MHB Tessa's questions at Yahoo Answers regarding related rates

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Related rates
AI Thread Summary
Tessa's questions focus on related rates in calculus, specifically involving a person observing a streetlight and the rising water level in a conical tank. For the first problem, the rate at which the angle of elevation of the person's line of sight changes, when 9 feet from the light, is calculated to be 4/15 radians per second. The second question involves determining how fast the water level is rising in a cone-shaped tank when the water is 12 feet deep, with the result being 20/(9π) feet per minute. Both problems utilize differentiation and geometric relationships to derive the rates. The thread provides detailed solutions and formulas for each scenario.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$\bbox[5px,border:2px solid red]{\left.\frac{dy}{dt} \right|_{y=12\text{ ft}}= 80\cdot \frac{40^2}{\pi\cdot20^2\cdot12^2}\, \frac{\text{ft}}{\text{min}}=\frac{20}{9\pi}\, \frac{\text{ft}}{\text{min}}}$$
 
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?
Code:
    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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top