Lecture 13: Differentiation Derivatives of Trigonometric Functions
|
|
- Arline Davis
- 7 years ago
- Views:
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
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 informationf(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 informationInverse 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 informationSection 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 informationAnswers 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 informationTo 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 information20. 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 informationLagrangian 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 informationThe 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 informationCourse 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 informationIntroduction 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 informationSample 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 informationExponential 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 informationElliptic 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 information5.3 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 informationTOPIC 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 informationSections 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 informationSection 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 informationCore 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 informationGiven 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 informationRight 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 information15.2. 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 informationChapter 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 informationMath 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 informationGeometry 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 informationAlgebra. 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 information6.1 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 informationThe 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 informationTrigonometry. 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 informationTrigonometric 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 informationThe 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 informationLecture 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 informationcorrect-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 informationCalculus 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 information2 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 informationCalculating 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 informationDifferentiability 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 informationUNIT 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 informationDifferentiation 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 informationMASSACHUSETTS 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 informationTrigonometry 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 informationUnit 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(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 informationCHAPTER 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 informationLines. 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 informationDear 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 informationTrigonometric 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 informationMath 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 informationLecture 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 informationSOLVING 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 informationANALYTICAL 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 informationApplications 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 informationGeorgia 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 informationOscillations. 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 informationMathematics 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 information4.3 & 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 informationSpring 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 informationAPPLIED 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 informationSection 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 informationSolving Quadratic Equations
9.3 Solving Quadratic Equations by Using the Quadratic Formula 9.3 OBJECTIVES 1. Solve a quadratic equation by using the quadratic formula 2. Determine the nature of the solutions of a quadratic equation
More informationMicroeconomic Theory: Basic Math Concepts
Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts
More informationRight 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 informationf(x + h) f(x) h as representing the slope of a secant line. As h goes to 0, the slope of the secant line approaches the slope of the tangent line.
Derivative of f(z) Dr. E. Jacobs Te erivative of a function is efine as a limit: f (x) 0 f(x + ) f(x) We can visualize te expression f(x+) f(x) as representing te slope of a secant line. As goes to 0,
More informationTrigonometry Review with the Unit Circle: All the trig. you ll ever need to know in Calculus
Trigonometry Review with the Unit Circle: All the trig. you ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. functions, identities and formulas, graphs:
More information88 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 informationPRE-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 informationSecond 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 informationChapter 11. Techniques of Integration
Chapter Techniques of Integration Chapter 6 introduced the integral. There it was defined numerically, as the limit of approximating Riemann sums. Evaluating integrals by applying this basic definition
More information2.2 Magic with complex exponentials
2.2. MAGIC WITH COMPLEX EXPONENTIALS 97 2.2 Magic with complex exponentials We don t really know what aspects of complex variables you learned about in high school, so the goal here is to start more or
More informationFigure 1.1 Vector A and Vector F
CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have
More informationGRE Prep: Precalculus
GRE Prep: Precalculus Franklin H.J. Kenter 1 Introduction These are the notes for the Precalculus section for the GRE Prep session held at UCSD in August 2011. These notes are in no way intended to teach
More informationSection 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 informationSemester 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 informationRules for Finding Derivatives
3 Rules for Fining Derivatives It is teious to compute a limit every time we nee to know the erivative of a function. Fortunately, we can evelop a small collection of examples an rules that allow us to
More informationHow 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 information1. Introduction circular definition Remark 1 inverse trigonometric functions
1. Introduction In Lesson 2 the six trigonometric functions were defined using angles determined by points on the unit circle. This is frequently referred to as the circular definition of the trigonometric
More informationEvaluating trigonometric functions
MATH 1110 009-09-06 Evaluating trigonometric functions Remark. Throughout this document, remember the angle measurement convention, which states that if the measurement of an angle appears without units,
More informationHomework # 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 informationSouth 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 informationThe Deadly Sins of Algebra
The Deadly Sins of Algebra There are some algebraic misconceptions that are so damaging to your quantitative and formal reasoning ability, you might as well be said not to have any such reasoning ability.
