How quickly is the distance from a baseball to first base changing as the baseball moves along the third baseline

A baseball is moving along the third base line at a constant velocity of 100 feet per second. If a baseball diamond is a 90-foot square, then how quickly is the distance from the ball to first base changing when:

the ball is halfway to third base;
the ball is at third base.

First, we give the general set up for the problem.

$\begin{align*} x &= \text{ the distance from the ball to first base}. \\ y &= \text{ the distance from the ball to home plate}. \end{align*}$

The following diagram illustrates the situation:

Rendered by QuickLaTeX.com

The quantity we are given then is

$\frac{dy}{dt} = 100 \text{ ft/s}.$

Further, $x^2 = y^2 + 90^2$ ; thus, $x = \sqrt{y^2 + 90^2}$ and so

$\frac{dx}{dy} = \frac{y}{\sqrt{y^2 + 90^2}}.$

Now, the two requested cases:

If the ball is halfway to third base, this means $y = 45$ feet (since the diamond is a square with sides of length 90 feet). So we have,
$\frac{dx}{dt} = \frac{dy}{dt} \cdot \frac{dx}{dy} = (100) \left( \frac{45}{\sqrt{45^2 + 90^2}}\right) = \frac{100}{\sqrt{5}} = 20 \sqrt{5} \text{ ft/s}.$
If the ball is at third base, this means $y = 90$ (since $y$ is the distance from the ball to home plate), thus we have,
$\frac{dx}{dt} = \frac{dy}{dt} \cdot \frac{dx}{dy} = (100) \left( \frac{90}{\sqrt{90^2 + 90^2}}\right) = \frac{100}{\sqrt{2}} = 50 \sqrt{2} \text{ ft/s}.$