 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x cos 2 x 6/25

44 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) cos 2 x = cos2 x + sin 2 x cos 2 x = 1 cos 2 x 6/25

45 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x = ( 1 cos x cos 2 x ) 2 = 1 cos 2 x 6/25

46 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x = ( 1 cos x cos 2 x ) 2 = sec 2 x = 1 cos 2 x 6/25

47 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x 7/25

48 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 x 7/25

49 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x 7/25

50 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 x ) 2 = 1 + ( sin x cos x = cos2 x cos 2 x + sin2 x cos 2 x 7/25

51 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x 7/25

52 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x 7/25

53 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x 7/25

54 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x 7/25

55 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x 7/25

56 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x = sec 2 x 7/25

57 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x = sec 2 x Most text books use the sec 2 x formula for the erivative of tan x, but Maple an other symbolic ifferentiating programs use the 1 + tan 2 x formula. 7/25

58 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. 8/25

59 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: 8/25

60 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: x sin u 8/25

61 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x 8/25

62 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x cos u 8/25

63 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x u cos u = sin u x x 8/25

64 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u u cos u = sin u x x 8/25

65 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u cos u = sin u x x 8/25

66 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x x sec u u cos u = sin u x x 8/25

67 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x 8/25

68 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x x csc u 8/25

69 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x u csc u = csc u cot u x x 8/25

70 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x x cot u u cos u = sin u x x u csc u = csc u cot u x x 8/25

71 Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x x cot u = csc2 u u ( ) x = 1 + cot 2 u u x u cos u = sin u x x u csc u = csc u cot u x x 8/25

72 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Example 46 Differentiating with Trig Functions Fin an simplify the inicate erivative(s) of each function. (a) (b) (c) Fin f (x) an f (x) for f (x) = x 2 cos (3x). Fin s t for s = cos t sin t + cos t. ( Fin C (x) for C(x) = tan e ) 1+x 2. 9/25

73 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) 10/25

74 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] 10/25

75 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) 10/25

76 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] 10/25

77 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) 10/25

78 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) = 2 cos (3x) 6x sin (3x) 6x sin (3x) 9x 2 cos (3x) 10/25

79 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) = 2 cos (3x) 6x sin (3x) 6x sin (3x) 9x 2 cos (3x) ( = 2 9x 2) cos (3x) 12x sin (3x) 10/25

80 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s t 11/25

81 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 11/25

82 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 11/25

83 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 11/25

84 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 1 = (sin t + cos t) 2 11/25

85 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 1 = (sin t + cos t) 2 This example illustrates the fact that when simplifying erivatives involving trig functions, you sometimes nee to use stanar trigonometric ientities. 11/25

86 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with 12/25

87 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) 12/25

88 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) 12/25

89 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) 12/25

90 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, 12/25

91 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) 12/25

92 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, 12/25

93 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) 12/25

94 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, 12/25

95 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) 12/25

96 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 12/25

97 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) 12/25

98 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) 12/25

99 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) ( = sec 2 1+x 2 e ) ( ) ( ) ( e 1+x x 2) 1/2 (2x) 12/25

100 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with where C(x) = f (g (h (k(x)))) f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) ( = sec 2 1+x 2 e ) ( ) ( ) ( e 1+x x 2) 1/2 (2x) ( = xe 1+x 2 sec 2 ) e 1+x x 2 12/25

101 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Example 47 Dampe Oscillations Consier the function q(t) = e 7t sin (24t) This function escribes ampe simple harmonic motion. It gives the position of a mass attache to a spring relative to the equilibrium (resting) position of the spring. A frictional force acts to graually slow the mass. (a) Fin q (t) an q (t) an explain their meaning in terms of the ampe oscillatory motion. (b) Note that q(0) = 0. This means that the initial position of the mass is at the equilibrium position of the spring. Fin the initial velocity of the mass. Also fin the velocity when the mass first returns to the equilibrium position. (c) Draw a graph of the function q(t). () Fin the first two times when the oscillating mass turns aroun. Show the corresponing points on the graph of q(t). (e) Show that the function q(t) satisfies the ifferential equation 2 q q + 14 t2 t + 625q = 0 13/25

102 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives 14/25

103 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) 14/25

104 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] 14/25

105 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] 14/25

106 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. 14/25

107 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives 14/25

108 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives q (t) 14/25

109 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) 14/25

110 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) ) ] sin (24t) 14/25

111 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) ) ] sin (24t) = e 7t [336 cos (24t) sin (24t)] 14/25

