Can someone help me solve this calculus problem?

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

Answer 1

(a) If x is his distance from first plate, and D is his distance from second plate,

dx/dt = -24 ft/sec (x is decreasing as he approaches 1st)

D = √(90^2+x^2)

dD/dt = 2x/2√(90^2+x^2)dx/dt = x/√(8100+x^2)dx/dt

At x=45

dD/dt = 45/√(8100+2025)*(-24) ft/sec solve that on your calculator (Negative answer, distance is decreasing.)

(b) Let F be the distance to 3rd base which will be

F = √((90-x)^2+90^2) = √(16200 - 180x + x^2)

dF/dt = (-180+2x)/(2√(16200 - 180x + x^2))*dx =

(-90+x)/√(16200 - 180x + x^2))*dx

so at x = 45

dF/dt = (-45)/(√16200 - 180*45 + 45^2) * (-24) ft/sec solve on your calculator. (Positive answer: distance increasing.)

I think that's how to do this.

Answer 2

Distance = sqrt(x^2+90^2)
velocity = v*x/sqrt(x^2+90^2) x=45 ft v=24ft/s

Answer 3

Twice as fast as half. For a and b