t what rate is his distance from second base decreasing when he is halfway to first base?

Question

A baseball diamond is a square with side $90ft$. A batter hits the ball and runs toward first base with a speed of $24ft/s$.

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?

I don't know how to do this problem, can you give me a hint

Answer 1

Hint:

A) Make a right triangle where the sides are: the distance to first base, distance to second base and the side connecting first and second base. Then differentiate to find related rates.

B) Make another triangle where the sides are: distance from starting point, distance from third base and the side connecting the starting point and third base.

Answer 2

a) So here is the diagram I draw to do this excercise.

Let $x$ be the distance between the batter and the home. Then $90-x$ is the distance between him and the first base. Since the batter runs halfway to the first base, his distance from the base is 45

The distance from the second base $(CD)$ d = $\sqrt{(90-x)^{2}+90^{2}}$

Differentiate both sides with respect to x:

$\dfrac{dd}{dt}= \dfrac{d(\sqrt{(90-x)^{2}+90^{2})}}{dt}=\dfrac{1}{2}[(90-x)^{2}+90^{2}]^{-\frac{1}{2}}\dfrac{d(90-x)^{2}}{dt}+\dfrac{d(90^{2})}{dt}$

$=\dfrac{1}{2}[(90-x)^{2}+90^{2}]^{-\frac{1}{2}}[2(90-x).\dfrac{d(90-x)}{dt}]$ $=\dfrac{1}{2}[(90-x)^{2}+90^{2}]^{-\frac{1}{2}}[2(90-x).(-1)]$

$=\dfrac{x-90}{\sqrt{(90-x)^{2}+90^{2}}}.\dfrac{dx}{dt}=\dfrac{45-90}{\sqrt{(90-45)^{2}+90^{2}}}.24=\dfrac{-1080}{\sqrt{(90-45)^{2}+90^{2}}}=\dfrac{-1080}{{45}\sqrt{5}}=-\dfrac{24}{\sqrt{5}}ft/s$

Is this the correct solution to the problem?

Answer 3

Everything is a function of time. The distance from home plate to first base is expressed by the expression: $$90-24t\ ft$$

The batter is going to be in the middle of this distance at the following time: $$90-24t=45\implies t=1.875\ s$$

By the Pythagorean theorem, the distance from the batter to second base is going to be the following function of time: $$D(t)=\sqrt{90^2+(90-24t)^2}\ ft$$

All you have to do now is find the rate at which the distance changes with respect to time and set it equal to $t=1.875\ s$. That's going to be the answer to a): $$D'(1.875)\ ft/s$$

Part b of the question is similar:

$$D_2(t)=\sqrt{90^2+(24t)^2}$$

And the answer: $$D'_2(1.875)\ ft/s$$