112 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) ) ] sin (24t) = e 7t [336 cos (24t) sin (24t)] The rate of change of velocity is acceleration, so this gives the acceleration of the mass at time t. 14/25

113 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) 15/25

114 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] 15/25

115 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 15/25

116 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when. 15/25

117 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. 15/25

118 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 15/25

119 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t 15/25

120 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 15/25

121 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v 24 15/25

122 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] 24 15/25

123 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/ /25

124 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/24 = /25

125 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/24 = This velocity is less than the initial velocity an in the opposite irection. 15/25

126 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown /25

127 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. As we saw in the last part, not only oes the amplitue ecrease, but so oes the velocity /25

128 Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. As we saw in the last part, not only oes the amplitue ecrease, but so oes the velocity. Further, note that where the graph is concave own, q (t) < 0, the mass is ecelerating. The velocity is getting less positive or more negative /25

Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and

Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and Lecture 9 : Derivatives of Trigonometric Functions (Please review Trigonometry uner Algebra/Precalculus Review on the class webpage.) In this section we will look at the erivatives of the trigonometric

More information

f(x) = a x, h(5) = ( 1) 5 1 = 2 2 1

f(x) = a x, h(5) = ( 1) 5 1 = 2 2 1 Exponential Functions an their Derivatives Exponential functions are functions of the form f(x) = a x, where a is a positive constant referre to as the base. The functions f(x) = x, g(x) = e x, an h(x)

More information

Inverse Trig Functions

Inverse Trig Functions Inverse Trig Functions c A Math Support Center Capsule February, 009 Introuction Just as trig functions arise in many applications, so o the inverse trig functions. What may be most surprising is that

More information

MAT1A01: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules

MAT1A01: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules MAT1A01: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 17 April 2013 Reminer Mats Learning Centre: C-Ring 512 My office: C-Ring 533A (Stats Dept corrior)

More information

Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations

Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of

More information

Answers to the Practice Problems for Test 2

Answers to the Practice Problems for Test 2 Answers to the Practice Problems for Test 2 Davi Murphy. Fin f (x) if it is known that x [f(2x)] = x2. By the chain rule, x [f(2x)] = f (2x) 2, so 2f (2x) = x 2. Hence f (2x) = x 2 /2, but the lefthan

More information

To differentiate logarithmic functions with bases other than e, use

To differentiate logarithmic functions with bases other than e, use To ifferentiate logarithmic functions with bases other than e, use 1 1 To ifferentiate logarithmic functions with bases other than e, use log b m = ln m ln b 1 To ifferentiate logarithmic functions with

More information

CHAPTER 5 : CALCULUS

CHAPTER 5 : CALCULUS Dr Roger Ni (Queen Mary, University of Lonon) - 5. CHAPTER 5 : CALCULUS Differentiation Introuction to Differentiation Calculus is a branch of mathematics which concerns itself with change. Irrespective

More information

The Quick Calculus Tutorial

The Quick Calculus Tutorial The Quick Calculus Tutorial This text is a quick introuction into Calculus ieas an techniques. It is esigne to help you if you take the Calculus base course Physics 211 at the same time with Calculus I,

More information

Lagrangian and Hamiltonian Mechanics

Lagrangian and Hamiltonian Mechanics Lagrangian an Hamiltonian Mechanics D.G. Simpson, Ph.D. Department of Physical Sciences an Engineering Prince George s Community College December 5, 007 Introuction In this course we have been stuying

More information

20. Product rule, Quotient rule

20. Product rule, Quotient rule 20. Prouct rule, 20.1. Prouct rule Prouct rule, Prouct rule We have seen that the erivative of a sum is the sum of the erivatives: [f(x) + g(x)] = x x [f(x)] + x [(g(x)]. One might expect from this that

More information

Pre-Calculus II. where 1 is the radius of the circle and t is the radian measure of the central angle.