More informationRELEASED. 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 informationMark Howell Gonzaga High School, Washington, D.C.
Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School,
More informationAP Calculus AB First Semester Final Exam Practice Test Content covers chapters 1-3 Name: Date: Period:
AP Calculus AB First Semester Final Eam Practice Test Content covers chapters 1- Name: Date: Period: This is a big tamale review for the final eam. Of the 69 questions on this review, questions will be
More informationTHE COMPLEX EXPONENTIAL FUNCTION
Math 307 THE COMPLEX EXPONENTIAL FUNCTION (These notes assume you are already familiar with the basic properties of complex numbers.) We make the following definition e iθ = cos θ + i sin θ. (1) This formula
More informationHOOKE S LAW AND SIMPLE HARMONIC MOTION
HOOKE S LAW AND SIMPLE HARMONIC MOTION Alexander Sapozhnikov, Brooklyn College CUNY, New York, alexs@brooklyn.cuny.edu Objectives Study Hooke s Law and measure the spring constant. Study Simple Harmonic
More informationDIFFRACTION AND INTERFERENCE
DIFFRACTION AND INTERFERENCE In this experiment you will emonstrate the wave nature of light by investigating how it bens aroun eges an how it interferes constructively an estructively. You will observe
More informationMathematics I, II and III (9465, 9470, and 9475)
Mathematics I, II and III (9465, 9470, and 9475) General Introduction There are two syllabuses, one for Mathematics I and Mathematics II, the other for Mathematics III. The syllabus for Mathematics I and
More informationDRAFT. Further mathematics. GCE AS and A level subject content
Further mathematics GCE AS and A level subject content July 2014 s Introduction Purpose Aims and objectives Subject content Structure Background knowledge Overarching themes Use of technology Detailed
More informationC B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N
Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a
More informationBasic numerical skills: EQUATIONS AND HOW TO SOLVE THEM. x + 5 = 7 2 + 5-2 = 7-2 5 + (2-2) = 7-2 5 = 5. x + 5-5 = 7-5. x + 0 = 20.
Basic numerical skills: EQUATIONS AND HOW TO SOLVE THEM 1. Introduction (really easy) An equation represents the equivalence between two quantities. The two sides of the equation are in balance, and solving
More information14.1. Basic Concepts of Integration. Introduction. Prerequisites. Learning Outcomes. Learning Style
Basic Concepts of Integration 14.1 Introduction When a function f(x) is known we can differentiate it to obtain its derivative df. The reverse dx process is to obtain the function f(x) from knowledge of
More informationboth double. A. T and v max B. T remains the same and v max doubles. both remain the same. C. T and v max
Q13.1 An object on the end of a spring is oscillating in simple harmonic motion. If the amplitude of oscillation is doubled, how does this affect the oscillation period T and the object s maximum speed
More informationThe Mathematics Diagnostic Test
The Mathematics iagnostic Test Mock Test and Further Information 010 In welcome week, students will be asked to sit a short test in order to determine the appropriate lecture course, tutorial group, whether
More information1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
More informationNonhomogeneous Linear Equations
Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form ay by cy G x where
More informationFind the length of the arc on a circle of radius r intercepted by a central angle θ. Round to two decimal places.
SECTION.1 Simplify. 1. 7π π. 5π 6 + π Find the measure of the angle in degrees between the hour hand and the minute hand of a clock at the time shown. Measure the angle in the clockwise direction.. 1:0.
More informationRIGHT TRIANGLE TRIGONOMETRY
RIGHT TRIANGLE TRIGONOMETRY The word Trigonometry can be broken into the parts Tri, gon, and metry, which means Three angle measurement, or equivalently Triangle measurement. Throughout this unit, we will
More information18.01 Single Variable Calculus Fall 2006
MIT OpenCourseWare http://ocw.mit.edu 8.0 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Unit : Derivatives A. What
More information