#### To determine

**To find:**

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

#### Answer

Distance is increasing at the rate of 13 ft/s

#### 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

⇒sdsdt=hdhdt+xdxdt

⇒ dsdt =1sh dhdt +x dxdt Now at 3 s

Since,

dhdt=5 ft/sec anddxdt=15 ft/s

x=3*15=45 ft and

h=3*5=15+45=60 ft Since balloon is at 45 feet when boy is at A

Now, from equation (1)

s2=h2+x2

⇒s=h2+x2= 602+452=3600+2025= 5625=75 ft

Therefore, s=75 ft

Now, from equation

dsdt =1sh dhdt +x dxdt

Substitute the values

dsdt=17560*5+45*15=175300+675=97575=13 ft/s

Therefore, Distance increasing at the rate of 13 ft/s

**Conclusion:**

Distance increasing at the rate of 13 ft/s