a baseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a…

a baseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 30 ft/s. (a) at what rate (in ft/s) is his distance from second base decreasing when he is halfway to first base? (round your answer to one decimal place.) ft/s (b) at what rate (in ft/s) is his distance from third base increasing at the same moment? (round your answer to one decimal place.) ft/s resources read it

a baseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 30 ft/s. (a) at what rate (in ft/s) is his distance from second base decreasing when he is halfway to first base? (round your answer to one decimal place.) ft/s (b) at what rate (in ft/s) is his distance from third base increasing at the same moment? (round your answer to one decimal place.) ft/s resources read it

Answer

Explanation:

Step1: Set up variables

Let $x$ be the distance of the runner from home - plate. The side - length of the square baseball diamond is $a = 90$ ft. The distance $y_1$ from the runner to second - base and the distance $y_2$ from the runner to third - base can be found using the Pythagorean theorem. The distance from the runner to second - base $y_1=\sqrt{(90)^2+(90 - x)^2}$, and the distance from the runner to third - base $y_2=\sqrt{(90)^2+x^2}$. The speed of the runner $\frac{dx}{dt}=- 30$ ft/s (negative because $x$ is increasing as the runner moves towards first - base).

Step2: Differentiate $y_1$ with respect to $t$

Using the chain - rule, if $y_1=\sqrt{(90)^2+(90 - x)^2}=(8100+(90 - x)^2)^{\frac{1}{2}}$, then $\frac{dy_1}{dt}=\frac{1}{2}(8100+(90 - x)^2)^{-\frac{1}{2}}\times2(90 - x)(-1)\frac{dx}{dt}$. When the runner is halfway to first - base, $x = 45$ ft. Substitute $x = 45$ ft and $\frac{dx}{dt}=-30$ ft/s into the formula for $\frac{dy_1}{dt}$. [ \begin{align*} y_1&=\sqrt{8100+(90 - 45)^2}=\sqrt{8100 + 2025}=\sqrt{10125}=100.623\ \frac{dy_1}{dt}&=\frac{- (90 - x)\frac{dx}{dt}}{\sqrt{(90)^2+(90 - x)^2}}\ &=\frac{-(90 - 45)\times(-30)}{\sqrt{8100+(90 - 45)^2}}\ &=\frac{45\times30}{\sqrt{8100 + 2025}}\ &=\frac{1350}{\sqrt{10125}}\ &\approx13.4 \end{align*} ]

Step3: Differentiate $y_2$ with respect to $t$

If $y_2=\sqrt{(90)^2+x^2}=(8100+x^2)^{\frac{1}{2}}$, then $\frac{dy_2}{dt}=\frac{1}{2}(8100+x^2)^{-\frac{1}{2}}\times2x\frac{dx}{dt}=\frac{x\frac{dx}{dt}}{\sqrt{(90)^2+x^2}}$. Substitute $x = 45$ ft and $\frac{dx}{dt}=-30$ ft/s into the formula for $\frac{dy_2}{dt}$. [ \begin{align*} y_2&=\sqrt{8100 + 45^2}=\sqrt{8100+2025}=\sqrt{10125}=100.623\ \frac{dy_2}{dt}&=\frac{45\times(-30)}{\sqrt{8100 + 45^2}}\ &=\frac{-1350}{\sqrt{10125}}\ &\approx - 13.4 \end{align*} ] The negative sign for $\frac{dy_2}{dt}$ is incorrect because we want the rate of increase. So the magnitude is considered.

Answer:

(a) $13.4$ ft/s (b) $13.4$ ft/s