Section 2.1 The Tangent and Velocity Problems

Section 2.1 The Tangent and Velocity Problems

2. A student bought a smartwatch that tracks the number of steps she walks throughout the day. The table shows the number of step recorded t minutes after 3:00 PM on the first day she wore the watch.

t(min) 0 10 20 30 40

Steps 3438 4559 5622 6536 7398

(a) Find the slopes of the secant lines corresponding to the given intervals of t. What do these slopes represent?

(i) [0,40] (ii) [10,20] (iii) [20,30]

(b) Estimate the student’s walking pace, in steps pre minute, at 3:20 PM by averaging the slopes of two secant lines.

Solution:

2 LIMITS AND DERIVATIVES

2.1 The Tangent and Velocity Problems

1. (a) Using  (15 250), we construct the following table:

  slope =  

5 (5 694) ⁶⁹⁴₅₋₁₅⁻²⁵⁰ = −⁴⁴⁴10 = −444 10 (10 444) ⁴⁴⁴₁₀⁻²⁵⁰₋₁₅ = −¹⁹⁴5 = −388 20 (20 111) ¹¹¹₂₀⁻²⁵⁰₋₁₅ = −¹³⁹5 = −278 25 (25 28) ²⁸₂₅⁻²⁵⁰₋₁₅ = −²²²10 = −222 30 (30 0) ⁰₃₀⁻²⁵⁰₋₁₅= −²⁵⁰15 = −166

(b) Using the values of  that correspond to the points closest to  ( = 10 and  = 20), we have

−388 + (−278)

2 = −333

(c) From the graph, we can estimate the slope of the tangent line at  to be⁻³⁰⁰₉ = −333.

2. (a) (i) On the interval [0 40], slope =7398 − 3438 40 − 0 = 99.

(ii) On the interval [10 20], slope =5622 − 4559

20 − 10 = 1063.

(iii) On the interval [20 30], slope =6536 − 5622 30 − 20 = 914.

The slopes represent the average number of steps per minute the student walked during the respective time intervals.

(b) Averaging the slopes of the secant lines corresponding to the intervals immediately before and after  = 20, we have 1063 + 914

2 = 9885

The student’s walking pace is approximately 99 steps per minute at 3:20PM.

° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c 75 3. The point P (2, −1) lies on the curve y = 1/(1 − x).

(a) If Q is the point (x, 1/(1 − x)), use your calculator to find the slope of the secant line P Q (correct to six decimal places) for the following values of x:

(i) 1.5 (ii) 1.9 (iii) 1.99 (iv) 1.999 (v) 2.5 (vi) 2.1 (vii) 2.01 (viii) 2.001

(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P (2, −1).

(c) Using the slope from part (b), find an equation of the tangent line to the curve at P (2, −1).

Solution:

2 LIMITS AND DERIVATIVES

2.1 The Tangent and Velocity Problems

1. (a) Using  (15 250), we construct the following table:

  slope =  

5 (5 694) ⁶⁹⁴₅₋₁₅⁻²⁵⁰ = −⁴⁴⁴10 = −444 10 (10 444) ⁴⁴⁴₁₀⁻²⁵⁰₋₁₅ = −¹⁹⁴5 = −388 20 (20 111) ¹¹¹₂₀⁻²⁵⁰₋₁₅ = −¹³⁹5 = −278 25 (25 28) ²⁸₂₅⁻²⁵⁰₋₁₅ = −²²²10 = −222 30 (30 0) ⁰₃₀⁻²⁵⁰₋₁₅= −²⁵⁰15 = −166

(b) Using the values of  that correspond to the points closest to  ( = 10 and  = 20), we have

−388 + (−278)

2 = −333

(c) From the graph, we can estimate the slope of the tangent line at  to be⁻³⁰⁰₉ = −333.

2. (a) Slope =²⁹⁴⁸₄₂^{− 2530}_{− 36} =⁴¹⁸₆ ≈ 6967 (b) Slope = ²⁹⁴⁸₄₂^{− 2661}_{− 38} =²⁸⁷₄ = 7175 (c) Slope =²⁹⁴⁸₄₂^{− 2806}

− 40 =¹⁴²₂ = 71 (d) Slope = ³⁰⁸⁰₄₄^{− 2948}

− 42 =¹³²₂ = 66

From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeatsminute after 42 minutes.

After being stable for a while, the patient’s heart rate is dropping.

