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Worksheet Related Rates - Activity 3.1.5

Activity 39.

A baseball diamond is \(90'\) square. A batter hits a ball along the third base line and runs to first base. At what rate is the distance between the ball and first base changing when the ball is halfway to third base, if at that instant the ball is traveling \(100\) feet/sec? At what rate is the distance between the ball and the runner changing at the same instant, if at the same instant the runner is \(1/8\) of the way to first base running at \(30\) feet/sec?
A diamond-shaped baseball field is shown with four black dots marking the bases. The top dot is labeled Second, the right dot First, the bottom dot Home, and the left dot Third. Blue line segments connect the bases in order around the diamond.
Figure 171. A baseball diamond.
Hint.
Let \(x\) denote the position of the ball along the third base line at time \(t\text{,}\) and \(z\) the distance from the ball to first base. Note that the basepaths meet at 90 degree angles.
Answer.
Let \(x\) denote the position of the ball at time \(t\) and \(z\) the distance from the ball to first base, as pictured below.
A tilted square is shown in gray, with one vertex connected to a blue right triangle. The blue triangle has a right angle at its bottom vertex, a leg labeled x on the left, and a base labeled 90 on the right. A dashed blue segment labeled z runs from the left vertex of the triangle to the right vertex where the triangle meets the square.
\(\left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{feet/sec} \text{.}\)
Let \(r\) be the runner’s position at time \(t\) and let \(s\) be the distance between the runner and the ball, as pictured in Figure 172.
A tilted square is shown in gray. Along its lower edge, two red points mark adjacent vertices. A short blue right-angle segment connects these two points, with the left segment labeled x and the right segment labeled r. A dashed blue segment labeled s runs between the two red points inside the angle.
Figure 172.
\(\left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec} \text{.}\)
Solution.
We let \(x\) denote the position of the ball at time \(t\) and \(z\) the distance from the ball to first base, as pictured below.
A tilted square is shown in gray, with one vertex connected to a blue right triangle. The blue triangle has a right angle at its bottom vertex, a leg labeled x on the left, and a base labeled 90 on the right. A dashed blue segment labeled z runs from the left vertex of the triangle to the right vertex where the triangle meets the square.
Figure 173.
By the Pythagorean Theorem, we know that \(x^2 + 90^2 = z^2\text{;}\) differentiating with respect to \(t\text{,}\) we have
\begin{equation*} 2x\frac{dx}{dt} = 2z\frac{dz}{dt}\text{.} \end{equation*}
At the instant the ball is halfway to third base, we know \(x = 45\) and \(\left. \frac{dx}{dt} \right|_{x = 45} = 100\text{.}\) Moreover, by Pythagoras, \(z^2 = 90^2 + 45^2\text{,}\) so \(z = 45\sqrt{5}\text{.}\) Thus,
\begin{equation*} 2 \cdot 45 \cdot 100 = 2 \cdot 45 \sqrt{5} \cdot \left. \frac{dz}{dt} \right|_{x = 45}\text{,} \end{equation*}
so
\begin{equation*} \left. \frac{dz}{dt} \right|_{x = 45} = \frac{100}{\sqrt{5}} \approx 44.7214 \ \text{feet/sec}\text{.} \end{equation*}
For the second question, we still let \(x\) represent the ball’s position at time \(t\text{,}\) but now we introduce \(r\) as the runner’s position at time \(t\) and let \(s\) be the distance between the runner and the ball. In this setting, as seen in the diagram in Figure 174.
A tilted square is shown in gray. Along its lower edge, two red points mark adjacent vertices. A short blue right-angle segment connects these two points, with the left segment labeled x and the right segment labeled r. A dashed blue segment labeled s runs between the two red points inside the angle.
Figure 174.
\(x\text{,}\) \(r\text{,}\) and \(s\) form the sides of a right triangle, so that
\begin{equation*} x^2 + r^2 = s^2\text{,} \end{equation*}
by the Pythagorean Theorem. Differentiating each side with respect to \(t\text{,}\) it follows that the three rates of change are related by the equation
\begin{equation*} 2x \frac{dx}{dt} + 2r \frac{dr}{dt} = 2s \frac{ds}{dt}\text{.} \end{equation*}
We are given that at the instant \(x = 45\text{,}\) \(r = \frac{90}{8}\text{,}\) so by Pythagoras, \(s = \frac{45}{4}\sqrt{17}\text{.}\) In addition, at this same instant we know that \(\left. \frac{dx}{dt} \right|_{x = 45} = 100\) and \(\left. \frac{dr}{dt} \right|_{x = 45} = 30\text{.}\) Applying this information,
\begin{equation*} 2 \cdot 45 \cdot 100 + 2 \cdot \frac{45}{4} \cdot 30 = 2 \cdot \frac{45}{4}\sqrt{17} \cdot \left. \frac{ds}{dt} \right|_{x = 45} \end{equation*}
and therefore
\begin{equation*} \left. \frac{ds}{dt} \right|_{x = 45} = \frac{430}{\sqrt{17}} \approx 104.2903 \ \text{feet/sec}\text{.} \end{equation*}
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