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Calculus 2 : Parametric Calculations

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Example Question #11 : Parametric Calculations

Given x=7t+9 and y=t-1 , what is the length of the arc from $0\leq t\leq1$ ?

Possible Answers:

$6\sqrt{2}$

$5\sqrt{2}$

$4\sqrt{2}$

$7\sqrt{2}$

$2\sqrt{2}$

Correct answer:

$5\sqrt{2}$

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=7t+9 and y=t-1 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)7t^{1-1}+(0)9=7$ and $\frac{dy}{dt}=(1)t^{1-1}-(0)1=1$ .

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

$L=\int_{0}^{1}\sqrt{(7)^{2}+(1)^{2}}dt$

$L=\int_{0}^{1}(\sqrt{49+1})dt$

$L=\int_{0}^{1}(\sqrt{50})dt$

$L=\int_{0}^{1}(\sqrt{25\times2})dt$

$L=\int_{0}^{1}(5\sqrt{2})dt$

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:

$L=(5t\sqrt{2})_{0}^{1}\textrm{}$

$L=5(1)\sqrt{2}-5(0)\sqrt{2}$

$L=5\sqrt{2}$

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Example Question #12 : Parametric Calculations

Given x=5t+2 and y=7t+10 , what is the arc length between $0\leqt\leq t\leq2$ ?

Possible Answers:

$7\sqrt{74}$

$5\sqrt{74}$

$3\sqrt{74}$

$2\sqrt{74}$

$\sqrt{74}$

Correct answer:

$2\sqrt{74}$

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=5t+2 and y=7t+10 ,we can use using the Power Rule
$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , to derive

$\frac{dx}{dt}=(1)5t^{1-1}+(0)2=5$ and

$\frac{dy}{dt}=(1)7t^{1-1}+(0)10=7$ .

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

$L=\int_{0}^{2}\sqrt{(5)^{2}+(7)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{25+49})dt$

$L=\int_{0}^{2}(\sqrt{74})dt$

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:

$L=[t\sqrt{74}]_{0}^{2}\textrm{}dt$

$L=2\sqrt{74}-0\sqrt{74}$

$L=2\sqrt{74}$

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Example Question #13 : Parametric Calculations

Given x=6t-4 and y=4t+6 , what is the arc length between $0\leqt\leq t\leq3$ ?

Possible Answers:

$4\sqrt{13}$

$8\sqrt{13}$

$7\sqrt{13}$

$6\sqrt{13}$

$5\sqrt{13}$

Correct answer:

$6\sqrt{13}$

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=6t-4 and y=4t+6 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)6t^{1-1}-(0)4=6$ and

$\frac{dy}{dt}=(1)4t^{1-1}+(0)6=4$ .

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

$L=\int_{0}^{3}\sqrt{(6)^{2}+(4)^{2}}dt$

$L=\int_{0}^{3}(\sqrt{36+16})dt$

$L=\int_{0}^{3}(\sqrt{52})dt$

$L=\int_{0}^{3}(\sqrt{13\times4})dt$

$L=\int_{0}^{3}(2\sqrt{13})dt$

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:

$L=[2t\sqrt{13}]_{0}^{3}\textrm{}dt$

$L=2(3)\sqrt{13}-2(0)\sqrt{13}$

$L=6\sqrt{13}$

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Example Question #1 : Functions, Graphs, And Limits

Given x=t+10 and y=2t-9 , what is the arc length between $0\leqt\leq t\leq4$ ?

Possible Answers:

$2\sqrt{5}$

$5\sqrt{5}$

$3\sqrt{5}$

$4\sqrt{5}$

$\sqrt{5}$

Correct answer:

$4\sqrt{5}$

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+10 and y=2t-9 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)t^{1-1}+(0)10=1$ and

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

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

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

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

$L=\int_{0}^{4}(\sqrt{5})dt$

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:

$L=[t\sqrt{5}]_{0}^{4}\textrm{}dt$

$L=(4)\sqrt{5}-(0)\sqrt{5}$

$L=4\sqrt{5}$

Report an Error

Example Question #5 : Parametric Form

Given x=4t-5 and y=6t+2 , what is the length of the arc from $0\leq t\leq2$ ?

Possible Answers:

$3\sqrt{13}$

$5\sqrt{13}$

$7\sqrt{13}$

$6\sqrt{13}$

$4\sqrt{13}$

Correct answer:

$4\sqrt{13}$

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=4t-5 and y=6t+2 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)4t^{1-1}-(0)5=4$ and

$\frac{dy}{dt}=(1)6t^{1-1}+(0)2=6$ .

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

$L=\int_{0}^{2}(\sqrt{(4)^{2}+(6)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{16+36})dt$

$L=\int_{0}^{2}(\sqrt{52})dt$

$L=\int_{0}^{2}(\sqrt{4\times13})dt$

$L=\int_{0}^{2}(2\sqrt{13})dt$

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:

$L=[2t\sqrt{13}]_{0}^{2}\textrm{}$

$L=2(2)\sqrt{13}-2(0)\sqrt{13}$

$L=4\sqrt{13}$

Report an Error

Example Question #14 : Parametric Calculations

Given x=9t+4 and y=4t-1 , what is the length of the arc from $0\leq t\leq3$ ?