Pre-Calculus II. where 1 is the radius of the circle and t is the radian measure of the central angle. Pre-Calculus II 4.2 Trigonometric Functions: The Unit Circle The unit circle is a circle of radius 1, with its center at the origin of a rectangular coordinate system. The equation of this unit circle

More information

Course outline, MA 113, Spring 2014 Part A, Functions and limits. 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems)

Course outline, MA 113, Spring 2014 Part A, Functions and limits. 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems) Course outline, MA 113, Spring 2014 Part A, Functions and limits 1.1 1.2 Functions, domain and ranges, A1.1-1.2-Review (9 problems) Functions, domain and range Domain and range of rational and algebraic

More information

Introduction to Integration Part 1: Anti-Differentiation

Introduction to Integration Part 1: Anti-Differentiation Mathematics Learning Centre Introuction to Integration Part : Anti-Differentiation Mary Barnes c 999 University of Syney Contents For Reference. Table of erivatives......2 New notation.... 2 Introuction

More information

Sample Problems. 10. 1 2 cos 2 x = tan2 x 1. 11. tan 2 = csc 2 tan 2 1. 12. sec x + tan x = cos x 13. 14. sin 4 x cos 4 x = 1 2 cos 2 x

Sample Problems. 10. 1 2 cos 2 x = tan2 x 1. 11. tan 2 = csc 2 tan 2 1. 12. sec x + tan x = cos x 13. 14. sin 4 x cos 4 x = 1 2 cos 2 x Lecture Notes Trigonometric Identities page Sample Problems Prove each of the following identities.. tan x x + sec x 2. tan x + tan x x 3. x x 3 x 4. 5. + + + x 6. 2 sec + x 2 tan x csc x tan x + cot x

More information

y = rsin! (opp) x = z cos! (adj) sin! = y z = The Other Trig Functions

y = rsin! (opp) x = z cos! (adj) sin! = y z = The Other Trig Functions MATH 7 Right Triangle Trig Dr. Neal, WKU Previously, we have seen the right triangle formulas x = r cos and y = rsin where the hypotenuse r comes from the radius of a circle, and x is adjacent to and y

More information

TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

More information

Exponential Functions: Differentiation and Integration. The Natural Exponential Function

Exponential Functions: Differentiation and Integration. The Natural Exponential Function 46_54.q //4 :59 PM Page 5 5 CHAPTER 5 Logarithmic, Eponential, an Other Transcenental Functions Section 5.4 f () = e f() = ln The inverse function of the natural logarithmic function is the natural eponential

More information

Elliptic Functions sn, cn, dn, as Trigonometry W. Schwalm, Physics, Univ. N. Dakota

Elliptic Functions sn, cn, dn, as Trigonometry W. Schwalm, Physics, Univ. N. Dakota Elliptic Functions sn, cn, n, as Trigonometry W. Schwalm, Physics, Univ. N. Dakota Backgroun: Jacobi iscovere that rather than stuying elliptic integrals themselves, it is simpler to think of them as inverses

More information

Core Maths C3. Revision Notes

Core Maths C3. Revision Notes Core Maths C Revision Notes October 0 Core Maths C Algebraic fractions... Cancelling common factors... Multipling and dividing fractions... Adding and subtracting fractions... Equations... 4 Functions...

More information

Given three vectors A, B, andc. We list three products with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B);

Given three vectors A, B, andc. We list three products with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B); 1.1.4. Prouct of three vectors. Given three vectors A, B, anc. We list three proucts with formula (A B) C = B(A C) A(B C); A (B C) =B(A C) C(A B); a 1 a 2 a 3 (A B) C = b 1 b 2 b 3 c 1 c 2 c 3 where the

More information

Sections 3.1/3.2: Introducing the Derivative/Rules of Differentiation

Sections 3.1/3.2: Introducing the Derivative/Rules of Differentiation Sections 3.1/3.2: Introucing te Derivative/Rules of Differentiation 1 Tangent Line Before looking at te erivative, refer back to Section 2.1, looking at average velocity an instantaneous velocity. Here

More information

5.3 SOLVING TRIGONOMETRIC EQUATIONS. Copyright Cengage Learning. All rights reserved.

5.3 SOLVING TRIGONOMETRIC EQUATIONS. Copyright Cengage Learning. All rights reserved. 5.3 SOLVING TRIGONOMETRIC EQUATIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Use standard algebraic techniques to solve trigonometric equations. Solve trigonometric equations

More information

15.2. First-Order Linear Differential Equations. First-Order Linear Differential Equations Bernoulli Equations Applications

