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 26 ft/s.
(a) At what rate is his distance from second base decreasing when he is halfway to first base?
ft/s
(b) At what rate is his distance from third base increasing at the same moment?
ft/s

Answer 1

Answer:

(a) Distance decreases at a rate of 11.628 ft/s.

(b) Distance increases at a rate of 11.628 ft/s.

Step-by-step explanation:

Refer to the diagram attached to at the bottom of this answer. We will call a the distance between the batter and 1st base, x the distance between the batter and 2nd base and y the distance between the batter and 3rd base.

The speed of the batter toward first base is 26 ft/s, which means:

$\frac{da}{dt}=26ft/s$

Since the diamond is a square, the angle between trajectories in each base is a square angle, so we can use Pythagoras theorem:

$x=\sqrt{90^2+(90-a)^2}$

$x=\sqrt{a^2-180a+16200}$

Finding the first derivative:

$\frac{dx}{dt}=\frac{a-90}{\sqrt{a^2-180a+16200}}\frac{da}{dt}$

When the batter is halfway to 1st, a = 45 ft, then:

$\frac{dx}{dt}=\frac{45-90}{\sqrt{45^2-(180)(45)+16200}}26=-\frac{26}{\sqrt5}=-11.628ft/s$

So the distance between the batter and 2nd base decreases at 11.628 ft/s

Now, using Pythagoras again to find y:

$y=\sqrt{90^2+b^2}$ and we know that $\frac{db}{dt}=26ft/s$

Finding the first derivative:

$\frac{dy}{dt}=\frac{b}{\sqrt{b^2+8100}}\frac{db}{dt}$

at  b = 45 ft

$\frac{dy}{dt}=\frac{45}{\sqrt{45^2+8100}}26=\frac{26}{\sqrt5}=11.628ft/s$

So the distance between the batter and 3rd base increases at 11.628 ft/s