   Chapter 2.R, Problem 79E

Chapter
Section
Textbook Problem

# A balloon is rising at a constant speed of 5 ft/s. A boy is cycling along a straight road at a speed of 15 ft/s. When he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 s later?

To determine

To find:

How fast is the distance between the boy and the balloon increasing 3 s later

Explanation

1. Concept:

Use differentiation

2. Formula:

i. Power rule:

ddxxn=nxn-1

ii. Difference rule:

ddxfx-gx=ddxfx-ddx(gx)

iii. Chain rule:

ddx(fgx=f'gx*g'(x)

3. Given:

dhst=5 ft/sec  anddxdt=15 ft/s

4. Calculation:

The relation between h, x, and s is

s2=h2+x2 → (1)

Differentiate with respect to t on both sides

Therefore,

ddt(s2)=ddt(h2+x2)

By using sum rule

ddt(s2)=ddt(h2)+ddt(x2)

By using power rule and chain rule

2sdsdt=2hdhdt+2xdxdt

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