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 31 ft/s. (a) At what rate is his distance from second base decreasing when he is halfway to first base

Answer 1

13.86 ft/sec

Explanation:

If we let x = distance batter has run at time t and D = distance from second base to the batter at time t, then we know
`(dx)/(dt)=31 ft/s` and we want
`(dD)/(dt)` when he is halfway (at x = 45).

Using Pythagoras theorem

`D^2=90^2+(90-x)^2\\\text{Differentiating with respect to t}\\2D(dD)/(dt)=0+2(90-x)(-1)(dx)/(dt)\\2D(dD)/(dt)=-2(90-x)(dx)/(dt)`

When x=45

`D^2=90^2+(90-x)^2\\D^2=90^2+(90-45)^2\\D^2=10125\\D=√(10125)`

`2D(dD)/(dt)=-2(90-x)(dx)/(dt)\\2√(10125)(dD)/(dt)=-2(90-45)(31)\\(dD)/(dt)=(-2(45)(31))/(2√(10125)) =-13.86ft/s`

Thus, when the batter is halfway to first base, the distance between second base and the batter is decreasing at the rate of about 13.86 ft/sec.