15.2. First-Order Linear Differential Equations. First-Order Linear Differential Equations Bernoulli Equations Applications 00 CHAPTER 5 Differential Equations SECTION 5. First-Orer Linear Differential Equations First-Orer Linear Differential Equations Bernoulli Equations Applications First-Orer Linear Differential Equations

More information

Section 6-3 Double-Angle and Half-Angle Identities

Section 6-3 Double-Angle and Half-Angle Identities 6-3 Double-Angle and Half-Angle Identities 47 Section 6-3 Double-Angle and Half-Angle Identities Double-Angle Identities Half-Angle Identities This section develops another important set of identities

More information

Chapter 7 Outline Math 236 Spring 2001

Chapter 7 Outline Math 236 Spring 2001 Chapter 7 Outline Math 236 Spring 2001 Note 1: Be sure to read the Disclaimer on Chapter Outlines! I cannot be responsible for misfortunes that may happen to you if you do not. Note 2: Section 7.9 will

More information

Learning Objectives for Math 165

Learning Objectives for Math 165 Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given

More information

Right Triangles A right triangle, as the one shown in Figure 5, is a triangle that has one angle measuring

Right Triangles A right triangle, as the one shown in Figure 5, is a triangle that has one angle measuring Page 1 9 Trigonometry of Right Triangles Right Triangles A right triangle, as the one shown in Figure 5, is a triangle that has one angle measuring 90. The side opposite to the right angle is the longest

More information

Solutions to Examples from Related Rates Notes. ds 2 mm/s. da when s 100 mm

Solutions to Examples from Related Rates Notes. ds 2 mm/s. da when s 100 mm Solutions to Examples from Relate Rates Notes 1. A square metal plate is place in a furnace. The quick temperature change causes the metal plate to expan so that its surface area increases an its thickness

More information

Math 230.01, Fall 2012: HW 1 Solutions

Math 230.01, Fall 2012: HW 1 Solutions Math 3., Fall : HW Solutions Problem (p.9 #). Suppose a wor is picke at ranom from this sentence. Fin: a) the chance the wor has at least letters; SOLUTION: All wors are equally likely to be chosen. The

More information

ALGEBRA 2/ TRIGONOMETRY

ALGEBRA 2/ TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA 2/ TRIGONOMETRY Friday, June 14, 2013 1:15 4:15 p.m. SAMPLE RESPONSE SET Table of Contents Practice Papers Question 28.......................

More information

Inverse Circular Function and Trigonometric Equation

Inverse Circular Function and Trigonometric Equation Inverse Circular Function and Trigonometric Equation 1 2 Caution The 1 in f 1 is not an exponent. 3 Inverse Sine Function 4 Inverse Cosine Function 5 Inverse Tangent Function 6 Domain and Range of Inverse

More information

G E O M E T R Y CHAPTER 9 RIGHT TRIANGLES AND TRIGONOMETRY. Notes & Study Guide

G E O M E T R Y CHAPTER 9 RIGHT TRIANGLES AND TRIGONOMETRY. Notes & Study Guide G E O M E T R Y CHAPTER 9 RIGHT TRIANGLES AND TRIGONOMETRY Notes & Study Guide 2 TABLE OF CONTENTS SIMILAR RIGHT TRIANGLES... 3 THE PYTHAGOREAN THEOREM... 4 SPECIAL RIGHT TRIANGLES... 5 TRIGONOMETRIC RATIOS...

More information

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:

correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were: Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that

More information

Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123

Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123 Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from

More information

Geometry Notes RIGHT TRIANGLE TRIGONOMETRY

Geometry Notes RIGHT TRIANGLE TRIGONOMETRY Right Triangle Trigonometry Page 1 of 15 RIGHT TRIANGLE TRIGONOMETRY Objectives: After completing this section, you should be able to do the following: Calculate the lengths of sides and angles of a right

More information

2 Integrating Both Sides

2 Integrating Both Sides 2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation

More information

Calculating Viscous Flow: Velocity Profiles in Rivers and Pipes

Calculating Viscous Flow: Velocity Profiles in Rivers and Pipes previous inex next Calculating Viscous Flow: Velocity Profiles in Rivers an Pipes Michael Fowler, UVa 9/8/1 Introuction In this lecture, we ll erive the velocity istribution for two examples of laminar

More information

6.1 Basic Right Triangle Trigonometry

6.1 Basic Right Triangle Trigonometry 6.1 Basic Right Triangle Trigonometry MEASURING ANGLES IN RADIANS First, let s introduce the units you will be using to measure angles, radians. A radian is a unit of measurement defined as the angle at

More information

Trigonometry. An easy way to remember trigonometric properties is:

Trigonometry. An easy way to remember trigonometric properties is: Trigonometry It is possible to solve many force and velocity problems by drawing vector diagrams. However, the degree of accuracy is dependent upon the exactness of the person doing the drawing and measuring.

