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 21 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?

Explanation:

Step1: Define variables

Let \(x\) be the distance of the runner from home - plate, \(y_1\) be the distance from second - base, and \(y_2\) be the distance from third - base. The side of the square \(a = 90\) ft. The speed of the runner \(\frac{dx}{dt}=21\) ft/s. When the runner is halfway to first base, \(x = 45\) ft.

Step2: Use the Pythagorean theorem for distance from second - base

By the Pythagorean theorem, \(y_1=\sqrt{x^{2}+90^{2}}\). Differentiate with respect to \(t\) using the chain rule. \(\frac{dy_1}{dt}=\frac{2x}{2\sqrt{x^{2}+90^{2}}}\cdot\frac{dx}{dt}\). Substitute \(x = 45\) ft and \(\frac{dx}{dt}=21\) ft/s. \(\frac{dy_1}{dt}=\frac{45}{\sqrt{45^{2}+90^{2}}}\cdot21=\frac{45}{\sqrt{2025 + 8100}}\cdot21=\frac{45}{\sqrt{10125}}\cdot21=\frac{45}{45\sqrt{5}}\cdot21=\frac{21}{\sqrt{5}}\approx9.33\) ft/s.

Step3: Use the Pythagorean theorem for distance from third - base

By the Pythagorean theorem, \(y_2=\sqrt{(90 - x)^{2}+90^{2}}\). Differentiate with respect to \(t\) using the chain rule. \(\frac{dy_2}{dt}=\frac{2(90 - x)(- 1)}{2\sqrt{(90 - x)^{2}+90^{2}}}\cdot\frac{dx}{dt}\). Substitute \(x = 45\) ft and \(\frac{dx}{dt}=21\) ft/s. \(\frac{dy_2}{dt}=\frac{-(90 - 45)}{\sqrt{(90 - 45)^{2}+90^{2}}}\cdot21=\frac{- 45}{\sqrt{2025+8100}}\cdot21=\frac{- 45}{45\sqrt{5}}\cdot21=-\frac{21}{\sqrt{5}}\approx - 9.33\) ft/s. The negative sign indicates that the distance from third - base is decreasing. The rate is \(|\frac{dy_2}{dt}|=\frac{21}{\sqrt{5}}\approx9.33\) ft/s.

Answer:

The rate at which the distance from second - base is changing is approximately \(9.33\) ft/s and the rate at which the distance from third - base is changing is approximately \(9.33\) ft/s.