Revolving Area and Related Rates

**Raj** · June 11th 2009, 03:10 PM

I have three problems that I need help with,

1. $f(x)=-x^3+4x$
The area under the curve from $x=0$ to $x=2$ is revolved about the x-axis. Calculate the volume of the surface of revolution.

I used the disk method and found the volume to be $\frac{1024\pi}{105}$ , can somone verify please?

2. $x^3y^4+xy^5=2$
Suppose an object is moving along with this curve. When the object arrived at (1,1) it was moving up with a speed of 4m/s. Calculate the speed and direction of the motion horizontally.

As well I have not had a problem like this before. I found the slope though to be $-4/9$ and the equation of the line to be $y=-4x/9+13/9$ . I suppose i am to either derive the equation or integrate, but i don't have a clue.

3. A boy is 20m away from a rising balloon. The angle of elevation is increasing at a rate of 3 degrees/sec. At the instant the balloon is 8m above the ground find its speed of ascent.

d = 20m
$d\theta/dt$ = 3 deg/sec
$dh/dt=$ ? when h = 8m

I do not know what equation i would be using, i tried using $tan\theta=h/20$ and my answer was 69.6m/s but i think that is wrong.

Help appreciated very much.

**apcalculus** · June 11th 2009, 03:31 PM

I'll get you started with the first problem. The method is called Disk Method and we use it whenever the region that's being revolved is adjacent to the rotational axis.

The formula is INTEGRAL Pi (RADIUS^2) dx from x=a to x=b.

The RADIUS is the distance from the curve to the rotational axis, in your case it is simply the y-coordinate, so R(x) = -x^3 + 4x

Set up the integral from x=0 to x=2.

Good luck!

**Raj** · June 11th 2009, 03:43 PM

Originally Posted by apcalculus

I'll get you started with the first problem. The method is called Disk Method and we use it whenever the region that's being revolved is adjacent to the rotational axis.

The formula is INTEGRAL Pi (RADIUS^2) dx from x=a to x=b.

The RADIUS is the distance from the curve to the rotational axis, in your case it is simply the y-coordinate, so R(x) = -x^3 + 4x

Set up the integral from x=0 to x=2.

Good luck!

$\int\pi r^2dx$ from x=2 to x=0

Are you saying the radius = $-x^3 + 4x$ ?

**apcalculus** · June 11th 2009, 03:46 PM

Originally Posted by Raj

$\int\pi r^2dx$ from x=2 to x=0

Are you saying the radius = $-x^3 + 4x$ ?

Correct, so you're squaring the cubic function.

Some low-quality demos: DISK METHOD DEMO GALLERY

**Soroban** · June 11th 2009, 04:36 PM

Hello, Raj!

Here's the second one . . .

2. . $x^3y^4 + xy^5 \:=\:2$
Suppose an object is moving along this curve.
When the object arrived at (1,1) it was moving up with a speed of 4 m/s.
Calculate the speed and direction of the motion horizontally.

Since this is a Related Rate problem,
. . you might expect to differentiate with respect to time.

When we do, we get: . $x^3\!\cdot\!4y^3\frac{dy}{dt} + 3x^2\!\cdot\!y^4\frac{dx}{dt} + x\!\cdot\!5y^4\frac{dy}{dt} + 1\!\cdot\!y^5\frac{dx}{dt} \:=\:0$

Re-arrange terms: . $3x^2y^4\frac{dx}{dt} + y^5\frac{dx}{dt} \;=\;-4x^3y^3\frac{dy}{dy} - 5xy^4\frac{dy}{dt}$

Factor: . $y^4(3x^2+y)\frac{dx}{dt} \;=\;-xy^3(4x^2 + 5y)\frac{dy}{dt}$

. . Hence: . $\frac{dx}{dt}\;=\;-\frac{x(4x^2+5y)}{y(3x^2+y)}\!\cdot\!\frac{dy}{dt}$ .[1]

We are given the point (1,1) and the object is moving up at 4 m/sec.
. . That is: . $x = 1,\;y = 1,\;\frac{dy}{dt} = 4$

Substitute into [1]: . $\frac{dx}{dt} \;=\;-\frac{1(4+5)}{1(3+1)}(4) \;=\;-9$

The object is moving at 9 m/sec to the left.

**Soroban** · June 11th 2009, 05:02 PM

Hello again, Raj!

3. A boy is 20 m away from a rising balloon.
The angle of elevation is increasing at a rate of 3 degrees/sec.
At the instant the balloon is 8 m above the ground, find its speed of ascent.

Code:

                  *
                * |
              *   |
            *     | h
          *       |
        * θ       |
      * - - - - - *
           20

We have: . $\tan\theta \:=\:\frac{h}{20} \quad\Rightarrow\quad h \:=\:20\tan\theta$

Differentiate with respect to time: . $\frac{dh}{dt} \:=\:20\!\cdot\!\sec^2\!\theta\!\cdot\!\frac{d\the ta}{dt}$ .[1]

We are given: . $\frac{d\theta}{dt} \:=\:3\text{ deg/sec} \:=\:\frac{\pi}{60}\text{ radians/sec}$

We are told that $h = 8$ .
The right triangle looks like this:

Code:

                      *
                  *   |
              *       | 8
          * θ         |
      * - - - - - - - *
              20

We have: . $opp = 8,\;adj = 20$
Pythagorus says: . $hyp = \sqrt{464} = 4\sqrt{29}$
Hence: . $\sec\theta \:=\:\frac{hyp}{adj} \:=\:\frac{4\sqrt{29}}{20} \:=\:\frac{\sqrt{29}}{5}$

Substitute into [1]: . $\frac{dh}{dt}\;=\;20\left(\frac{\sqrt{29}}{5}\righ t)^2\!\left(\frac{\pi}{60}\right) \;=\;\frac{29\pi}{75}\text{ m/sec}$

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