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Calculus 2 : Parametric

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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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Calculus 2 : Parametric

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

All Calculus 2 Resources

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