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Calculus 2 : Parametric, Polar, and Vector

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=7t^{2}-12$ and $x=5t^{2}+2$ .

Possible Answers:

$-\frac{7}{5}$

$\frac{7}{5}$

$\frac{5}{7}$

None of the above

$-\frac{5}{7}$

Correct answer:

$\frac{7}{5}$

Explanation:

Given equations for and in terms of , we can find the derivative of parametric equations as follows:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ , as the terms will cancel out.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ and given $y=7t^{2}-12$ and $x=5t^{2}+2$ :

$\frac{dy}{dt}=(2)7t^{2-1}-(0)12=14t$

$\frac{dx}{dt}=(2)5t^{2-1}+(0)2=10t$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t}{10t}=\frac{7}{5}$ .

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Example Question #1 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=9t^{2}-13$ and $x=t^{2}+15$ .

Possible Answers:

$-\frac{1}{9}$

None of the above

$\frac{1}{9}$

Correct answer:

Explanation:

Given equations for and in terms of , we can find the derivative of parametric equations as follows:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ , as the terms will cancel out.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ and given $y=9t^{2}-13$ and $x=t^{2}+15$ :

$\frac{dy}{dt}=(2)9t^{2-1}-(0)13=18t$

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

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{18t}{2t}=9$

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Example Question #2 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if y=2t-9 and x=5t+10 .

Possible Answers:

$\frac{2}{5}$

None of the above

$-\frac{2}{5}$

$\frac{5}{2}$

$-\frac{5}{2}$

Correct answer:

$\frac{2}{5}$

Explanation:

Since we have two equations y=2t-9 and x=5t+10 , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms cancel out by standard rules of division of fractions.

In order to find the derivatives of and , let's use the Power Rule

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

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

$x=5t+10\rightarrow\frac{dx}{dt}=(1)5t^{1-1}+(0)10=5$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{5}$ .

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Example Question #4 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if y=7t+4 and x=7t-8 .

Possible Answers:

$\frac{1}{7}$

Correct answer:

Explanation:

Since we have two equations y=7t+4 and x=7t-8 , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ since the terms cancel out by standard rules of division of fractions.

In order to find the derivatives of and , let's use the Power Rule

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

$y=7t+4\rightarrow \frac{dy}{dt}=(1)7t^{1-1}+(0)4=7$

$x=7t-8\rightarrow\frac{dx}{dt}=(1)7t^{1-1}-(0)8=7$

Therefore,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{7}{7}=1$ .

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Example Question #1 : Parametric Calculations

Solve for $\frac{dy}{dx}$ if $y=5t^{2}+6t-9$ and $x=2t^{2}-6t+12$ .

Possible Answers:

$\frac{5t+3}{2t-3}$

None of the above

$-\frac{4}{10}$

$\frac{5t-3}{2t-3}$

$\frac{4}{10}$

Correct answer:

$\frac{5t+3}{2t-3}$

Explanation:

Since we have two equations $y=5t^{2}+6t-9$ and $x=2t^{2}-6t+12$ , we can find $\frac{dy}{dx}$ by dividing the derivatives of the two equations - thus:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ (since the terms cancel out by standard rules of division of fractions).

In order to find the derivatives of and , let's use the Power Rule

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

$y=5t^{2}+6t-9\rightarrow \frac{dy}{dt}=(2)5t^{2-1}+(1)6t^{1-1}-(0)9=10t+6$

$x=2t^{2}-6t-12\rightarrow \frac{dx}{dt}=(2)2t^{2-1}-(1)6t^{1-1}-(0)12=4t-6$

Therefore, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10t+6}{4t-6}=\frac{2(5t+3)}{2(2t-3)}=\frac{5t+3}{2t-3}$ .

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Example Question #2 : Parametric Calculations

Given x=5t-1 and y=2t+9 , what is the length of the arc from $0\leq t\leq3$ ?

Possible Answers:

$2\sqrt{29}$

$4\sqrt{29}$

$\sqrt{29}$

$3\sqrt{29}$

Correct answer:

$3\sqrt{29}$

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-1 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

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

$L=\int_{0}^{3}\sqrt{(5)^{2}+(2)^{2}}dt$

$L=\int_{0}^{3}(\sqrt{25+4})dt$

$L=\int_{0}^{3}(\sqrt{29})dt$

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:

$L=(t\sqrt{29})_{0}^{3}\textrm{}$

$L=3\sqrt{29}-0\sqrt{29}$

$L=3\sqrt{29}$

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Example Question #143 : Parametric

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

Possible Answers:

$2\sqrt{2}$

$7\sqrt{2}$

$6\sqrt{2}$

$4\sqrt{2}$

$5\sqrt{2}$

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 #144 : Parametric

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

Possible Answers:

$7\sqrt{74}$

$2\sqrt{74}$

$5\sqrt{74}$

$3\sqrt{74}$

$\sqrt{74}$

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 #145 : Parametric

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

Possible Answers:

$5\sqrt{13}$

$8\sqrt{13}$

$4\sqrt{13}$

$7\sqrt{13}$

$6\sqrt{13}$

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 #3 : 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:

$3\sqrt{5}$

$4\sqrt{5}$

$2\sqrt{5}$

$5\sqrt{5}$

$\sqrt{5}$

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:

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

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

$L=4\sqrt{5}$

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #4 : Parametric Calculations

Example Question #1 : Parametric Calculations

Example Question #2 : Parametric Calculations

Example Question #4 : Parametric Calculations

Example Question #1 : Parametric Calculations

Example Question #2 : Parametric Calculations

Example Question #143 : Parametric

Example Question #144 : Parametric

Example Question #145 : Parametric

Example Question #3 : Functions, Graphs, And Limits

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