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:

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

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

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

Using the Pythagorean identity:

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

and

\(\displaystyle \frac{dx}{dt}=-r\)

and our original equation, we obtain:

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

Multiply through by $x^2$:

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

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

Now we may plug in the given data:

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

to get:

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

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

\(\displaystyle V=\frac{1}{3}\pi R^2H\)

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:

\(\displaystyle V=\frac{1}{3}\pi x^2y\)

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

\(\displaystyle \frac{x}{y}=\frac{R}{H}\)

Hence:

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

And so we have:

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

Differentiating with respect to $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{R^2}{H^2}\pi y^2\frac{dy}{dt}\)

And so we have:

\(\displaystyle \frac{dy}{dt}=\frac{dV}{dt}\cdot\frac{H^2}{\pi R^2y^2}\)

Now, plugging in the given data:

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

We have:

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

 

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

We have: \(\displaystyle \:\tan\theta \,=\,\tfrac{12}{x} \quad\Rightarrow\quad x \,=\,12\cot\theta\)

Differentate with respect to time:
\(\displaystyle \;\;\;\tfrac{dx}{dt} \:=\:\text{-}12\csc^2\theta\tfrac{d\theta}{dt} \quad\Rightarrow\quad \tfrac{d\theta}{dt} \:=\:\text{-}\tfrac{1}{12}\sin^2\theta\tfrac{dx}{dt}\)

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

The angle of elevation is increasing at \(\displaystyle \tfrac{4}{15}\) radians/sec.