2 Limits and Derivatives

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1 2 Limits and Derivatives 2.7 Tangent Lines, Velocity, and Derivatives A tangent line to a circle is a line tat intersects te circle at exactly one point. We would like to take tis idea of tangent line and apply it to oter curves. y = f(x) P (a, f(a)) Te word tangent is derived from te Latin word tangens, wic means toucing. Tus a tangent to a curve is a line tat touces te curve. In oter words, a tangent line sould ave te same direction as te curve at te point of contact. Our goal will be to find an equation of te tangent line to a curve given by y = f(x) at te point P (a, f(a)). Note tat wen te x-coordinate is a, te y-coordinate is necessarily given by y = f(a). Let s begin by recalling te formula for te equation of a line. Point-slope Equation of a Line Te point slope formula for te equation 1

2 of a line is y y 1 = m(x x 1 ) were (x 1, y 1 ) is a point on te line, and m is te slope of te line. Tis is a good start. For our tangent line, we ave a point, P (a, f(a)). So we can let x 1 = a and y 1 = f(a). Next we need to find te slope of te tangent line, m tan. Slope of a Line Te formula for te slope of a line is m = y 2 y 1 x 2 x 1, were (x 1, y 1 ) and (x 2, y 2 ) are two points on te line. In order to use tis formula for slope, we need to know te coordinates of two points on a line. So far, we ave only one set of coordinates, x 1 = a, y 1 = f(a). We are going to get around tis problem by introducing a secant line. Tis will be a line troug te point P and anoter point Q(x, f(x)) on te curve. Te idea is tat te slope of te secant line will be almost te same as te slope of te tangent line provided tat te point Q is close to P. 2

3 Q(x, f(x)) Secant Line y = f(x) Tangent Line P (a, f(a)) Te secant line contains te point Q(x, f(x)) and te point P (a, f(a)). Terefore te slope of te secant line is given by m sec = rise run = Again, we are going to use te slope of te secant line as an approximation for te slope of te tangent line. To get a really good approximation, we want to make Q as close to P as possible. Tis is acieved by finding te limit of te slope of te secant line, m sec, as P approaces Q. Te slope of te tangent line, m tan, is ten defined as te limit of te slope of te secant line, m sec, as Q approaces P. Tat is, m tan Q P m sec To make Q approac P, we can let x approac a. Moreover, a formula for m sec was given above. We now give te following definition for te slope of te tangent line. 3

4 Definition Te slope of te tangent line to a curve y = f(x) at (a, f(a) is given by te formula m tan x a Tis was sort of te wole point of studying limits in te previous capter: we wanted to be able to make tis limit calculation. We can now do an example were we find te equation of te tangent line to a specific function. Example 1 Find te equation of te tangent line to te curve f(x) = x 2 at te point (1, 1). Solution Te point-slope formula for a line is y y 1 = m(x x 1 ). For tis example, x 1 = 1 and y 1 = 1. Note tat f(x) = x 2 and a = 1. We can now use te formula to find te slope of te tangent line, m tan. m tan x a x 1 f(x) f(1) x 1 x x 1 x 1 x 1 (x 1)(x + 1) (x 1) x 1 (x + 1) = = 2 We ave x 1 = 1, y 1 = 1 and m tan = 2. Terefore, wen substituting into te point-slope equation for te line we get y y 1 = m(x x 1 ) 4

5 y 1 = 2(x 1) y = 2x 1 Example 2 Find te slope of te tangent line at te given value of x = a using te formula given above for m tan. 1. f(x) = x 2 3x + 1, a = 2 2. f(x) = x, a = 4 3. f(x) = x 3, a = 3 Solution 1. f(x) = x 2 3x + 1, a = 2 2. f(x) = x, a = 4 m tan = x a lim = f(x) f(2) lim x 2 x 2 (x 2 3x + 1) ((2) 2 3(2) + 1) x 2 x 2 (x 2 3x + 1) ( 1) x 2 (x 2) x 2 3x + 2 x 2 (x 2) (x 1)(x 2) x 2 (x 2) (x 1) = 2 1 = 1 x 2 m tan = x a lim f(x) f(4) x 4 x 4 x 4 x 4 x 4 5

