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Calculus 2 : Parametric Calculations

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Example Question #6 : Parametric Form

Find the length of the following parametric curve

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$ , $0\leq t \leq 2 \pi$ .

Possible Answers:

$e^{2 \pi}$

$e^{t}cos(t)$

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

sin(t)+cos(t)

Correct answer:

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

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 product rule,

$\frac{d}{dt}(a*b)=a\frac{db}{dt}+b\frac{da}{dt}$ , when and are functions of ,

the trigonometric rule,

$\frac{d}{dt}[cos(t)]=-sin(t)$ and $\frac{d}{dt}[sin(t)]=cos(t)$

and exponential rule,

$\frac{d}{dt}[e^{t}]=e^{t}$ to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

In this case

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$

$\frac{dx}{dt}=\frac{d}{dt}e^{t}cos(t)=e^{t}\frac{d}{dt}cos(t)+cos(t)\frac{d}{dt}e^{t}$

$=e^{t}*(-sin(t))+cos(t)*e^{t}=e^{t}cos(t)-e^{t}sin(t)$

$\frac{dy}{dt}=\frac{d}{dt}e^{t}sin(t)=e^{t}\frac{d}{dt}sin(t)+sin(t)\frac{d}{dt}e^{t}$

$=e^{t}cos(t)+sin(t)*e^{t}=e^{t}cos(t)+e^{t}sin(t)$

$\alpha=0, \beta=2\pi$

The length of this curve is

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

Using the identity $(a\pm b)^{2}=a^{2}\pm2ab+b^{2}$

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

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

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

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

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

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

$=\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}$

Using the exponential rule, $\int e^{t}dt}=e^{t}$

$\sqrt{2}\int_{0}^{2\pi }e^{t}dt}=\sqrt{2}e^{t}|_{0}^{2\pi}=\sqrt{2}\left ( e^{2 \pi}-e^{0}}\right )$

Using the exponential rule, $e^{0}=1$ , gives us the final solution

$L=\sqrt{2}\left ( e^{2 \pi}-1}\right )$

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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:

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

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

$\sqrt{122}-8$

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:

$10\pi$

$5\pi$

$5\pi-5$

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

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

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

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Example Question #24 : Parametric Calculations

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

Possible Answers:

$L=2\sqrt{5}$

$L=4\sqrt{5}$

$L=3\sqrt{5}$

$L=5\sqrt{5}$

$L=\sqrt{5}$

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 #25 : Parametric Calculations

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

Possible Answers:

$\sqrt{85}$

$4\sqrt{85}$

$2\sqrt{85}$

$5\sqrt{85}$

$3\sqrt{85}$

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 #21 : Parametric Calculations

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

Possible Answers:

$4\sqrt{13}$

$5\sqrt{13}$

$2\sqrt{13}$

$3\sqrt{13}$

$\sqrt{13}$

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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Calculus 2 : Parametric Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #6 : Parametric Form

Example Question #21 : Parametric Calculations

Example Question #22 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #24 : Parametric Calculations

Example Question #25 : Parametric Calculations

Example Question #21 : Parametric Calculations

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