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djverab [1.8K]
3 years ago
6

Match the parametric equations with the verbal descriptions of the surfaces by putting the letter of the verbal description to t

he left of the letter of the parametric equation.
1. \mathbf{r} \left( u, v \right) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^{2} \mathbf{k}
2. \mathbf{r} \left( u, v \right) = u \mathbf{i} + u \cos v \mathbf{j} + u \sin v \mathbf{k}
3. \mathbf{r} \left( u, v \right) = u \mathbf{i} + \cos v \mathbf{j} + \sin v \mathbf{k}
4. \mathbf{r} \left( u, v \right) = u \mathbf{i} + v \mathbf{j} + \left( 2u - 3v \right) \mathbf{k}


A. circular cylinder
B. circular paraboloid
C. cone
D. plane
Mathematics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

1. B

2. C

3. A

4. D

Step-by-step explanation:

The parametric equations of the circular cylinder are:

x(u,v)=a\cos v\\y(u,v)=a\sin v\\z(u,v)=u

If the orientation of the cylinder is changed to have the height u along the x-axis, the parametric equations of the cylinder match:

3. \mathbf{r} \left( u, v \right) = u \mathbf{i} + \cos v \mathbf{j} + \sin v \mathbf{k}

The parametric equations of the circular paraboloid are:

x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=u^2

Using the units vectors the parametric equations match:

1. \mathbf{r} \left( u, v \right) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + u^{2} \mathbf{k}

The parametric equations of the cone are:

x(u,v)=au\cos v\\y(u,v)=au\sin v\\z(u,v)=u

Using the units vectors  and rotating the base of the cone from z=0 to x=0  the parametric equations match:

2. \mathbf{r} \left( u, v \right) = u \mathbf{i} + u \cos v \mathbf{j} + u \sin v \mathbf{k}

The equation left is the equation of a plane:

4. \mathbf{r} \left( u, v \right) = u \mathbf{i} + v \mathbf{j} + \left( 2u - 3v \right) \mathbf{k}

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The correct answer is:  [B]:  " \frac{(-3)(4)}{(6)} " . 
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Consider choice [A]:  \frac{(3)(4)}{(6)} ; 
                              = \frac{12}{6} ; 
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Consider choice [B]:  \frac{(-3)(4)}{(6)} ; 
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Consider choice [C]:  \frac{(-3)(-4)}{(6)} ; 
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3 years ago
Find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.
amm1812

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Step-by-step explanation:

For this case we are interested on the region shaded on the figure attached.

And we can find the volume with the method of rings.

The area on this case is given by:

A(x) = \pi [f(x)]^2 = \pi r^2 = \pi [3x]^2 = 9\pi x^2

And the volume is given by the following formula:

V= \int_{a}^b A(x) dx

For our case our limits are x=0 and x=2 so we have this:

V = \pi \int_{0}^{2} x^2 dx

And if we solve the integral we got this:

V= \pi [\frac{x^3}{3}]\Big|_0^{2}

And after evaluate we got this:

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3 years ago
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