6 3. f(x) = x 3, a = 3 x 4 x 2 (x 4) x 4 x 2 (x 4) ( ) x + 2 x + 2 ( x) 2 (2) 2 x 4 (x 4)( x + 2) x 4 x 4 (x 4)( x + 2) 1 = x 4 x + 2 m tan x a x 3 f(x) f(2) x 3 x 3 x 3 (3) 3 x = 1 4 x 3 (x 3)(x 2 + 3x + (3) 3 ) (x 3) x 3 (x 2 + 3x + 9) = (3) 2 + 3(3) + 9 = 27 Example 3 Find te tangent line to te function f(x) = 1/x at (2, 1 2 ). Solution First we find te slope m tan were f(x) = 1/x and a = 2. m tan = lim x a = f(x) f(2) lim x 2 x 2 = 1 lim x 2 x 2 x 2 Multiply te numerator and denominator by 2x to clear te fraction. x 2 ( 1 x 1 2 )2x (x 2)2x 6

7 (2 x) x 2 (x 2)2x (x 2) x 2 (x 2)2x 1 x 2 2x 1 = 2 2 = 1 4 We can use te point-slope formula wit x 1 = 2, y 1 = f(2) = 1/2, and m tan = 1/4. Tis gives y y 1 = m tan (x x 1 ) y 1/2 = 1 (x 2) 4 y = x Velocity and Rate of Cange Suppose we drop a rock from a cliff in te Hig Sierras. Tere is an equation tat we can use to determine te position of a falling object after t seconds. If s(t) is te position in meters of a falling object after t seconds ten we ave te formula s(t) = 4.9t 2 m. For example, after 1 second, te rock as fallen s(1) = 4.9(1) 2 = 4.9 m. After 2 seconds, te rock as fallen s(2) = 4.9(2) 2 = 4.9(4) = 19.6 m. In fact, tis formula can be used to estimate te eigt of a cliff. We can make te approximation tat s(t) = 5t 2 m. Ten we can drop a rock off of a cliff and record ow long it takes for te rock to fall. Say te rock takes 5 seconds to fall. In tat case te distance tat te rock fell is approximately s(10) = 5 (5) 2 = 125 m. So te distance from te top of te cliff to te base is about 125 meters. Now let s talk about velocity. Using tis formula for position, or te distance tat a falling object as traveled, we can calculate te average velocity 7

8 of an object. Suppose we are interested in knowing ow fast a falling object is going after it as fallen a seconds. We ave te following formula for average velocity. v avg = s(t 2) s(t 1 ) t 2 t 1 So te units for average velocity will be meters per second in our example. To find ow fast te falling object is going after it as traveled a seconds, we can let t 1 = a and t 2 = t. Tis gives v avg = s(t) s(a) t a Common sense tells us tat to get te best approximation possible, we sould make te interval over wic we find te average velocity as small as possible. We would like to take te limit of te average velocity as t approaces a. Tis is ow we define instantaneous velocity. Te instantaneous velocity of of a particle at time t = a wit position function s(t) is given by s(t) s(a) v(a). t a t a Example 4 Suppose tat a ball is dropped from te upper deck of te CN Tower, 450 meters above te ground. Te position function of te ball is given by s(t) = 4.9t 2 meters after 2 seconds. Wat is te velocity of te ball after 5 seconds? Solution formula. We can find a formula for velocity at time 2 seconds using te v(2) = s(t) s(2) lim t 2 t 2 = 4.9t 2 4.9(2) 2 lim t 2 t 2 = 4.9(t 2 (2) 2 ) lim t 2 t 2 8

