Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #671 : Calculus Ii

Find the length of the following parametric curve 

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

Possible Answers:

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

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

\(\displaystyle \frac{1}{6}\)

\(\displaystyle 1\)

\(\displaystyle 2\sqrt{3}\)

Correct answer:

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

Explanation:

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

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

\(\displaystyle x=\frac{1}{3}t^{3}\),   \(\displaystyle y=\frac{1}{2}t^{2}\)

In this case

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

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

\(\displaystyle \alpha=0, \beta=1\)

The length of this curve is

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

\(\displaystyle =\int_{0}^{1 }\sqrt{t^{4}+t^{2}}dt\)

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

Using the identity \(\displaystyle \sqrt{a*b}=\sqrt{a}\sqrt{b}\)

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

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

Using a u-substitution

Let \(\displaystyle u=t^{2}+1\)

 \(\displaystyle du=2tdt\Rightarrow dt=2t/du\)

and changing the bounds

\(\displaystyle u=t^{2}+1=(1)^{2}+1=2\)

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

\(\displaystyle L=\frac{1}{2}\int_{1}^{2}\sqrt{u}du\)

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

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

 

 

 

 

Example Question #672 : Calculus Ii

Find the length of the following parametric curve 

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

Possible Answers:

\(\displaystyle \sqrt{122}-8\)

\(\displaystyle 7\)

\(\displaystyle 0\)

\(\displaystyle \sqrt{130}-5\sqrt{7}\)

\(\displaystyle 4\left [ 65\sqrt{65}-2\sqrt{2} \right ]\)

Correct answer:

\(\displaystyle 4\left [ 65\sqrt{65}-2\sqrt{2} \right ]\)

Explanation:

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

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

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

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

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

In this case, the length of this curve is

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

\(\displaystyle =\int_{1}^{8 }\sqrt{144t^{2}+144t^{4}}dt\)

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

Using the identity \(\displaystyle \sqrt{a*b}=\sqrt{a}\sqrt{b}\)

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

using a u-substitution

\(\displaystyle u=1+t^{2}\)

\(\displaystyle du=2tdt\Rightarrow dt=2t/du\)

and changing the bounds

\(\displaystyle u=1+t^{2}=1+8^{2}=65\)

\(\displaystyle u=1+t^{2}=1+1^{2}=2\)

\(\displaystyle L=\frac{12}{2}\int_{2}^{65 }\sqrt{u}du\)

\(\displaystyle =\frac{12}{2}\left [\frac{2}{3} \right ]\left [ 65^{\frac{3}{2}}-2^{\frac{3}{2}}\right ]\)

\(\displaystyle =4\left [ \sqrt{65^{3}}-\sqrt{2^{3}} \right ]\)

\(\displaystyle =4\left [ \sqrt{65^{2}*65}-\sqrt{2^{2}*2} \right ]\)

\(\displaystyle =4\left [ \sqrt{65^{2}}\sqrt{65}-\sqrt{2^{2}}\sqrt{2} \right ]\)

\(\displaystyle =4\left [ 65\sqrt{65}-2\sqrt{2} \right ]\)

Example Question #673 : Calculus Ii

Find the length of the following parametric curve 

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

Possible Answers:

\(\displaystyle 5\pi-5\)

\(\displaystyle 0\)

\(\displaystyle 1\)

\(\displaystyle 10\pi\)

\(\displaystyle 5\pi\)

Correct answer:

\(\displaystyle 10\pi\)

Explanation:

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

We then use the following trigonometric rules, 

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

where \(\displaystyle a\) and \(\displaystyle b\) are constants.

