Answer
$\frac{dh}{dt}=-5.423$
Work Step by Step
Step 1: Information
We can infer that the triangle made by the line from home to third base, third base to the runner's position and the runner's position to home that it forms a right-angled triangle.
Let $x$ be the distance between the runner and third base
Let $h$ be the distance between the runner and home plate
$\frac{dx}{dt}=-25$
$\frac{dh}{dt}=\frac{dx}{dt}\times\frac{dh}{dx}$
$h=\sqrt {x^2+90^2}$
$h$ can also be written as $(x^2+90^2)^\frac{1}{2}$
Step 2: Derive $(x^2+90^2)^\frac{1}{2}$ with respect to $x$
$\frac{dh}{dt}=x(x^2+90^2)^{-\frac{1}{2}}$
Step 3: Multiply the derivatives
$\frac{dh}{dt}=\frac{dx}{dt}\times\frac{dh}{dx}$
$\frac{dh}{dt}=-25\times x(x^2+90^2)^{-\frac{1}{2}}$
$\frac{dh}{dt}=-25 x(x^2+90^2)^{-\frac{1}{2}}$
Step 4: Substitute the value of $x$
$\frac{dh}{dt}=-25 (20)(20^2+90^2)^{-\frac{1}{2}}$
$\frac{dh}{dt}=-5.423$