9 4.9(t 2)(t + 2) t 2 t 2 4.9(t + 2) = 4.9(2 + 2) = 4.9(4) = 19.6 m/s t 2 Example 5 Te position function of a particle is given by te equation s = t 3 were t is measured in seconds and s in meters. Find te velocity at time t = 2 seconds. Solution formula. We can find a formula for velocity at time 2 seconds using te v(2) = s(t) s(2) lim t 2 t 2 = (t 3 ) ((2) 3 ) lim t 2 t 2 t 2 (t 2)(t 2 + 2t + 4) t 2 t 2 (t 2 + 2t + 4) = (2) 2 + 2(2) + 4 = 12 m/s Derivatives In te previous subsection we gave a formula for te slope of te tangent line, m tan. Te slope of te tangent line is equal to wat is called te derivative of a function f at a number a. Te slope of te tangent line to a curve at a point (a, f(a)) equals te derivative. However, tere are many oter applications for te derivative of a function. Te bulk of tis capter will be spent studying te derivative as a pure mat entity. Definition Te derivative of a function f at a number a, denoted by f (a), is given by te following limit, provided te limit exists. f (a) x a 9

10 Example 6 Find te derivative of f(x) = x 2 + 2x + 5 at a = 2. Solution f (2) x 2 f(x) f(2) x 2 x 2 ((x) 2 + 2(x) + 5) ((2) 2 + 2(2) + 5) x 2 x 2 x 2 + 2x 8 x 2 x 2 (x + 4)(x 2) (x 2) x 2 (x + 4) = = 6 Te derivative can be different for different values of a. We would like to be able to find a general formula for te derivative f (a) and ten evaluate it for different values of a. Example 7 Let f(x) = x Compute a formula for f (a) using te definition of derivative. 2. Calculate f ( 1), f (0) and f (1). Solution 1. f (a) x a x 2 a 2 x a x a ()(x + a) x a (x + a) = a + a = 2a. 2. f ( 1) = 2( 1) = 1, f (0) = 2(0) = 0, and f (1) = 2(1) = 2. 10

11 f(x) = x 2 slope= 2 ( 1, 1) (0, 0) (1, 1) slope=2 slope=0 Derivative equals slope of te tangent line. Te slope of te tangent line can take different values for different points on te grap. We see tat for f(x) = x 2, we ave f ( 1) = 2, f (0) = 0, and f (1) = 2. Tere is an equivalent formula for derivative. If we make te substitution =, ten x = a +, and as x a, 0. So tat f (a) x a f(a + ) f(a) An Equivalent Definition for Derivative f at a number x = a is given by Te derivative of a function f (a) f(a + ) f(a) For certain functions, tis is an easier formula to work wit. Example 8 Find te derivative using te equivalent definition of derivative of f(x) = x 2 at a = 2. Solution 11

12 f (a) f(2 + ) f(2) (4 + ) (4 + ) = = 4 (2 + ) 2 (2) 2 Example 9 Find a formula for f (a) for f(x) = x 2 using te equivalent definition of derivative. Solution f (a) f(a + ) f(a) a 2 + 2a + 2 a 2 (a + ) 2 a 2 2a + 2 (2a + ) (2a + ) = 2a + 0 = 2a We see tat te derivative of a function depends on te number a and is terefore itself a function. Rater tat use te letter a, we can use te letter x. Definition Te derivative of a function f denoted f is given by f f(x + ) f(x) (x), provided tat te limit exists. Example 10 Find a formula for f (x) using te equivalent definition of derivative, were f(x) = x. Solution f f(x + ) f(x) (x) 1 ( ) x + x 12

