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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #674 : Calculus Ii

Given x=-t+3 and y=3-2y , what is the arc length between $0\leq t\leq 2$

Possible Answers:

$L=2\sqrt{5}$

$L=\sqrt{5}$

$L=4\sqrt{5}$

$L=3\sqrt{5}$

$L=5\sqrt{5}$

Correct answer:

$L=2\sqrt{5}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=-t+3 and y=3-2y , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(1)(-1)t^{1-1}+(0)3=-1$ and

$\frac{dy}{dt}=(0)3-(1)2y^{1-1}=-2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}(\sqrt{(-1)^{2}+(-2)^{2}})dt$

$L=\int_{0}^{2}(\sqrt{1+4})dt$

$L=\int_{0}^{2}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{5}]_{0}^{2}\textrm{}$

$L=(2)\sqrt{5}-(0)\sqrt{5}$

$L=2\sqrt{5}$

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Example Question #675 : Calculus Ii

Given x=2t+15 and y=4+9y , what is the arc length between $0\leq t\leq 3$ ?

Possible Answers:

$3\sqrt{85}$

$\sqrt{85}$

$2\sqrt{85}$

$5\sqrt{85}$

$4\sqrt{85}$

Correct answer:

$3\sqrt{85}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=2t+15 and y=4+9y , wwe can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(1)2t^{1-1}+(0)15=2$ and

$\frac{dy}{dt}=(0)4+(1)9y^{1-1}=9$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{3}(\sqrt{(2)^{2}+(9)^{2}})dt$

$L=\int_{0}^{3}(\sqrt{4+81})dt$

$L=\int_{0}^{3}(\sqrt{85})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{85}]_{0}^{3}\textrm{}$

$L=(3)\sqrt{85}-(0)\sqrt{85}$

$L=3\sqrt{85}$

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Example Question #676 : Calculus Ii

Given x=2-3t and y=9+2t , what is the arc length between $1\leq t\leq 3$ ?

Possible Answers:

$5\sqrt{13}$

$4\sqrt{13}$

$\sqrt{13}$

$3\sqrt{13}$

$2\sqrt{13}$

Correct answer:

$2\sqrt{13}$

Explanation:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=2-3t and y=9+2t , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(0)2-(1)3t^{1-1}=-3$ and

$\frac{dy}{dt}=(0)9+(1)2t^{1-1}=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{1}^{3}(\sqrt{(-3)^{2}+(2)^{2}})dt$

$L=\int_{1}^{3}(\sqrt{9+4})dt$

$L=\int_{1}^{3}(\sqrt{13})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{13}]_{1}^{3}\textrm{}$

$L=(3)\sqrt{13}-(1)\sqrt{13}$

$L=2\sqrt{13}$

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Example Question #1 : Polar

Rewrite the polar equation

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

in rectangular form.

Possible Answers:

$y ^{4} = \left (x^{2} +y^{2} \right )^{2}$

$y ^{3} = \left (x^{2} +y^{2} \right )^{2}$

$y ^{4} = \left (x^{3} +y^{3} \right )^{2}$

$y ^{3} = \left (x^{2} +y^{2} \right )^{4}$

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

Correct answer:

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

Explanation:

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

$r^{2} \cdot r = r^{2} \sin ^{2} \theta$

$r^{3} =\left ( r \sin \theta \right ) ^{2}$

$\left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}$

$\left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{3} = y ^{4}$

or $y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

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Example Question #2 : Polar

Rewrite in polar form:

$x^{2} = 3xy + 1$

Possible Answers:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$

$r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Correct answer:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Explanation:

$x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

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Example Question #1 : Polar

Rewrite the polar equation

$r ^{2} + 1 = \cos \theta$

in rectangular form.

Possible Answers:

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

$\sqrt{x^{2} + y ^{2}} =\frac{ y}{{x^{2} + y ^{2}} +1}$

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

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

$\sqrt{x^{2} + y ^{2}} =\frac{x+ y}{{x^{2} + y ^{2}} +1}$

Correct answer:

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

Explanation:

$r ^{2} + 1 = \cos \theta$

$r\left ( r ^{2} + 1 \right )= r \cos \theta$

$\sqrt{x^{2} + y ^{2}} \left ( {x^{2} + y ^{2}} +1 \right ) = x$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

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Example Question #1 : Polar

Give the polar form of the equation of the line with intercepts (0,4), (-6,0) .

Possible Answers:

$r = 12\tan \theta$

$r = \frac{12}{2 \sin \theta - 3 \cos \theta }$

$r = \frac{12}{3 \sin \theta + 2 \cos \theta }$

$r = \frac{12}{ \tan \theta }$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

Correct answer:

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

Explanation:

This line has slope $\frac{4-0}{0- (-6)} = \frac{2}{3}$ and -intercept (0.4) , so its Cartesian equation is $y = \frac{2}{3}x + 4$ .

By substituting, we can rewrite this:

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

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

$3 r \sin \theta - 2 r \cos \theta = 12$

$r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

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Example Question #5 : Polar

Give the rectangular coordinates of the point with polar coordinates

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

Possible Answers:

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

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

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

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

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

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 #6 : Polar

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

Possible Answers:

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

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

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

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

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

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 #7 : Polar

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

Find its corresponding (x,y) coordinate.

Possible Answers:

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

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

(0,3)

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

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

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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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #674 : Calculus Ii

Example Question #675 : Calculus Ii

Example Question #676 : Calculus Ii

Example Question #1 : Polar

Example Question #2 : Polar

Example Question #1 : Polar

Example Question #1 : Polar

Example Question #5 : Polar

Example Question #6 : Polar

Example Question #7 : Polar

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