3. (a)  = 1

1 − ,  (2 −1)

 ( 1(1 − ))  

(i) 15 (15 −2) 2

(ii) 19 (19 −1111 111) 1111 111 (iii) 199 (199 −1010 101) 1010 101 (iv) 1999 (1999 −1001 001) 1001 001 (v) 25 (25 −0666 667) 0666 667 (vi) 21 (21 −0909 091) 0909 091 (vii) 201 (201 −0990 099) 0990 099 (viii) 2001 (2001 −0999 001) 0999 001

(b) The slope appears to be 1.

(c) Using  = 1, an equation of the tangent line to the curve at  (2 −1) is  − (−1) = 1( − 2), or

 =  − 3.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c 67

6. If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by y = 10t − 1.86t².

1

(a) Find the average velocity over the given time intervals: (i) [1,2] (ii) [1,1.5] (iii) [1,1.1] (iv) [1,1.01] (v) [1,1.001]

(b) Estimate the instantaneous velocity when t = 1.

Solution:

68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES

4. (a)  = cos ,  (05 0)

   

(i) 0 (0 1) −2

(ii) 04 (04 0309017) −3090170 (iii) 049 (049 0031411) −3141076 (iv) 0499 (0499 0003142) −3141587

(v) 1 (1 −1) −2

(vi) 06 (06 −0309017) −3090170 (vii) 051 (051 −0031411) −3141076 (viii) 0501 (0501 −0003142) −3141587

(b) The slope appears to be −.

(c)  − 0 = −( − 05) or  = − +¹2.

(d)

5. (a)  = () = 10 − 49². At  = 15,  = 10(15) − 49(15)²= 3975. The average velocity between times 15 and 15 + is

ave=(15 + ) − (15) (15 + ) − 15 =

10(15 + ) − 49(15 + )²

− 3975



=15 + 10 − 11025 − 147 − 49²− 3975

 = −47 − 49²

 = −47 − 49, if  6= 0

(i) [15 2]:  = 05, ave= −715 ms (ii) [15 16]:  = 01, ave= −519 ms (iii) [15 155]:  = 005, ave= −4945 ms (iv) [15 151]:  = 001, ave= −4749 ms (b) The instantaneous velocity when  = 15 ( approaches 0) is −47 ms.

6. (a)  = () = 10 − 186². At  = 1,  = 10(1) − 186(1)²= 814. The average velocity between times 1 and 1 +  is

ave=(1 + ) − (1) (1 + ) − 1 =

10(1 + ) − 186(1 + )²

− 814

 =628 − 186²

 = 628 − 186, if  6= 0.

(i) [1 2]:  = 1, ave= 442ms (ii) [1 15]:  = 05, ave= 535ms (iii) [1 11]:  = 01, ave= 6094ms (iv) [1 101]:  = 001, ave= 62614ms

(v) [1 1001]:  = 0001, ave= 627814ms

(b) The instantaneous velocity when  = 1 ( approaches 0) is 628 ms.

7. (a) (i) On the interval [1 3], ave=(3) − (1)

3 − 1 = 107 − 14

2 = 93

2 = 465ms.

(ii) On the interval [2 3], ave=(3) − (2)

3 − 2 = 107 − 51

1 = 56ms.

(iii) On the interval [3 5], ave=(5) − (3)

5 − 3 = 258 − 107

2 = 151

2 = 755ms.

(iv) On the interval [3 4], ave=(4) − (3)

4 − 3 = 177 − 107

1 = 7ms.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

7. The table shows the position of a motorcyclist after accelerating from rest.

t(seconds) 0 1 2 3 4 5 6

s(meters) 0 1.5 6.3 14.2 24.1 38.0 53.9 (a) Find the average velocity for each time period:

(i) [2,4] (ii) [3,4] (iii) [4,5] (iv) [4,6]

(b) Use the graph of s as a function of t to estimate the instantaneous velocity when t = 3.

Solution:

(a) (i) On the interval [2, 4], vavg= ^s(4)−s(2)₄₋₂ =^24.1−6.3₄₋₂ = 8.9 m/s.

(ii) On the interval [3, 4], vavg= ^s(4)−s(3)₄₋₃ = ^24.1−14.2₄₋₃ = 9.9 m/s.

(iii) On the interval [4, 5], vavg= ^s(5)−s(4)₅₋₄ = ^38.0−24.1₅₋₄ = 13.9 m/s.

(iv) On the interval [4, 6], v_avg= ^s(6)−s(4)₆₋₄ =^53.9−24.1₆₋₄ = 14.9 m/s.

(b) Using the points (2, 6.3) and (4, 24.1) from the approximate tangent line, the instantaneous velocity at t = 3 is about ^24.1−6.3₄₋₂ = 8.9 m/s.

2