Question

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 19 ft/s.

Answer 1

So if you check the picture below, each side of the diamond has 90ft, so half-way from first to second is 45ft over the "x" line

a)

in this case, notice that "y" is static, is a constant, it doesn't change, whilst the  "r" distance is and the "x" as well, when the runner is going to 1st base

$\bf r^2=x^2+y^2\implies r^2=x^2+90^2\impliedby \textit{since "y" is a constant} \\\\\\ 2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+0\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}}{r}\quad \begin{cases} x=45\\ \frac{dx}{dt}=19\\ r=45\sqrt{5} \end{cases} \\\\\\ \cfrac{dr}{dt}=\cfrac{19\cdot 45}{45\pm\sqrt{5}}\implies \cfrac{dr}{dt}=-\cfrac{19}{\sqrt{5}}\impliedby \begin{array}{llll} \textit{we used }-\sqrt{5}\\ \textit{because the rate is}\\ negative \end{array}$

b)

well on this case, the distance "y" from home plate to 3rd base, isn't changing either, is a constant and is also 90ft as well

$\bf r^2=x^2+y^2\implies r^2=x^2+90^2\impliedby \textit{since "y" is a constant} \\\\\\ 2r\cfrac{dr}{dt}=2x\cfrac{dx}{dt}+0\implies \cfrac{dr}{dt}=\cfrac{x\frac{dx}{dt}}{r}\quad \begin{cases} x=45\\ \frac{dx}{dt}=19\\ r=45\sqrt{5} \end{cases} \\\\\\ \cfrac{dr}{dt}=\cfrac{19\cdot 45}{45\pm\sqrt{5}}\implies \cfrac{dr}{dt}=\cfrac{19}{\sqrt{5}}\impliedby \begin{array}{llll} \textit{we used }+\sqrt{5}\\ \textit{because the rate is}\\ positive \end{array}$