What is the Distance Rate?

Question

A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s.

At what rate is his distance from second base changing when he is halfway to first base?

At what rate is his distance from third base changing at the same moment?

Answer 1

Let

L = 90 ft be the side of the diamond,

x(t) be the distance the runner covered after some time t,

x'(t) = 24 ft/s be the derivative of distance, which is the given speed,

d2(t) be the distance between the runner and second base at time t,

d3(t) be the distance between the runner and third base at time t.

As a matter of notation here, the quote denotes derivative with respect to time, i.e.,

for any function f, f' = df/dt

For the sake of conciseness I will drop the dependence on (t).

From Pythagorean theorem

d2 = √(L² + (L - x)²)

d3 = √(L² + x²)

The questions concern the rate of change in d2 and d3, i.,e., their derivatives.

In case you need help with the derivatives, let me do the d3'(t) step by step

d3' = ((L² + x²)^1/2)' =

(1/2) (L² + x²)^-1/2 (L² + x²)' =

(1/2) (L² + x²)^-1/2 (x²)' =

(1/2) (L² + x²)^-1/2 2xx' =

(L² + x²)^-1/2 xx'

d2' = -(L² + (L - x)²)^-1/2 (L - x) x'

We are to evaluate d2' and d3' at the time t1 when the runner is half way to the first base, i.e.,

when x(t1) = L/2.

At that point in time, L - x(t1) = x(t1) and therefore d2'(t1) = -d3'(t1).

So we need to do just one calculation.

Substituting actual numbers

d3'(t1) = (90² + 45²)^-1/2 × 45 × 24 = 10.7 ft/s

d2'(t1) = -d3'(t1) = -10.7 ft/s