# Practice Problems for Module 13

Covering Sections 10.1 and 10.2

Directions.
The following are review problems. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link for the full video page to find related videos.
1. Convert the given parametric equations to Cartesian equations and find the corresponding graph.
1. $$\displaystyle x=-t^2 \text{ , } y=t+1 \text{ , } -3 \leq t \leq 3$$

$$\displaystyle x = -(y-1)^2$$

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2. $$\displaystyle x = \sqrt{t} \text{ , } y = 1 - t$$​

$$\displaystyle x = \sqrt{1 -y} \text{ , } x \geq 0$$

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2. Eliminate the parameter to obtain the Cartesian equation and find the graph given by the following.
1. $$\displaystyle x=6\cos(\theta)\text{ , } y=7\sin(\theta) \text{ , } -\pi/2 \leq t \leq \pi/2$$​

$$\displaystyle \frac{x}{36} + \frac{y}{49} = 1$$

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2. $$\displaystyle x = \sin(t) \text{ , } y = \csc(t) \text{ , } 0 < t < \pi/2$$​

$$\displaystyle y=\frac{1}{x}$$

Video errata: There should be an open circle at the point $$(1,1)$$.

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3. ​Find the exact length of the curve given by $$x = e^t + e^{-t}$$, $$y= 5 - 2t$$, $$0 \leq t \leq 2$$.

$$\displaystyle e^2 - e^{-2}$$

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4. Find the integral that represents the surface area obtained by rotating the following curve about the $$x$$-axis.$x= t\cos(t), \quad y=t\sin(t), \quad 0\leq t \leq \pi / 2$

$$\displaystyle 2\pi\int\limits_0^{\pi/2} \! t\sin(t)\sqrt{t^2+1}\,dt$$

Video Errata: The Surface Area formula and the answer both need to be multiplied by $$2\pi$$. Also the $$t$$ inside the square root should be $$t^2$$.

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5. Find the exact area of the surface obtained by rotating the given curve about the $$y$$-axis.
$x= 9t^2, \quad y=9t - 3t^3, \quad 0 \leq t \leq 2$

$$\displaystyle 162\pi\left(\frac{8}{3}+\frac{32}{5} \right)$$

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