Question

A balloon is 200 feet off the ground and rising vertically at the constant rate of 15 feet per second. An automobile passed beneath it traveling along a straight line at the constant rate of 66 miles per hour. How fast is the distance between them changing one second later?

Answer 1

Answer:

The distance between them changing 33.7 feet/second.

Step-by-step explanation:

The above situation forms a right angled triangle of base "a" and height "b".

Let "h" be the hypotenuse.

So, $h^2 = ax^2  + b^2$

Differentiating  with respect to the time "t"

$2h\frac{dh}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt}$

When we divide both sides by 2, we get

$h\frac{dh}{dt} = a\frac{da}{dt} + b\frac{db}{dt}$

Here we have to find $\frac{dh}{dt}$

Given:  $\frac{da}{dt} = 66, \frac{db}{dt}  = 15$

We are given the horizontal distance = a =  66 ft  and vertical distance = 200 + 15 = 215 ft

h² =  215² + 66² = 50581

Taking square root on both sides, we get

h = 225 ft

Plug in these values in the derivative, we get

$h\frac{dh}{dt} = a\frac{da}{dt} + b\frac{db}{dt}$

225$\frac{dh}{dt}$=  (66×66) + (215×15)

(225) $\frac{dh}{dt}$  = 7581

$\frac{dh}{dt}$ =  7581 ÷ 225

$\frac{dh}{dt}$ = 33.69    

$\frac{dh}{dt}$ = 33.7 feet/second.

So the distance between them changing 33.7 feet/second.