Answer
The distance between the boy and the balloon is increasing at a rate of $~~13~ft/s$
Work Step by Step
Let point $A$ be the point on the ground directly under the balloon.
Then, the boy, the balloon, and point A form a triangle.
Let $x$ be the horizontal length of the triangle and let $y$ be the height of the balloon. Let $L$ be the hypotenuse of the triangle, which is the distance between the boy and the balloon.
After 3 seconds, $x = (15~ft/s)(3~s) = 45~ft$
After 3 seconds, $y = (5~ft/s)(3~s)+45~ft = 60~ft$
We can find $L$ when $x = 45~ft$ and $y = 60~ft$:
$L^2 = x^2+y^2$
$L = \sqrt{x^2+y^2}$
$L = \sqrt{(45~ft)^2+(60~ft)^2}$
$L = 75~ft$
We can find $\frac{dL}{dt}$:
$L^2 = x^2+y^2$
$2L\frac{dL}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}$
$L\frac{dL}{dt} = x\frac{dx}{dt}+y\frac{dy}{dt}$
$\frac{dL}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{L}$
$\frac{dL}{dt} = \frac{(45~ft)(15~ft/s)+(60~ft)(5~ft/s)}{75~ft}$
$\frac{dL}{dt} = 13~ft/s$
The distance between the boy and the balloon is increasing at a rate of $~~13~ft/s$