In this case

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

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

The length of this curve is

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

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

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

Using the identity \(\displaystyle \sqrt{a*b}=\sqrt{a}\sqrt{b}\)

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

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

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

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

Using the rule of integration for constants

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

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

Example Question #674 : Calculus Ii

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

Possible Answers:

\(\displaystyle L=2\sqrt{5}\)

\(\displaystyle L=\sqrt{5}\)

\(\displaystyle L=4\sqrt{5}\)

\(\displaystyle L=3\sqrt{5}\)

\(\displaystyle L=5\sqrt{5}\)

Correct answer:

\(\displaystyle L=2\sqrt{5}\)

Explanation:

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

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

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

for all ,

to derive

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

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

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

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

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

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

Now, using the Power Rule for Integrals

for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{5}]_{0}^{2}\textrm{}\)

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

\(\displaystyle L=2\sqrt{5}\)

Example Question #675 : Calculus Ii

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

Possible Answers:

\(\displaystyle 3\sqrt{85}\)

\(\displaystyle \sqrt{85}\)

\(\displaystyle 2\sqrt{85}\)

\(\displaystyle 5\sqrt{85}\)

\(\displaystyle 4\sqrt{85}\)

Correct answer:

\(\displaystyle 3\sqrt{85}\)

Explanation:

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

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

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

 for all ,

to derive 

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

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

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

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

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

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

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{85}]_{0}^{3}\textrm{}\)

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

\(\displaystyle L=3\sqrt{85}\)

Example Question #676 : Calculus Ii

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

Possible Answers:

\(\displaystyle 5\sqrt{13}\)

\(\displaystyle 4\sqrt{13}\)

\(\displaystyle \sqrt{13}\)

\(\displaystyle 3\sqrt{13}\)

\(\displaystyle 2\sqrt{13}\)

Correct answer:

\(\displaystyle 2\sqrt{13}\)

Explanation:

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

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

 for all ,

to derive 

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

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

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

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

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

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

Now, using the Power Rule for Integrals

 for all ,

we can determine that:

\(\displaystyle L=[t\sqrt{13}]_{1}^{3}\textrm{}\)

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

\(\displaystyle L=2\sqrt{13}\)

 

Example Question #1 : Polar

Rewrite the polar equation 

\(\displaystyle r = \sin ^{2} \theta\)

in rectangular form.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

\(\displaystyle r = \sin ^{2} \theta\)

\(\displaystyle r^{2} \cdot r = r^{2} \sin ^{2} \theta\)

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

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

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

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

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

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

Example Question #2 : Polar

Rewrite in polar form:

\(\displaystyle x^{2} = 3xy + 1\)

Possible Answers:

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

\(\displaystyle r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }\)

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

\(\displaystyle r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }\)

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

Correct answer:

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

Explanation:

\(\displaystyle x^{2} = 3xy + 1\)

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

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

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

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

\(\displaystyle r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }\)

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

Example Question #1 : Polar

Rewrite the polar equation 

\(\displaystyle r ^{2} + 1 = \cos \theta\)

in rectangular form.

Possible Answers:

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} }\)

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ y}{{x^{2} + y ^{2}} +1}\)

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ y}{x^{2} + y ^{2}}\)

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}\)

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{x+ y}{{x^{2} + y ^{2}} +1}\)

Correct answer:

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}\)

Explanation:

\(\displaystyle r ^{2} + 1 = \cos \theta\)

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

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

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}\)

Example Question #671 : Calculus Ii

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

Possible Answers:

\(\displaystyle r = \frac{12}{2 \sin \theta - 3 \cos \theta }\)

\(\displaystyle r = \frac{12}{3 \sin \theta - 2 \cos \theta }\)

\(\displaystyle r = 12\tan \theta\)

\(\displaystyle r = \frac{12}{ \tan \theta }\)

\(\displaystyle r = \frac{12}{3 \sin \theta + 2 \cos \theta }\)

Correct answer:

\(\displaystyle r = \frac{12}{3 \sin \theta - 2 \cos \theta }\)

Explanation:

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

By substituting, we can rewrite this:

\(\displaystyle r \sin \theta = \frac{2}{3}r \cos \theta + 4\)

\(\displaystyle r \sin \theta - \frac{2}{3}r \cos \theta = 4\)

\(\displaystyle 3 r \sin \theta - 2 r \cos \theta = 12\)

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

\(\displaystyle r = \frac{12}{3 \sin \theta - 2 \cos \theta }\)

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