More information

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =

Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) = Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a

More information

The Method of Partial Fractions Math 121 Calculus II Spring 2015

The Method of Partial Fractions Math 121 Calculus II Spring 2015 Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

More information

Differentiability of Exponential Functions

Differentiability of Exponential Functions Differentiability of Exponential Functions Philip M. Anselone an John W. Lee Philip Anselone (panselone@actionnet.net) receive his Ph.D. from Oregon State in 1957. After a few years at Johns Hopkins an

More information

Lecture L25-3D Rigid Body Kinematics

Lecture L25-3D Rigid Body Kinematics J. Peraire, S. Winall 16.07 Dynamics Fall 2008 Version 2.0 Lecture L25-3D Rigi Boy Kinematics In this lecture, we consier the motion of a 3D rigi boy. We shall see that in the general three-imensional

More information

Trigonometric Functions: The Unit Circle

Trigonometric Functions: The Unit Circle Trigonometric Functions: The Unit Circle This chapter deals with the subject of trigonometry, which likely had its origins in the study of distances and angles by the ancient Greeks. The word trigonometry

More information

Calculus Refresher, version 2008.4. c 1997-2008, Paul Garrett, garrett@math.umn.edu http://www.math.umn.edu/ garrett/

Calculus Refresher, version 2008.4. c 1997-2008, Paul Garrett, garrett@math.umn.edu http://www.math.umn.edu/ garrett/ Calculus Refresher, version 2008.4 c 997-2008, Paul Garrett, garrett@math.umn.eu http://www.math.umn.eu/ garrett/ Contents () Introuction (2) Inequalities (3) Domain of functions (4) Lines (an other items

More information

The Derivative. Philippe B. Laval Kennesaw State University

The Derivative. Philippe B. Laval Kennesaw State University The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition

More information

1 Derivatives of Piecewise Defined Functions

1 Derivatives of Piecewise Defined Functions MATH 1010E University Matematics Lecture Notes (week 4) Martin Li 1 Derivatives of Piecewise Define Functions For piecewise efine functions, we often ave to be very careful in computing te erivatives.

More information

Geometry Mathematics Curriculum Guide Unit 6 Trig & Spec. Right Triangles 2016 2017

Geometry Mathematics Curriculum Guide Unit 6 Trig & Spec. Right Triangles 2016 2017 Unit 6: Trigonometry and Special Right Time Frame: 14 Days Primary Focus This topic extends the idea of triangle similarity to indirect measurements. Students develop properties of special right triangles,

More information

Trigonometry Hard Problems

Trigonometry Hard Problems Solve the problem. This problem is very difficult to understand. Let s see if we can make sense of it. Note that there are multiple interpretations of the problem and that they are all unsatisfactory.

More information

CHAPTER 8: DIFFERENTIAL CALCULUS

CHAPTER 8: DIFFERENTIAL CALCULUS CHAPTER 8: DIFFERENTIAL CALCULUS 1. Rules of Differentiation As we ave seen, calculating erivatives from first principles can be laborious an ifficult even for some relatively simple functions. It is clearly

More information

UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

UNIT 1: ANALYTICAL METHODS FOR ENGINEERS UNIT : ANALYTICAL METHODS FOR ENGINEERS Unit code: A/60/40 QCF Level: 4 Credit value: 5 OUTCOME 3 - CALCULUS TUTORIAL DIFFERENTIATION 3 Be able to analyse and model engineering situations and solve problems

More information

(1.) The air speed of an airplane is 380 km/hr at a bearing of. Find the ground speed of the airplane as well as its

(1.) The air speed of an airplane is 380 km/hr at a bearing of. Find the ground speed of the airplane as well as its (1.) The air speed of an airplane is 380 km/hr at a bearing of 78 o. The speed of the wind is 20 km/hr heading due south. Find the ground speed of the airplane as well as its direction. Here is the diagram:

More information

Differentiation and Integration

Differentiation and Integration This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have

More information

Lines. We have learned that the graph of a linear equation. y = mx +b

Lines. We have learned that the graph of a linear equation. y = mx +b Section 0. Lines We have learne that the graph of a linear equation = m +b is a nonvertical line with slope m an -intercept (0, b). We can also look at the angle that such a line makes with the -ais. This

More information

Chapter 6 Trigonometric Functions of Angles

Chapter 6 Trigonometric Functions of Angles 6.1 Angle Measure Chapter 6 Trigonometric Functions of Angles In Chapter 5, we looked at trig functions in terms of real numbers t, as determined by the coordinates of the terminal point on the unit circle.

