Question

Question #1973d

Answer 1

Let `x` be the distance between the batter and the home base, and let `s` be the distance between the batter and the third base.

By Pythagorean Theorem,

`s^2=x^2+90^2`

by implicitly differentiating with respect to `t`,

`=> 2s{ds}/{dt}=2x{dx}/{dt}`

by dividing by `2s`,

`=> {ds}/{dt}=x/s{dx}/{dt}`

Since

`x=90/2=45`, `{dx}/{dt}=31`, and `s=sqrt{90^2+45^2}=45sqrt{5}`,

we have

`{ds}/{dt}=45/{45sqrt{5}}cdot31=31/sqrt{5} approx 13.9` ft/s

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I hope that this was helpful.