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A baseball diamond is a square with side length $90$ ft. A batter hits the ball and runs toward first base with a speed of $f(t)$ ft/s after $t$ seconds. At what rate is the batter's distance to second base decreasing when the batter is halfway to first base? This is not a duplicate, because the speed is varying in this question compared to the one you guys linked it to.

modimagni
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1 Answers1

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Hint:

Using the pythagorean theorem, you should be able to find that your distance $D$ to base 2 is

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

for $x$ the distance of the batter to base 1 and $b$ the distance between base 1 and base 2 (make a drawing if it is hard for you to conceptualize).

Then, you want to find the rate of change of $x$ relative to time, that is derivate $D$ relative to time: $$\frac{dD}{dt} = \frac{x-90}{\sqrt{(90-x)^2 + b^2}} \frac{dx}{dt}$$

Now, you need to understand what is $x$ and$\frac{dx}{dt}$, and replace it in the equation. Then you can evaluate it by finding $t$ when $x = 45$ft.

Also, make sure that you are able to find $\frac{dD}{dt}$ on your own.

Community
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Olivier
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  • By the way, I don't think this question is a duplicate since it involves differentiation. In the other question, the speed is constant. Here we have speed varying in function of time and the position as a function of time is possibly found by the integration of the speed function (with $x = 0$ when $t = 0$). I think this question lacks details and context though. – Olivier Nov 25 '13 at 22:59
  • What do you mean by understanding what x and dx/dt, from my understanding x=45. dx/dt would be f(t) ft/sec. Thanks for the help! – modimagni Nov 26 '13 at 02:11
  • @user111392 You got it. – Olivier Nov 26 '13 at 03:59