Possible Answers:

$6\sqrt{97}$

$4\sqrt{97}$

$5\sqrt{97}$

$7\sqrt{97}$

$3\sqrt{97}$

Correct answer:

$3\sqrt{97}$

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=9t+4 and y=4t-1 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)9t^{1-1}+(0)4=9$ and

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

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

$L=\int_{0}^{3}(\sqrt{(9)^{2}+(4)^{2}}dt$

$L=\int_{0}^{3}(\sqrt{81+16})dt$

$L=\int_{0}^{3}(\sqrt{97})dt$

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:

$L=[t\sqrt{97}]_{0}^{3}\textrm{}$

$L=(3)\sqrt{97}-(0)\sqrt{97}$

$L=3\sqrt{97}$

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Example Question #15 : Parametric Calculations

Given x=t-11 and y=7t+6 , what is the length of the arc from $0\leq t\leq4$ ?

Possible Answers:

$35\sqrt{2}$

$30\sqrt{2}$

$15\sqrt{2}$

$20\sqrt{2}$

$25\sqrt{2}$

Correct answer:

$20\sqrt{2}$

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-11 and y=7t+6 , we can use using the Power Rule

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

$\frac{dx}{dt}=(1)t^{1-1}-(0)11=1$ and

$\frac{dy}{dt}=(1)7t^{1-1}+(0)6=7$ .

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

$L=\int_{0}^{4}(\sqrt{(1)^{2}+(7)^{2}}dt$

$L=\int_{0}^{4}(\sqrt{1+49})dt$

$L=\int_{0}^{4}(\sqrt{50})dt$

$L=\int_{0}^{4}(\sqrt{25\times2})dt$

$L=\int_{0}^{4}(5\sqrt{2})dt$

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:

$L=[5t\sqrt{2}]_{0}^{4}\textrm{}$

$L=5(4)\sqrt{2}-5(0)\sqrt{2}$

$L=20\sqrt{2}$

Report an Error

Example Question #11 : Parametric Calculations

Eliminate the parameter from $x=t^{3}+3t^{2}-3t-12$ and y=t-8 to write this system as one equation.

Possible Answers:

$x=(y+8)^{3}+3(y+8)^{2}-3(y+8)-12$

$x=3y^{2}$

$x=3y^{2}+6y-3$

$x=\frac{y^{2}+6y-3}{3y^{2}}$

$x=6y^{2}+6y-3$

Correct answer:

$x=(y+8)^{3}+3(y+8)^{2}-3(y+8)-12$

Explanation:

To eliminate the parameter from and , we will solve the equation for and substitute the new expression into the equation. We could also solve the equation for and substitute the new expression into the equation, depending on which is easier.

For our equations, $x=t^{3}+3t^{2}-3t-12$ and y=t-8 , it is easiest to solve the equation for , giving us t=y+8 .

Substituting our new expression for into the equation, we get

$x=(y+8)^{3}+3(y+8)^{2}-3(y+8)-12$

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Example Question #17 : Parametric Calculations

Eliminate the parameter from x=ln(t) and y=ln(4t) to write this system as one equation.

Possible Answers:

$x=e^{y}$

x=ln(4y)

x=y

y=ln(t)

$x=ln(\frac{1}{4})+y$

Correct answer:

$x=ln(\frac{1}{4})+y$

Explanation:

For our equations, x=ln(t) and y=ln(4t) , we will rearrange the equation

y=ln(4t)

To eleiminate the on the right side of the equation, we will take the exponential of both sides of the equation

$e^{y}=e^{ln(4t)}$

Using the exponential identity $e^{ln(x)}=x$

$e^{y}=4t$

$\frac{1}{4}e^{y}=t$

Substituting this value of into the equation, we have

$x=ln(\frac{1}{4}e^{y})$

Using the logarithmic identity, ln(xy)=ln(x)+ln(y)

$ln(\frac{1}{4}e^{y})=ln(\frac{1}{4})+ln(e^{y})$

The using the identity, $ln(e^{x})=x$

$ln(\frac{1}{4})+ln(e^{y})=ln(\frac{1}{4})+y$

Giving us the final expression

$x=ln(\frac{1}{4})+y$

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Example Question #18 : Parametric Calculations

Eliminate the parameter from $x=e^{2t}$ and $y=e^{5t}$ .

Possible Answers:

y=x

$y=x^{10}$

$y=x^{2.5}$

$y=e^{10x}$

y=t

Correct answer:

$y=x^{10}$

Explanation:

For our equations, $x=e^{2t}$ and $y=e^{5t}$ , we will rearrange the equation.

$x=e^{2t}$

To eliminate the exponential from the right side of the equation, we will take the of both sides of the equation.

$ln(x)=ln(e^{2t})$

Using the logarithmic identity, $ln(e^{x})=x$

ln(x)=2t

$\frac{1}{2}ln(x)=t$

Substituting this value of into the equation, we have

$y=e^{5*\frac{1}{2}ln(x)}=e^{10ln(x)}$

Using the logarithmic identity $cln(x)=ln(x^{c})$ , where is a constant

$e^{10ln(x)}=e^{ln(x^{10})}=x^{10}$

Therefore $y=x^{10}$ .

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Calculus 2 : Parametric Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Parametric Calculations

Example Question #12 : Parametric Calculations

Example Question #13 : Parametric Calculations

Example Question #1 : Functions, Graphs, And Limits

Example Question #5 : Parametric Form

Example Question #14 : Parametric Calculations

Example Question #15 : Parametric Calculations

Example Question #11 : Parametric Calculations

Example Question #17 : Parametric Calculations

Example Question #18 : Parametric Calculations

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