Tessa's questions at Yahoo Answers regarding related rates

[MarkFL]

Here are the questions:

CALC 1 word problems?

Q1.

A person 6 ft tall is watching a streetlight 18 ft high while walking toward it at a speed of 5 ft/sec (see figure 3.53 in the book). At what rate is the angle of elevation of the person's line of sight changing with respect to time when the person is 9 ft from the base of the light?
(Let x(t) denote the distance from the lamp at time t and \theta the angle of elevation at time t)

d(theta)/dt = ? radians/sec

Q2.

Model a water tank by a cone 40 ft high with a circular base of radius 20 feet at the top. Water is flowing into the tank at a constant rate of 80 cubic ft/min . How fast is the water level rising when the water is 12 ft deep?
(Let x(t) denote the radius of the top circle of water at time t and y(t) denote the height of the water at time t.)

dy/dt = ? ft/min

Please help, thank you!

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Tessa,

Q1.) Let's work this problem in general terms so that we then get a formula into which we can plug the given data. Let:

$h$ = the height of the person.

$s$ = the height of the streetlight.

$r$ = the rate at which the person is walking towards the streetlight.

$\theta$ = the angle of inclination of the person's line of sight at time $t$ to the streetlight.

$x$ = the persons horizontal distance at time $t$ from the person to the streetlight.

Now, we may state:

$$\tan(\theta)=\frac{s-h}{x}$$

Differentiating with respect to time $t$, we obtain:

$$\sec^2(\theta)\frac{d\theta}{dt}=\frac{h-s}{x^2}\frac{dx}{dt}$$

Using the Pythagorean identity:

$$\tan^2(u)+1=\sec^2(u)$$

and

$$\frac{dx}{dt}=-r$$

and our original equation, we obtain:

$$\left(\left(\frac{s-h}{x} \right)^2+1 \right)\frac{d\theta}{dt}=\frac{r(s-h)}{x^2}$$

Multiply through by $x^2$:

$$\left((s-h)^2+x^2 \right)\frac{d\theta}{dt}=r(s-h)$$

$$\frac{d\theta}{dt}=\frac{r(s-h)}{(s-h)^2+x^2}$$

Now we may plug in the given data:

$$x=9\text{ ft},\,r=5\frac{\text{ft}}{\text{s}},\,h=6\text{ ft},\,s=18\text{ ft}$$

to get:

$$\bbox[5px,border:2px solid red]{\left.\frac{d\theta}{dt} \right|_{x=9\text{ ft}}=\frac{5(18-6)}{(18-6)^2+9^2}\,\frac{\text{rad}}{\text{s}}=\frac{4}{15}\,\frac{\text{rad}}{\text{s}}}$$

Q2.) Let $R$ be the radius of the tank at the top and $H$ be the height of the tank.

$$V=\frac{1}{3}\pi R^2H$$

Now let $y$ be the depth of water in the tank at time $t$ and $x$ be the radius of the surface at that time. The the volume $V$ of water at time $t$ is given by:

$$V=\frac{1}{3}\pi x^2y$$

By similarity, we know that at any time, we must have:

$$\frac{x}{y}=\frac{R}{H}$$

Hence:

$$x^2=\left(\frac{R}{H}y \right)^2$$

And so we have:

$$V=\frac{1}{3}\pi\left(\frac{R}{H}y \right)^2y=\frac{R^2}{3H^2}\pi y^3$$

Differentiating with respect to $t$, we obtain:

$$\frac{dV}{dt}=\frac{R^2}{H^2}\pi y^2\frac{dy}{dt}$$

And so we have:

$$\frac{dy}{dt}=\frac{dV}{dt}\cdot\frac{H^2}{\pi R^2y^2}$$

Now, plugging in the given data:

$$\frac{dV}{dt}=80\frac{\text{ft}^3}{\text{min}},\,H=40\text{ ft},\,R=20\text{ ft},\,y=12\text{ ft}$$

We have:

$$\bbox[5px,border:2px solid red]{\left.\frac{dy}{dt} \right|_{y=12\text{ ft}}= 80\cdot \frac{40^2}{\pi\cdot20^2\cdot12^2}\, \frac{\text{ft}}{\text{min}}=\frac{20}{9\pi}\, \frac{\text{ft}}{\text{min}}}$$

 

[soroban]

Hello, Tessa!

MarkFL's explanation is excellent.
I'll solve it directly.

(Q1) A person 6 ft tall is watching a streetlight 18 ft high
while walking toward it at a speed of 5 ft.sec.
At what rate is the angle of elevation of the person's
line of sight changing when the person is 9 ft from
the base of the light?

    C *
      |   * 
   12 |       *
      |           *
      |           θ   *
    E * - - - - - - - - - * A
      |         x         |
    6 |                   | 6
      |                   |
    D *-------------------* B
      : - - - - x - - - - :

The person is [math]AB = 6.[/math]
The streetlight is [math]CD = 18.[/math]
His distance is [math]BD = AE = x.\;\;\tfrac{dx}{dt} = \text{-}5[/math] ft/sec.
Let [math]\theta = \angle CAE.[/math]

We have: [math]\:\tan\theta \,=\,\tfrac{12}{x} \quad\Rightarrow\quad x \,=\,12\cot\theta[/math]

Differentate with respect to time:
[math]\;\;\;\tfrac{dx}{dt} \:=\:\text{-}12\csc^2\theta\tfrac{d\theta}{dt} \quad\Rightarrow\quad \tfrac{d\theta}{dt} \:=\:\text{-}\tfrac{1}{12}\sin^2\theta\tfrac{dx}{dt}[/math]

When [math]x = 9,\:\sin\theta \,=\,\tfrac{4}{5}[/math]
We have: [math]\;\tfrac{d\theta}{dt} \:=\:\text{-}\tfrac{1}{12}\left(\tfrac{4}{5}\right)^2(\text{-}5) \:=\:\tfrac{4}{15}[/math]

The angle of elevation is increasing at [math]\tfrac{4}{15}[/math] radians/sec.