13 1 ( ) x + + x x + x x + + x 1 1 = ( ) (x + ) x x + + x ( x + + x ) 1 x + + x 1 x x = 1 2 x Finding te derivative directly from te definition is long and difficult. We will introduce some rules for finding te derivative tat will make te job muc easier. Example 11 Te limit below represents te derivative of some function f at some number a. State suc an f and a. f (a) sec() 1 Solution Te formula used for f (a) appears to be te one involving. We compare. f (a) sec() 1 and f (a) f(a + ) f(a) Let s only look at te numerator of te two fractions. f(a + ) f(a) and sec() 1 Te terms f(a+) and sec() sould agree. Note tat sec() = sec(0+). We sould ave f(a + ) = sec(0 + ). Terefore, f(x) = sec(x) and a = 0. It sould follow tat f(0) = 1. We see tat f(a) = sec(0) = 1. Answer: f(x) = sec x, a = 0 Example 12 Te limit below represents te derivative of some function f at some number a. State suc an f and a. f cos(x) 1/2 (a) x π/3 x π/3 13

14 Solution We make a comparison. f (a) x a and f cos(x) 1/2 (a) x π/3 x π/3 We ave x a and x π/3. We conclude tat a = π/3. We can compare just te numerators of te fractions: and cos(x) 1/2. We conclude tat f(x) = cos x. Moreover, we sould ave f(a) = cos(a) = cos(π/3) = 1/2. Answer: f(x) = cos x and a = π/3. 14

15 Mat 180 Homework 2.7 Te Tangent Lines, Velocity, and Derivatives Sketc a plausible tangent line at te given point. 1. At x = π 2. At x = 0 3. At x = 0 4. At x = 1 15

16 5. Estimate te slope of te tangent line to te curve at (a) x = 0 (b) x = 0.5 (c) x = 1 (d) x = Estimate te slope of te tangent line to te curve at (a) x = 1 (b) x = 2 (c) x = 3 7. List te points A, B, C, and D in order of increasing slope of te tangent line. 16

17 Find te equation of te tangent line to te curve at te given point. 8. f(x) = x 2 2, (1, 1) 9. f(x) = x 2 2, (0, 2) 10. f(x) = 1/(x 2), (3, 1) 11. f(x) = x, (1, 1) 12. If a ball is trown into te air wit velocity of 40 ft/s, its eigt (in feet) after t seconds is given by y = 40 16t 2. find te velocity wen t = Te displacement (in meters) of a particle moving in a straigt line is given by te equation of motion s = 1/t 2, were t is measured in seconds. find te velocity of te particle at times t = a, t = 1, t = 2, and t = 3. For tese omework problems, use eiter definition of derivative given in tis section. or f (a) f(a + ) f(a) f (a) x a For te functions f given below, find a formula for f (a). Use te formula for f (a) to find f (0), f (5) and f (8). 14. f(x) = x f(x) = x 2 + x f(x) = 3x + 1. Eac limit represents te derivative of some function f at some number a. State suc an f and a in eac case. 17

18 17. f (1 + ) 10 1 (a) 18. f (a) f (a) x 5 2 x 32 x f tan x 1 (a) x π/4 x π/4 21. f (a) cos(π + ) f (a) t 1 t 4 + t 2 t 1 18

19 Mat 180 Homework Te Tangent Lines, Velocity, and Derivatives Solutions Tere is no tangent line at x = (a) 1 (b) 1/2 (c) 1/2 (d) 1 19

20 6. (a) 1 (b) 0 (c) 1 7. C, B, A, D 8. y = 2x 3 9. y = y = x y = x Te ball will be falling toward te ground at a velocity of 64 ft/s. 13. v(a) = 2/a 3 m/s, v(1) = 2 m/s, v(2) = 1/4 m/s, v(3) = 2/27 m/s. 14. f (a) = 3a 2. f (0) = 0, f (5) = 75, f (8) = f (a) = 2a + 1. f (0) = 1, f (5) = 11, f (8) = f(x) = 17. f(x) = x 10, a = x + 1. f (0) = 3/2, f (5) = 3/8, f (8) = 3/ f(x) = 4 x, a = f(x) = 2 x, a = f(x) = tan x, a = π/4 21. f(x) = cos x, a = π 22. f(t) = t 4 + t, a = 1 20

f(x) f(a) x a Our intuition tells us that the slope of the tangent line to the curve at the point P is m P Q =

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