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

Answer:

13.86 ft/sec

Step-by-step 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 $\frac{dx}{dt}=31 ft/s$ and we want $\frac{dD}{dt}$ when he is halfway (at x = 45).

Using Pythagoras theorem

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

When x=45

$D^2=90^2+(90-x)^2\\D^2=90^2+(90-45)^2\\D^2=10125\\D=\sqrt{10125}$

$2D\frac{dD}{dt}=-2(90-x)\frac{dx}{dt}\\2\sqrt{10125}\frac{dD}{dt}=-2(90-45)(31)\\\frac{dD}{dt}=\frac{-2(45)(31)}{2\sqrt{10125}} =-13.86ft/s$

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.