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 14 10/27/2008 MOMENT GENERATING FUNCTIONS

MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 14 10/27/2008 MOMENT GENERATING FUNCTIONS MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2008 Lecture 14 10/27/2008 MOMENT GENERATING FUNCTIONS Contents 1. Moment generating functions 2. Sum of a ranom number of ranom variables 3. Transforms

More information

Dear Accelerated Pre-Calculus Student:

Dear Accelerated Pre-Calculus Student: Dear Accelerated Pre-Calculus Student: I am very excited that you have decided to take this course in the upcoming school year! This is a fastpaced, college-preparatory mathematics course that will also

More information

Mechanical Vibrations

Mechanical Vibrations Mechanical Vibrations A mass m is suspended at the end of a spring, its weight stretches the spring by a length L to reach a static state (the equilibrium position of the system). Let u(t) denote the displacement,

More information

Unit 6 Trigonometric Identities, Equations, and Applications

Unit 6 Trigonometric Identities, Equations, and Applications Accelerated Mathematics III Frameworks Student Edition Unit 6 Trigonometric Identities, Equations, and Applications nd Edition Unit 6: Page of 3 Table of Contents Introduction:... 3 Discovering the Pythagorean

More information

Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = e x. y = exp(x) if and only if x = ln(y)

Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = e x. y = exp(x) if and only if x = ln(y) Lecture 3 : The Natural Exponential Function: f(x) = exp(x) = Last day, we saw that the function f(x) = ln x is one-to-one, with domain (, ) and range (, ). We can conclude that f(x) has an inverse function

More information

TOPIC 3: CONTINUITY OF FUNCTIONS

TOPIC 3: CONTINUITY OF FUNCTIONS TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let

More information

Math Placement Test Practice Problems

Math Placement Test Practice Problems Math Placement Test Practice Problems The following problems cover material that is used on the math placement test to place students into Math 1111 College Algebra, Math 1113 Precalculus, and Math 2211

More information

ANALYTICAL METHODS FOR ENGINEERS

ANALYTICAL METHODS FOR ENGINEERS UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations

More information

Calculus I Notes. MATH /Richards/

Calculus I Notes. MATH /Richards/ Calculus I Notes MATH 52-154/Richards/10.1.07 The Derivative The derivative of a function f at p is denoted by f (p) and is informally defined by f (p)= the slope of the f-graph at p. Of course, the above

More information

Basic Integration Formulas and the Substitution Rule

Basic Integration Formulas and the Substitution Rule Basic Integration Formulas and the Substitution Rule The second fundamental theorem of integral calculus Recall from the last lecture the second fundamental theorem of integral calculus. Theorem Let f(x)

More information

Items related to expected use of graphing technology appear in bold italics.

Items related to expected use of graphing technology appear in bold italics. - 1 - Items related to expected use of graphing technology appear in bold italics. Investigating the Graphs of Polynomial Functions determine, through investigation, using graphing calculators or graphing

More information

Applications of Second-Order Differential Equations

Applications of Second-Order Differential Equations Applications of Second-Order Differential Equations Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration

More information

MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.

MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4. MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin

More information

Oscillations. Vern Lindberg. June 10, 2010

Oscillations. Vern Lindberg. June 10, 2010 Oscillations Vern Lindberg June 10, 2010 You have discussed oscillations in Vibs and Waves: we will therefore touch lightly on Chapter 3, mainly trying to refresh your memory and extend the concepts. 1

More information

Trigonometric Functions and Triangles

Trigonometric Functions and Triangles Trigonometric Functions and Triangles Dr. Philippe B. Laval Kennesaw STate University August 27, 2010 Abstract This handout defines the trigonometric function of angles and discusses the relationship between

More information

Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11}

Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11} Mathematics Pre-Test Sample Questions 1. Which of the following sets is closed under division? I. {½, 1,, 4} II. {-1, 1} III. {-1, 0, 1} A. I only B. II only C. III only D. I and II. Which of the following

More information

How to Avoid the Inverse Secant (and Even the Secant Itself)

How to Avoid the Inverse Secant (and Even the Secant Itself) How to Avoi the Inverse Secant (an Even the Secant Itself) S A Fulling Stephen A Fulling (fulling@mathtamue) is Professor of Mathematics an of Physics at Teas A&M University (College Station, TX 7783)

More information

Chapter 6: Periodic Functions

Chapter 6: Periodic Functions Chapter 6: Periodic Functions In the previous chapter, the trigonometric functions were introduced as ratios of sides of a triangle, and related to points on a circle. We noticed how the x and y values

More information

Spring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations

Spring Simple Harmonic Oscillator. Spring constant. Potential Energy stored in a Spring. Understanding oscillations. Understanding oscillations Spring Simple Harmonic Oscillator Simple Harmonic Oscillations and Resonance We have an object attached to a spring. The object is on a horizontal frictionless surface. We move the object so the spring

More information

You can solve a right triangle if you know either of the following: Two side lengths One side length and one acute angle measure

You can solve a right triangle if you know either of the following: Two side lengths One side length and one acute angle measure Solving a Right Triangle A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. Every right triangle has one right angle, two acute angles, one hypotenuse, and two legs. To solve

More information

4.1 Radian and Degree Measure

4.1 Radian and Degree Measure Date: 4.1 Radian and Degree Measure Syllabus Objective: 3.1 The student will solve problems using the unit circle. Trigonometry means the measure of triangles. Terminal side Initial side Standard Position

More information

Lab M1: The Simple Pendulum

Lab M1: The Simple Pendulum Lab M1: The Simple Pendulum Introduction. The simple pendulum is a favorite introductory exercise because Galileo's experiments on pendulums in the early 1600s are usually regarded as the beginning of

More information

Georgia Department of Education Kathy Cox, State Superintendent of Schools 7/19/2005 All Rights Reserved 1

Georgia Department of Education Kathy Cox, State Superintendent of Schools 7/19/2005 All Rights Reserved 1 Accelerated Mathematics 3 This is a course in precalculus and statistics, designed to prepare students to take AB or BC Advanced Placement Calculus. It includes rational, circular trigonometric, and inverse

More information

SOLVING TRIGONOMETRIC EQUATIONS

SOLVING TRIGONOMETRIC EQUATIONS Mathematics Revision Guides Solving Trigonometric Equations Page 1 of 17 M.K. HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C2 Edexcel: C2 OCR: C2 OCR MEI: C2 SOLVING TRIGONOMETRIC

More information

Trigonometry Lesson Objectives

Trigonometry Lesson Objectives Trigonometry Lesson Unit 1: RIGHT TRIANGLE TRIGONOMETRY Lengths of Sides Evaluate trigonometric expressions. Express trigonometric functions as ratios in terms of the sides of a right triangle. Use the

More information

4. Important theorems in quantum mechanics

4. Important theorems in quantum mechanics TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 1 TILLEGG 4 4. Important theorems in quantum mechanics Before attacking three-imensional potentials in the next chapter, we shall in chapter 4 of this

More information

Right Triangle Trigonometry

Right Triangle Trigonometry Section 6.4 OBJECTIVE : Right Triangle Trigonometry Understanding the Right Triangle Definitions of the Trigonometric Functions otenuse osite side otenuse acent side acent side osite side We will be concerned

More information

Second Order Linear Differential Equations

Second Order Linear Differential Equations CHAPTER 2 Second Order Linear Differential Equations 2.. Homogeneous Equations A differential equation is a relation involving variables x y y y. A solution is a function f x such that the substitution

More information

88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a

88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a 88 CHAPTER. VECTOR FUNCTIONS.4 Curvature.4.1 Definitions and Examples The notion of curvature measures how sharply a curve bends. We would expect the curvature to be 0 for a straight line, to be very small

More information

PRE-CALCULUS GRADE 12

PRE-CALCULUS GRADE 12 PRE-CALCULUS GRADE 12 [C] Communication Trigonometry General Outcome: Develop trigonometric reasoning. A1. Demonstrate an understanding of angles in standard position, expressed in degrees and radians.

