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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 #2 : Polar

Give the rectangular coordinates of the point with polar coordinates

$\left ( -4 , \frac{\pi }{3}\right )$ .

Possible Answers:

$\left ( -2 \sqrt{3} , -2\right )$

$\left ( -2, -2 \sqrt{3} \right )$

$\left ( -2 \sqrt{3} , 2\right )$

$\left ( 2 \sqrt{3} , -2\right )$

$\left ( 2, -2 \sqrt{3} \right )$

Correct answer:

$\left ( -2, -2 \sqrt{3} \right )$

Explanation:

$x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$

$y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}$

The point will have rectangular coordinates $\left (- 2, -2 \sqrt{3} \right )$ .

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Example Question #1 : Polar Form

What would be the equation of the parabola $\small y=x^{2}$ in polar form?

Possible Answers:

$\small r=cot\theta cos\theta$

$\small \small \small r=tan\theta sin\theta$

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

$\small \small r=cot\theta csc\theta$

$\small r=\theta ^{2}$

Correct answer:

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

Explanation:

We know $\small y=rsin\theta$ and $\small x=rcos\theta$ .

Subbing that in to the equation $\small y=x^{2}$ will give us $\small rsin\theta=r^{2}cos^{2}\theta$ .

Multiplying both sides by $\small 1/(rcos^{2}\theta)$ gives us

$\small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta$ .

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Example Question #1 : Polar Form

A point in polar form is given as $\left(3,\frac{\pi}{3}\right)$ .

Find its corresponding (x,y) coordinate.

Possible Answers:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

(0,3)

$\left(\frac{3\sqrt3}{2},\frac{3}{2}\right)$

$\left(\frac{3}{2},\frac{3}{2}\right)$

$\left(3,\frac{3\sqrt3}{2}\right)$

Correct answer:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

Explanation:

To go from polar form to cartesion coordinates, use the following two relations.

$x=rcos(\theta)$

$y=rsin(\theta)$

In this case, our is and our $\theta$ is $\frac{\pi}{3}$ .

Plugging those into our relations we get

$x=\frac{3}{2}$

$y=\frac{3\sqrt3}{2}$ ,

which gives us our (x,y) coordinate.

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Example Question #1 : Polar

What is the magnitude and angle (in radians) of the following cartesian coordinate?

(3,4)

Give the answer in the format below.

$(r,\theta)$

Possible Answers:

(12,0.707)

(5,0.643)

(5,0.927)

(4,0.927)

(3,1.04)

Correct answer:

(5,0.927)

Explanation:

Although not explicitly stated, the problem is asking for the polar coordinates of the point (3,4) . To calculate the magnitude, , calculate the following:

$r=\sqrt{x^2+y^2}$

r=5

To calculate $\theta$ , do the following:

$\theta=tan^{-1}\left(\frac{y}{x}\right)$ in radians. (The problem asks for radians)

$\theta = 0.927$

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Example Question #6 : Polar

What is the following coordinate in polar form?

(-2,-5)

Provide the angle in degrees.

Possible Answers:

(16,112)

$(\sqrt{29},202)$

$(\sqrt{29},68)$

(10,248)

$(\sqrt{29},248)$

Correct answer:

$(\sqrt{29},248)$

Explanation:

To calculate the polar coordinate, use

$r=\sqrt{x^2+y^2}$

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

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Example Question #1 : Polar Form

What is the equation $y=2x^{2}$ in polar form?

Possible Answers:

$r=\frac{1}{2}\cos \theta}{\tan \theta$

$r=\frac{1}{2}\sin \theta}{\tan \theta$

$r=\frac{1}{2}\tan \theta}{\sec \theta$

$r=\frac{1}{2}\sin \theta}{\sec \theta$

$r=\frac{1}{2}\sin \theta}{\cos \theta$

Correct answer:

$r=\frac{1}{2}\tan \theta}{\sec \theta$

Explanation:

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

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

$\frac{\tan \theta}{\cos \theta}=2r$

$\tan \theta}{\sec \theta=2r$

$\frac{1}{2}\tan \theta}{\sec \theta=r$

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Example Question #1 : Polar Form

What is the equation $y=\frac{6}{x}$ in polar form?

Possible Answers:

$r=\sqrt\frac{6}{\cos \theta}$

$r=\sqrt\frac{6}{\tan \theta}$

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\sin \theta}$

None of the above

Correct answer:

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{6}{x}$ , then $r\sin \theta=\frac{6}{r\cos \theta}$ . Multiplying both sides by $r\cos \theta$ ,

$r^{2}\sin \theta cos \theta=6$

$r^{2}=\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

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Example Question #5 : Polar Form

What is the equation $y=\frac{7}{3}x^{2}$ in polar form?

Possible Answers:

$r=\frac{3}{7}\tan \theta\sec \theta$

$r=\frac{3}{7}\cos \theta\sin \theta$

$r=\frac{7}{3}\cos \theta\sin \theta$

$r=\frac{7}{3}\tan \theta\sec \theta$

None of the above

Correct answer:

$r=\frac{3}{7}\tan \theta\sec \theta$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{7}{3}x^{2}$ , then $r\sin \theta=\frac{7}{3}{r^{2}\cos^{2} \theta}$ . Simplifying accordingly,

$\tan \theta=\frac{7}{3}{r\cos\theta}$

$\tan \theta\sec \theta =\frac{7}{3}r$

$\frac{3}{7}\tan \theta\sec \theta =r$

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Example Question #6 : Polar Form

Given x=3t+15 and y=15t-6 , what is in terms of (rectangular form)?

Possible Answers:

y=15x-81

y=5x-81

y=5x-3

y=15x-3

y=5x-15

Correct answer:

y=5x-81

Explanation:

Knowing that x=3t+15 and y=15t-6 , we can isolate in both equations as follows:

$x=3t+15\rightarrow t=\frac{x-15}{3}$

$y=15t-6\rightarrow t=\frac{y+6}{15}$

Since both of these equations equal , we can set them equal to each other:

$\frac{y+6}{15}=\frac{x-15}{3}$

$y+6=15(\frac{x-15}{3})$

y+6=5({x-15})

y+6=5x-75

y=5x-81

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Example Question #172 : Parametric, Polar, And Vector

Convert the following function into polar form:

f(x)=2x^2+5x+2+2y^2

Possible Answers:

$2r^2+5r\sin(\theta)+2$

$r^2+1+5r\cos(\theta)$

$r^2+\sin(\theta)$

$2r^2+5r\cos(\theta )+2$

Correct answer:

$2r^2+5r\cos(\theta )+2$

Explanation:

The following formulas were used to convert the function from polar to Cartestian coordinates:

$x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2$

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

f(x)=2x^2+5x+2+2y^2

f(x)=(x^2+y^2)2+5x+2

$f(x)=2r^2+5rcos(\theta)+2$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Polar

Example Question #1 : Polar Form

Example Question #1 : Polar Form

Example Question #1 : Polar

Example Question #6 : Polar

Example Question #1 : Polar Form

Example Question #1 : Polar Form

Example Question #5 : Polar Form

Example Question #6 : Polar Form

Example Question #172 : Parametric, Polar, And Vector

All Calculus 2 Resources

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