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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 #12 : Parametric Calculations

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

Possible Answers:

$3\sqrt{74}$

$7\sqrt{74}$

$5\sqrt{74}$

$2\sqrt{74}$

$\sqrt{74}$

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:

$7\sqrt{13}$

$6\sqrt{13}$

$8\sqrt{13}$

$4\sqrt{13}$

$5\sqrt{13}$

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 : Parametric Form

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

Possible Answers:

$\sqrt{5}$

$3\sqrt{5}$

$4\sqrt{5}$

$5\sqrt{5}$

$2\sqrt{5}$

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:

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

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

$L=4\sqrt{5}$

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Example Question #2 : Parametric Form

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

Possible Answers:

$3\sqrt{13}$

$7\sqrt{13}$

$5\sqrt{13}$

$4\sqrt{13}$

$6\sqrt{13}$

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:

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

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

$L=4\sqrt{13}$

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Example Question #151 : Parametric

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

Possible Answers:

$7\sqrt{97}$

$4\sqrt{97}$

$6\sqrt{97}$

$5\sqrt{97}$

$3\sqrt{97}$

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 #151 : Parametric

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

Possible Answers:

$15\sqrt{2}$

$25\sqrt{2}$

$30\sqrt{2}$

$20\sqrt{2}$

$35\sqrt{2}$

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:

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

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

$L=20\sqrt{2}$

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Example Question #152 : Parametric

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

Possible Answers:

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

$x=3y^{2}$

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

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

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

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 #13 : Parametric Calculations

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

Possible Answers:

y=ln(t)

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

x=ln(4y)

x=y

$x=e^{y}$

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 #152 : Parametric

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

Possible Answers:

y=x

y=t

$y=x^{10}$

$y=x^{2.5}$

$y=e^{10x}$

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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Example Question #3 : Parametric, Polar, And Vector Functions

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:

$e^{2 \pi}$

sin(t)+cos(t)

$e^{t}cos(t)$

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #12 : Parametric Calculations

Example Question #13 : Parametric Calculations

Example Question #1 : Parametric Form

Example Question #2 : Parametric Form

Example Question #151 : Parametric

Example Question #151 : Parametric

Example Question #152 : Parametric

Example Question #13 : Parametric Calculations

Example Question #152 : Parametric

Example Question #3 : Parametric, Polar, And Vector Functions

All Calculus 2 Resources

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