Chapter 4.1, Problem 13E

(a)

To determine

To find: The rate at which the distance from plane to station is incresing when distance is $2$ miles.

Answer to Problem 13E

The rate at which distance from plan to station is increasing at $250\sqrt{3}$ mi/h.

Explanation of Solution

Given information:

Given that the plane is flying horizontally at an altitude of 1 mile and speed is $500$ miles/h.

Calculation:

Draw the figure of given situation.

  

From the given figure , by the pythagorean theorem.

  $y^2=1+x^2\mathrm{.......}\left(1\right)$

Differentiate $\left(1\right)$ with respect to $t$

  $\begin{array}{c}2y\frac{dy}{dt}=2x\frac{dx}{dt}\\ \frac{dy}{dt}=\frac{x}{y}\frac{dx}{dt}\end{array}$

Since $y=2$ miles, then the equation $y^2=1+x^2$ is,

  $\begin{array}{l}4=1+x^2\\ x^2=3\\ x=\sqrt{3}\end{array}$

And since, the speed of the plane is, $\frac{dx}{dt}=500$ miles/hr

Substitute the values of $x=\sqrt{3}$ miles, $y=2$ miles, and $\frac{dx}{dt}=500$ miles/hr in the equation $\frac{dy}{dt}=\frac{\sqrt{3}}{2}.500$ mi/h

  $=250\sqrt{3}$ mi/h

Therefore, the rate at which distance from plan to station is increasing at $250\sqrt{3}$ mi/h

(b)

To determine

To find: The $\frac{dx}{dt}$ when $\frac{dy}{dt}=3$ and $x=12$ .

Answer to Problem 13E

The coordinate of line 3 and line 4 are not same..

Explanation of Solution

Given information:

Line3: $x=4t,$

  $y=3-6t,$

  $z=-5+2t$

Line 4: $x=8-2t$

  $y=-9+3t,$

  $z=6-t$

Calculation:

Differentiating equation with repect to $t$

  $\begin{array}{c}\frac{dy}{dt}=\frac{1}{2\sqrt{2x+1}}.\left(2.1+0\right).\frac{dx}{dt}\\ =\frac{1}{2\sqrt{2x+1}}.\left(2\right).\frac{dx}{dt}\\ =\frac{1}{\sqrt{2x+1}}\frac{dx}{dt}\end{array}$

Thus,

  $\begin{array}{c}\frac{dy}{dt}=\frac{1}{\sqrt{2x+1}}\frac{dx}{dt}\\ \end{array}$

Here, $\frac{dy}{dt}=5,x=12$

Calculate $\frac{dx}{dt}$

  $\begin{array}{c}\frac{dy}{dt}=\frac{1}{\sqrt{2x+1}}\frac{dx}{dt}\\ 5=\frac{1}{\sqrt{2.12+1}}\frac{dx}{dt}\\ 5=\frac{1}{\sqrt{25}}\frac{dx}{dt}\\ 5=\frac{1}{5}\frac{dx}{dt}\\ 25=\frac{dx}{dt}\end{array}$

Therefore, $\frac{dy}{dt}=1$ .