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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #95 : Polar

Calculate the polar form hypotenuse of the following cartesian equation: $y=\frac{2x^2}{7}$

Possible Answers:

$\frac{7\tan(\theta)}{2\cos(\theta)}$

$\frac{7\tan(\theta)}{2\cos^2(\theta)}$

$\frac{7\tan(\theta)}{2\sin(\theta)}$

$\frac{7\tan(\theta)}{2\sin^2(\theta)}$

Correct answer:

$\frac{7\tan(\theta)}{2\cos(\theta)}$

Explanation:

In a cartesian form, the primary parameters are and . In polar form, they are and $\theta$

is the hypotenuse, and $\theta$ is the angle created by .

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

$y=\frac{2x^2}{7} = r\sin(\theta)$

$x= r\cos(\theta)$

$y=\frac{2(r\cos(\theta))^2}{7}$

$y=\frac{2r^2\cos^2(\theta)}{7} = r\sin(\theta)$

$r = \frac{7\tan(\theta)}{2\cos(\theta)}$

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Example Question #771 : Calculus Ii

Calculate the polar form hypotenuse of the following cartesian equation: y^2=100x^8

Possible Answers:

$\sqrt[3]{\frac{\tan(\theta)}{\cos(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

$\frac{\tan(\theta)}{10\cos^3(\theta)}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos(\theta)}}$

Correct answer:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Explanation:

In a cartesian form, the primary parameters are and . In polar form, they are and $\theta$

is the hypotenuse, and $\theta$ is the angle created by .

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

y^2=100x^8

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

$r^3 = \frac{\tan(\theta)}{10\cos^3(\theta)}$

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

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Example Question #97 : Polar

What is the polar form of $y=\frac{4}{3}x^{2}$

Possible Answers:

$r=\tan\frac{3}{4} \theta\sec\theta$

$r=\frac{4}{3}\tan \theta\sec\theta$

$r=\frac{3}{4}\tan \theta\sec\theta$

$r=-\frac{4}{3}\tan \theta\sec\theta$

$r=-\frac{3}{4}\tan \theta\sec\theta$

Correct answer:

$r=\frac{3}{4}\tan \theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=\frac{4}{3}x^{2}$ , then:

$r\sin\theta=\frac{4}{3}(r\cos \theta)^{2}$

$r\sin\theta=\frac{4}{3}r^{2}\cos^{2} \theta$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=\frac{4}{3}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{4}{3}r$

$\tan \theta\sec\theta=\frac{4}{3}r$

$r=\frac{3}{4}\tan \theta\sec\theta$

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Example Question #91 : Polar Form

What is the polar form of $y=-x^{2}$ ?

Possible Answers:

$r=-\frac{1}{\tan \theta\sec\theta}$

$r=-\frac{\tan \theta}{\sec\theta}$

$r=\frac{1}{\tan \theta\sec\theta}$

$r=-\tan \theta\sec\theta$

$r=\tan \theta\sec\theta$

Correct answer:

$r=-\tan \theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=-x^{2}$ , then:

$r\sin\theta=-(r\cos \theta)^{2}$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=-r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=-r$

$\tan \theta\sec\theta=-r$

$r=-\tan \theta\sec\theta$

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Example Question #92 : Polar Form

What is the polar form of y=6+14x ?

Possible Answers:

$r=\frac{\sin\theta-14\cos \theta}{6}$

$r=-\frac{6}{\sin\theta-14\cos \theta}$

$r=\frac{6}{\sin\theta+14\cos \theta}$

$r=\frac{6}{\sin\theta-14\cos \theta}$

$r=-\frac{6}{\sin\theta+14\cos \theta}$

Correct answer:

$r=\frac{6}{\sin\theta-14\cos \theta}$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given y=6+14x , then:

$r\sin\theta=6+14(r\cos \theta)$

$r\sin\theta-14(r\cos \theta)=6$

$r(\sin\theta-14\cos \theta)=6$

$r=\frac{6}{\sin\theta-14\cos \theta}$

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Example Question #93 : Polar Form

What is the polar form of y=2-9x ?

Possible Answers:

$r=-\frac{2}{\sin\theta-9\cos \theta}$

$r=\frac{2}{\sin\theta-9\cos \theta}$

$r=\frac{2}{\sin\theta+9\cos \theta}$

$r=-\frac{2}{\sin\theta+9\cos \theta}$

$r=\frac{\sin\theta-9\cos \theta}{2}$

Correct answer:

$r=\frac{2}{\sin\theta+9\cos \theta}$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given y=2-9x , then:

$r\sin\theta=2-9(r\cos \theta)$

$r\sin\theta+9(r\cos \theta)=2$

$r(\sin\theta+9\cos \theta)=2$

$r=\frac{2}{\sin\theta+9\cos \theta}$

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Example Question #1 : Graphing Polar Form

Which of the following substitutions will help solve the following integral?

$\int\frac{\sqrt{25x^2-4}}{x}dx$

Possible Answers:

$x = \frac{2}{5}sec\theta$

u = 25x^2 - 4

$u = \frac{1}{x}$

$x = \frac{2}{5} tan\theta$

$x = \frac{2}{5}sin\theta$

Correct answer:

$x = \frac{2}{5}sec\theta$

Explanation:

As we can see in this integral, there is no reverse chain-rule u-substitution possible. The logical step is to use a trigonometric substitution. If one recalls that trig substitutions of the type x^2-a^2 could be solved with the substitution $\frac{2}{5}sec\theta$ , then the answer is easily seen. However, we can also use a right triangle: Screen_shot_2015-04-18_at_6.33.31_pm

And thus we have:

$sec \theta = \frac{5x}{2}$

or:

$x = \frac{2}{5}sec\theta$

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Example Question #21 : Parametric, Polar, And Vector Functions

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

R_sinx

R_cosx

R_cosx_1

R_cos2x

Faker_cosx

Correct answer:

R_cosx

Explanation:

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

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Example Question #7 : Parametric Form

Draw the graph of $r=sin\theta$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R_sinx_1

R_cosx

R_sinx

Faker_cosx

R_sin2x

Correct answer:

R_sinx

Explanation:

Between and $\frac{\pi }{2}$ , the radius approaches from .

From $\frac{\pi }{2}$ to $\pi$ the radius goes from to .

Between $\pi$ and $\frac{3\pi }{2}$ , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From $\frac{3\pi }{2}$ and $2\pi$ , the curve is redrawn in the second quadrant as the radius approaches from .

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Example Question #22 : Parametric, Polar, And Vector Functions

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R_sin2x

R_cos2x

R_sin2x

R2_cos2x

R_cosx_1

Correct answer:

R_cos2x

Explanation:

Because this function has a period of $\pi$ , the x-intercepts of the graph y=cos(2x) happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #95 : Polar

Example Question #771 : Calculus Ii

Example Question #97 : Polar

Example Question #91 : Polar Form

Example Question #92 : Polar Form

Example Question #93 : Polar Form

Example Question #1 : Graphing Polar Form

Example Question #21 : Parametric, Polar, And Vector Functions

Example Question #7 : Parametric Form

Example Question #22 : Parametric, Polar, And Vector Functions

All Calculus 2 Resources

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