Answer to Question #174358 in Calculus for Holden Giles Cabrito

Question

A softball diamond has the shape of a square wide sides 60 ft long. If 

a player is running from second base to third at a speed of 24 ft/sec, at 

what rate is his distance from home plate changing when she is 20 ft 

from the third ?

Answer 1

Solution:

From base 1 to base 2 then to base 3 is a right-angled triangle.

When the runner is 20 feet from base 3, they are 40 feet from base 2

Letting the distance between the runner and the home plate be s we have:

s² = 60² + 40² = 5200

so,

$s=20\sqrt{13}$ feet

Letting the distance from base 1 to base 2 be x, from base 2 to base 3 be y we have:

s² = x² + y²

so,

$2s\frac{ds}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$

Now, with $s=20\sqrt{13}$ , x = 60, $\frac{dx}{dt}=0$ , $\frac{dy}{dt}=24$ and y = 40 we have:

$(40\sqrt{13})\frac{ds}{dt}=2(60)(0)+2(40)(24)$

so, $(40\sqrt{13})\frac{ds}{dt}=$ 1920 => $\frac{ds}{dt}=\frac{1920}{(40\sqrt{13})}$

or, $\frac{ds}{dt}=\frac{(48\sqrt{13})}{13}$ => 13.3 $(\frac{ft}{sec})$