More information

respect to x and we did not know what y is. In this case x and y are unknown variables because everything is with respect to t.

respect to x and we did not know what y is. In this case x and y are unknown variables because everything is with respect to t. .8 Related Rates Section.8 Notes Page This section involves word problems dealing with how something is changing with respect to time. We will first find a formula that is used for the problem and then

More information

1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. t + ") # x (t) = A! n. t + ") # v(0) = A!

1.10 Using Figure 1.6, verify that equation (1.10) satisfies the initial velocity condition. t + ) # x (t) = A! n. t + ) # v(0) = A! 1.1 Using Figure 1.6, verify that equation (1.1) satisfies the initial velocity condition. Solution: Following the lead given in Example 1.1., write down the general expression of the velocity by differentiating

More information

4.3 & 4.8 Right Triangle Trigonometry. Anatomy of Right Triangles

4.3 & 4.8 Right Triangle Trigonometry. Anatomy of Right Triangles 4.3 & 4.8 Right Triangle Trigonometry Anatomy of Right Triangles The right triangle shown at the right uses lower case a, b and c for its sides with c being the hypotenuse. The sides a and b are referred

More information

APPLIED MATHEMATICS ADVANCED LEVEL

APPLIED MATHEMATICS ADVANCED LEVEL APPLIED MATHEMATICS ADVANCED LEVEL INTRODUCTION This syllabus serves to examine candidates knowledge and skills in introductory mathematical and statistical methods, and their applications. For applications

More information

Section 5-9 Inverse Trigonometric Functions

Section 5-9 Inverse Trigonometric Functions 46 5 TRIGONOMETRIC FUNCTIONS Section 5-9 Inverse Trigonometric Functions Inverse Sine Function Inverse Cosine Function Inverse Tangent Function Summar Inverse Cotangent, Secant, and Cosecant Functions

More information

Introduction to Calculus

Introduction to Calculus Introduction to Calculus Contents 1 Introduction to Calculus 3 11 Introduction 3 111 Origin of Calculus 3 112 The Two Branches of Calculus 4 12 Secant and Tangent Lines 5 13 Limits 10 14 The Derivative

More information

Homework # 3 Solutions

Homework # 3 Solutions Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8

More information

Semester 2, Unit 4: Activity 21

Semester 2, Unit 4: Activity 21 Resources: SpringBoard- PreCalculus Online Resources: PreCalculus Springboard Text Unit 4 Vocabulary: Identity Pythagorean Identity Trigonometric Identity Cofunction Identity Sum and Difference Identities

More information

Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations

Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a

More information

ALGEBRA 2/TRIGONOMETRY

ALGEBRA 2/TRIGONOMETRY ALGEBRA /TRIGONOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA /TRIGONOMETRY Friday, June 14, 013 1:15 to 4:15 p.m., only Student Name: School Name: The possession

More information

RELEASED. Student Booklet. Precalculus. Fall 2014 NC Final Exam. Released Items

RELEASED. Student Booklet. Precalculus. Fall 2014 NC Final Exam. Released Items Released Items Public Schools of North arolina State oard of Education epartment of Public Instruction Raleigh, North arolina 27699-6314 Fall 2014 N Final Exam Precalculus Student ooklet opyright 2014

More information

Right Triangles and SOHCAHTOA: Finding the Measure of an Angle Given any Two Sides (ONLY for ACUTE TRIANGLES Why?)

Right Triangles and SOHCAHTOA: Finding the Measure of an Angle Given any Two Sides (ONLY for ACUTE TRIANGLES Why?) Name Period Date Right Triangles and SOHCAHTOA: Finding the Measure of an Angle Given any Two Sides (ONLY for ACUTE TRIANGLES Why?) Preliminary Information: SOH CAH TOA is an acronym to represent the following

More information

A C O U S T I C S of W O O D Lecture 3

A C O U S T I C S of W O O D Lecture 3 Jan Tippner, Dep. of Wood Science, FFWT MU Brno jan. tippner@mendelu. cz Content of lecture 3: 1. Damping 2. Internal friction in the wood Content of lecture 3: 1. Damping 2. Internal friction in the wood

More information

South Carolina College- and Career-Ready (SCCCR) Pre-Calculus

South Carolina College- and Career-Ready (SCCCR) Pre-Calculus South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know

More information