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§1.5 Interpreting, estimating, and using the derivative

在运动物体位置以外的情境中,函数的导数度量的是什么?

1. Introduction

Introduction

2. Units of the derivative function

Units of the derivative function

3. Toward more accurate derivative estimates

Toward more accurate derivative estimates

4. Travel time at a given speed

With T(s)=50sT(s) = \frac{50}{s} hours for a 5050 mile trip at ss mph, compute T(75)T(75) as a fraction of an hour.

5. Interpreting a population derivative

The function y=P(t)y = P(t) measures the population of a city (in thousands) at the start of year tt, with t=0t = 0 corresponding to 2020. The statement P′(2)=21.37P'(2) = 21.37 means that

6. Meaning of a negative time derivative

Driving 5050 miles takes T(s)=50sT(s) = \frac{50}{s} hours at speed ss mph, and T′(60)≈−0.0139T'(60) \approx -0.0139. What does this value say?

7. Interpreting a car value derivative

Let V(m)V(m) be the value in dollars of a car after mm miles of driving. The statement V′(10250)=−0.89V'(10250) = -0.89 means that

8. Interpreting a water volume derivative

Let W(h)W(h) measure the amount of water (in liters) in a tank filled to a depth of hh meters. The statement W′(0.75)=3.43W'(0.75) = 3.43 means that

9. Interpreting a cooling rate

Let S(t)S(t) be the temperature in degrees Celsius of a soda in a refrigerator at time tt in minutes. The statement S′(20)=−0.527S'(20) = -0.527 means that after 2020 minutes, one more minute should lower the soda's temperature by about 0.5270.527 degrees Celsius.

10. Composite units on a derivative

Let C(s)C(s) measure the rate at which a cyclist burns calories (in calories per hour) when riding at speed ss kilometers per hour. The statement C′(19)=52.1C'(19) = 52.1 means that

11. Central difference estimate

A function satisfies f(1)=2.5f(1) = 2.5, f(2)=3.25f(2) = 3.25, and f(3)=3.625f(3) = 3.625. Using a central difference at a=2a = 2, the estimate of f′(2)f'(2) is

12. Central difference for the potato temperature

A potato's temperature in degrees Fahrenheit is sampled at t=10,20,30t = 10, 20, 30 minutes with values 94.794.7, 141.7141.7, 167.6167.6. Use a central difference to estimate F′(20)F'(20) in degrees per minute, rounded to three decimals.

13. Central difference later in the baking

The same potato has temperature readings F(30)=167.6F(30) = 167.6 and F(50)=197.9F(50) = 197.9 degrees Fahrenheit. Use a central difference to estimate F′(40)F'(40) in degrees per minute, rounded to three decimals.

14. Comparing two later cooling-rate estimates

A baking potato's temperature increases less and less as time goes on: central difference estimates gave F′(20)≈3.645F'(20) \approx 3.645 and F′(40)≈1.515F'(40) \approx 1.515 degrees per minute. Without further calculation, which should be greater, F′(50)F'(50) or F′(60)F'(60)?

15. Predicting with a derivative value

If F(46)=192.5F(46) = 192.5 degrees and F′(46)=1.39F'(46) = 1.39 degrees per minute, then F(47)F(47) should be about 193.89193.89 degrees.

16. Describing increasing at a decreasing rate

On the time interval [10,40][10, 40] minutes, the potato's temperature readings rise throughout, but the central difference estimates of its rate of change fall from 3.6453.645 toward 1.5151.515 degrees per minute while staying positive. Which sentence describes this behavior?

17. Central difference from three samples

Fuel consumption data gives f(80)=0.015f(80) = 0.015, f(90)=0.02f(90) = 0.02, and f(100)=0.027f(100) = 0.027 liters per kilometer at speeds of 8080, 9090, and 100100 km/h. The central difference estimate of f′(90)f'(90) is

18. Interpreting a function value in fuel context

For the fuel consumption function ff (liters per kilometer at speed ss in km/h), the statement f(80)=0.015f(80) = 0.015 means that

19. Interpreting the derivative of a consumption rate

For the fuel consumption function ff with f′(90)≈0.0006f'(90) \approx 0.0006 liters per kilometer per km/h: driving at 9191 km/h instead of 9090 km/h should raise the per-kilometer consumption by about 0.00060.0006 liters per kilometer.

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