Chapter 5.6, Problem 19E

a.

To determine

To find: the speed of the top of the ladder sliding down the wall.

Answer to Problem 19E

The value of $\frac{dy}{dt}$ is $-\text{12 feet/sec}$ .

Explanation of Solution

Given information:

Hypotenuse is 13 feet.

Base $x$ is 12 feet.

  $\frac{dx}{dt}=5\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the speed of the top of the ladder sliding down the wall at that moment.

Use Pythagoras theorem to find the value of y .

  $\begin{array}{l}{z}^2=x^2+y^2\\ y^2=z^2-x^2\\ y=\sqrt{z^2-x^2}\\ y={\left(z^2-x^2\right)}^{\frac{1}{2}}\end{array}$ .

From the given figure, we can see that the distance y is decreasing. So, $\frac{dy}{dt}$ is negative.

  $y={\left(z^2-x^2\right)}^{\frac{1}{2}}$

Put the value of $z=13$ in above equation.

  $y={\left[{\left(13\right)}^2-x^2\right]}^{\frac{1}{2}}$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dy}{dt}=-\frac{1}{2}{\left[169-x^2\right]}^{\frac{1}{2}-1}·\left(2x\frac{dx}{dt}\right)\\ \frac{dy}{dt}=-\frac{1}{2}{\left[169-x^2\right]}^{-\frac{1}{2}}·\left(2x\frac{dx}{dt}\right)\\ \frac{dy}{dt}=-\frac{1}{2\sqrt{169-x^2}}·2x\frac{dx}{dt}\end{array}$

Putting the value of $x=12$ and $\frac{dx}{dt}=5$ ,

  $\begin{array}{l}\frac{dy}{dt}=-\frac{1}{\sqrt{169-{\left(12\right)}^2}}·\left(12\times 5\right)\\ \frac{dy}{dt}=-\frac{1}{\sqrt{169-144}}·\left(60\right)\\ \frac{dy}{dt}=-\frac{1}{\sqrt{25}}·\left(60\right)\\ \frac{dy}{dt}=-\frac{1}{5}·60\\ \frac{dy}{dt}=-\text{12 feet/sec}\end{array}$

Hence, the value $\frac{dy}{dt}$ is $-\text{12 feet/sec}$ .

b.

To determine

To find: the rate at the area of the triangle formed by the ladder.

Answer to Problem 19E

The value of $\frac{d\text{A}}{dt}$ is $-59.5{\text{ feet}}^2\text{/sec}$ .

Explanation of Solution

Given information:

Hypotenuse is 13 feet.

Base $x$ is 12 feet.

  $\frac{dx}{dt}=5\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rate at the area of the triangle formed by the ladder.

We know that,

Area of a triangle is $\text{A}=\frac{1}{2}\times \text{length}\times \text{breadth}$ ,

From the given figure, we can see that length = x and breadth = y

Therefore,

  $\text{A}=\frac{1}{2}xy$ .

Differentiate with respect to t .

  $\frac{d\text{A}}{dt}=\frac{1}{2}\left[y\frac{dx}{dt}+x\frac{dy}{dt}\right]\mathrm{.....}\left(1\right)$ using chain rule.

First, we have to find the value of y .

From the given diagram,

y $z=13$ 

  $x=12$

  $\begin{array}{l}{z}^2=x^2+y^2\\ y=\sqrt{z^2-x^2}\\ y=\sqrt{{\left(13\right)}^2-{\left(12\right)}^2}\\ y=\sqrt{167-144}\\ y=\sqrt{25}\\ y=5\end{array}$

Putting the value of $x=12$ , $y=5$ and $\frac{dy}{dt}=-12$ in equation (1),

  $\begin{array}{l}\frac{d\text{A}}{dt}=\frac{1}{2}\left[\left(5\right)\left(5\right)+\left(12\right)\left(-12\right)\right]\\ \frac{d\text{A}}{dt}=\frac{1}{2}\left[25-144\right]\\ \frac{d\text{A}}{dt}=\frac{1}{2}\left[-119\right]\\ \frac{d\text{A}}{dt}=\frac{-119}{2}\\ \frac{d\text{A}}{dt}=-59.5{\text{ feet}}^2\text{/sec}\end{array}$

Hence, the value of $\frac{d\text{A}}{dt}$ is $-59.5{\text{ feet}}^2\text{/sec}$ .

c.

To determine

To find: the rate at the angle between the ladder.

Answer to Problem 19E

The value of $\frac{d\theta }{dt}$ is $-1\text{ rad/sec}$ .

Explanation of Solution

Given information:

Hypotenuse is 13 feet.

Base $x$ is 12 feet.

  $\frac{dx}{dt}=5\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rate at the angle between the ladder.

From trigonometry,

  $\begin{array}{l}\sin \theta =\frac{y}{z}=\frac{5}{13}\\ \cos \theta =\frac{x}{z}=\frac{12}{13}\\ \text{tan}\theta =\frac{y}{x}=\frac{5}{12}\end{array}$ .

Take any of them.

  $\sin \theta =\frac{y}{z}=\frac{1}{13}y$

Differentiate with respect to t .

  $\begin{array}{l}\cos \theta \frac{d\theta }{dt}=\frac{1}{13}·\frac{dy}{dt}\\ \frac{d\theta }{dt}=\frac{1}{13\left(\cos \theta \right)}·\frac{dy}{dt}\end{array}$

Putting the value of $\cos \theta =\frac{12}{13}$ and $\frac{dy}{dt}=-12$ in above equation,

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{1}{13\times \frac{12}{13}}\times \left(-12\right)\\ \frac{d\theta }{dt}=\frac{13\times \left(-12\right)}{13\times 12}\\ \frac{d\theta }{dt}=-1\text{ rad/sec}\end{array}$

Hence, the value of $\frac{d\theta }{dt}$ is $-1\text{ rad/sec}$ .