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 22 ft/s. At what rate is his distance from second base decreasing when he is halfway to first base?

Answer 1

Label the four points A, B, C, D where A = home plate, B = first base, etc.
Then AC = 90sqrt2 feet.
If the runner is at a point X, distant x from A, along AB, then say his distance from C = y.
We have triangle ACX. Use cosine rule to find y in terms of x
y^2 = AC^2 + x^2 - 180sqrt2. x. cos45
y^2 = 16200 + x^2 -180x. (since cos45 = 1/sqrt2)
When he is half way, x = 45 feet.
y^2 = 16200 + 2025 - 8100
y^2 = 10125
y = 100.62 ft.
Now comes the velocity part.
Differentiate the equation for y.
2ydy/dt = 2xdx/dt - 180 dx/dt
dy/dt = [(x - 90) dx/dt] / y
But dx/dt is constant at 22 ft per sec. So dx/dt = 22.
At the halfway, i.e. when x = 45, dx/dt = 22 and y = 100.62 from above.
So at the halfway point, dy/dt = - 45 times 22 /100.62
dy/dt = -9.84. (Rate at which XC is decreasing)
So his distance from C, or 2nd base, is decreasing at 9.84 ft/sec when he is halfway to first base.

You can use the same method for finding the rate of increase from 3rd base.
A bit easier since the angle is 90 deg not 45 deg.
XD^2 = AD^2 + x^2
or y^2 = 8100 + x^2. When x = 45 y^2 = 10125 so y = 100.62 again.
Differentiate the distance equation to get the velocity equation.
This time 2ydy/dt = 2xdx/dt.
dy/dt = xdx/dt / y
Again dx/dt = 22 and x = 45 so dy/dt = (22 times 45 ) / y
When y = 100.62, dy/dt = 9.84.
This time the answer is positive, so the rate of distance increase from 3rd base is 9.84 ft/sec, the same as the rate of DEcrease from 2nd base.