Chapter 4.1, Problem 18E

(a)

To determine

To find: The rate is distance from seconds base decreasing.

Answer to Problem 18E

The distance from the second base is decreasing with rate of $\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$

Explanation of Solution

Given information:

The given expression is baseball is a square $90$ ft. and toward first base with a speed of $24\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$ .

Calculation:

Calculate the expression.

From the figure, the man hits the ball at a point $H$ and runs towards the point $F$ .

  

Figure (1)

We have been given that $HF=FS=ST=TH=90ft$

The distance between man and second base at time

  $\begin{array}{l}S{P}^2=F{S}^2+P{F}^2\\ Y^2={\left(90\right)}^2+{\left(90-X\right)}^2\mathrm{.............}\left(1\right)\end{array}$

Differentiating with respect to time

  $\begin{array}{c}2y\frac{dy}{dt}=-2\left(90-x\right)\frac{dx}{dt}\\ \frac{dy}{dt}=-\frac{\left(90-x\right)}{y}\left(\frac{dx}{dt}\right)\mathrm{..........}\left(2\right)\\ putx=45ftfrom\left(1\right)\\ y^2={\left(90\right)}^2+{\left(90-45\right)}^2\\ y^2=10125\\ y=\sqrt{10125}=45\sqrt{5}ft\\ from\left(2\right)wehave\\ \frac{dy}{dt}=-\frac{\left(90-45\right)}{45\sqrt{5}}\left(24\right)\\ \frac{dy}{dt}=-\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.\end{array}$

Therefore, the distance from the second base is decreasing with rate of $\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$

(b)

To determine

To find: The rate is distance from third base increasing.

Answer to Problem 18E

The distance from the third base is increasing with rate of $\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$ .

Explanation of Solution

Given information:

The given expression is baseball is a square $90$ ft. and toward first base with a speed of $24\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$ .

Calculation:

Calculate the expression.

From the figure (1),

  

Figure (1)

  $\begin{array}{c}T{P}^2=T{H}^2+H{P}^2\\ z^2={\left(90\right)}^2+x^2\mathrm{...........}\left(3\right)\\ 2z\frac{dz}{dt}=2x\frac{dx}{dt}\\ \frac{dz}{dt}=\left(\frac{x}{z}\right)\frac{dx}{dt}\mathrm{.............}\left(4\right)\\ substitutethevalueofx=45from\left(3\right)\\ z^2={\left(90\right)}^2+{45}^2\\ z^2=10125\\ z=\sqrt{10125}=45\sqrt{5}ft\\ \frac{dz}{dt}=\left(\frac{45}{45\sqrt{5}}\right)\left(24\right)\\ \frac{dz}{dt}=\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.\end{array}$

Therefore the distance from the third base is increasing with rate of $\frac{24}{\sqrt{5}}\raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s.$}\right.$