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Calculus 2 : Calculus II

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 #21 : Parametric Calculations

Find the length of the following parametric curve

$x=\frac{1}{3}t^{3}$ , $y=\frac{1}{2}t^{2}$ , $0\leq t \leq 1$ .

Possible Answers:

$2\sqrt{3}$

$\frac{1}{5}\left ( 6\sqrt{6}-1\right )$

$\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

$\frac{1}{6}$

Correct answer:

$\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the power rule $\frac{d}{dt}t^{c}=ct^{c-1}$ , where is a constant, to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

$x=\frac{1}{3}t^{3}$ , $y=\frac{1}{2}t^{2}$

In this case

$\frac{dx}{dt}=\frac{d}{dt}(\frac{1}{3}t^{3})=t^{2}$

$\frac{dy}{dt}=\frac{d}{dt}(\frac{1}{2}t^{2})=t$

$\alpha=0, \beta=1$

The length of this curve is

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

$=\int_{0}^{1 }\sqrt{t^{4}+t^{2}}dt$

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

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

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

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

Using a u-substitution

Let $u=t^{2}+1$

$du=2tdt\Rightarrow dt=2t/du$

and changing the bounds

$u=t^{2}+1=(1)^{2}+1=2$

$u=t^{2}+1=(0)^{2}+1=1$

$L=\frac{1}{2}\int_{1}^{2}\sqrt{u}du$

$L=\frac{1}{2}*\frac{2}{3}\left ( 2^{\frac{3}{2}}-1^{\frac{3}{2}}\right )$

$L=\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

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Example Question #22 : Parametric Calculations

Find the length of the following parametric curve

$x=6t^{2}$ , $y=4t^{^{3}}$ , $1\leq t \leq 8$ .

Possible Answers:

$4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

$\sqrt{122}-8$

$\sqrt{130}-5\sqrt{7}$

Correct answer:

$4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

$x=6t^{2}$ , $y=4t^{^{3}}$ , $1\leq t \leq 8$ .

We use the power rule $\frac{d}{dt}t^{c}=ct^{c-1}$ , where is a constant, to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

$\frac{dx}{dt}=\frac{d}{dt}(6t^{2})=12t$

$\frac{dx}{dt}=\frac{d}{dt}(4t^{3})=4t^{2}$

In this case, the length of this curve is

$L=\int_{1}^{8 }\sqrt{\left (12t \right )^{2}+\left (12t^{2} \right )^{2}}dt$

$=\int_{1}^{8 }\sqrt{144t^{2}+144t^{4}}dt$

$=\int_{1}^{8 }\sqrt{144t^{2}(1+t^{2})}dt$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$\int_{1}^{8 }\sqrt{144t^{2}(1+t^{2})}dt=\int_{1}^{8 }\sqrt{144t^{2}}\sqrt{1+t^{2}}dt=\int_{1}^{8 }\12t\sqrt{1+t^{2}}dt$

using a u-substitution

$u=1+t^{2}$

$du=2tdt\Rightarrow dt=2t/du$

and changing the bounds

$u=1+t^{2}=1+8^{2}=65$

$u=1+t^{2}=1+1^{2}=2$

$L=\frac{12}{2}\int_{2}^{65 }\sqrt{u}du$

$=\frac{12}{2}\left [\frac{2}{3} \right ]\left [ 65^{\frac{3}{2}}-2^{\frac{3}{2}}\right ]$

$=4\left [ \sqrt{65^{3}}-\sqrt{2^{3}} \right ]$

$=4\left [ \sqrt{65^{2}*65}-\sqrt{2^{2}*2} \right ]$

$=4\left [ \sqrt{65^{2}}\sqrt{65}-\sqrt{2^{2}}\sqrt{2} \right ]$

$=4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

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Example Question #23 : Parametric Calculations

Find the length of the following parametric curve

x=cos(5t) , y=sin(5t) , $0\leq t \leq 2\pi$ .

Possible Answers:

$5\pi$

$10\pi$

$5\pi-5$

Correct answer:

$10\pi$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We then use the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case

$\frac{dx}{dt}=\frac{d}{dt}(cos(5t))=-5sin(5t)$ ,

$\frac{dy}{dt}=\frac{d}{dt}(sin(5t))=5cos(5t)$ , $\alpha=0, \beta=2\pi$

The length of this curve is

$L=\int_{0}^{2\pi }\sqrt{\left (-5sin(5t) \right )^{2}+\left (5cos(5t) \right )^{2}}dt$

$=\int_{0}^{2\pi }\sqrt{25sin^{2}(5t)+25cos^{2}(5t)}dt$

$=\int_{0}^{2\pi }\sqrt{25(sin^{2}(5t)+cos^{2}(5t))}dt$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$\int_{0}^{2\pi }\sqrt{25(sin^{2}(5t)+cos^{2}(5t))}dt=\int_{0}^{2\pi }\sqrt{25}\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt$

$=\int_{0}^{2\pi }5\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt$

Using the trigonometric identity $sin^{2}(at)+cos^{2}(at)=1$ where is a constant

$\int_{0}^{2\pi }5\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt=5\int_{0}^{2\pi }dt$

Using the rule of integration for constants

$\int_{a}^{b}dt=t|_{a}^{b}=(b-a)$

$5\int_{0}^{2\pi }dt=5(2\pi -0)=10\pi$

Report an Error

Example Question #22 : Parametric Calculations

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 #23 : Parametric Calculations

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

Possible Answers:

$\sqrt{85}$

$5\sqrt{85}$

$2\sqrt{85}$

$4\sqrt{85}$

$3\sqrt{85}$

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 #24 : Parametric Calculations

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

Possible Answers:

$5\sqrt{13}$

$\sqrt{13}$

$2\sqrt{13}$

$3\sqrt{13}$

$4\sqrt{13}$

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 #161 : Parametric, Polar, And Vector

Rewrite the polar equation

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

in rectangular form.

Possible Answers:

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

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

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

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

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

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 #1 : Polar Form

Rewrite in polar form:

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

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:

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

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

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

$r = 12\tan \theta$

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Parametric Calculations

Example Question #22 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #22 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #24 : Parametric Calculations

Example Question #161 : Parametric, Polar, And Vector

Example Question #1 : Polar Form

Example Question #1 : Polar

Example Question #1 : Polar

All Calculus 2 Resources

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