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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 #601 : Calculus Ii

Which of the following parametric equations parametrizes an ellipse? (Assume that $0\leq t\leq 2\pi$ )

Possible Answers:

$x(t)=\sqrt{2}\cos t, y(t)=\sqrt{2}\cos t$

$x(t)=\cos^2 t, y(t)=\sin^2 t$

$x(t)=\sqrt{2}\cos t, y(t)=\sqrt{3}\cos t$

$x(t)=2\cos t, y(t)=3\sin t$

Correct answer:

$x(t)=2\cos t, y(t)=3\sin t$

Explanation:

The parametric equations

$x(t)=2\cos t, y(t)=3\sin t$

describe an ellipse because we have

$\frac{x(t)^2}{4}+\frac{y(t)^2}{9}=\cos^2 t+\sin^2 t=1$

which means

$\frac{x(t)^2}{4}+\frac{y(t)^2}{9}=1$

which is the equation for an ellipse.

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Example Question #602 : Calculus Ii

Given $x=t^{2}-5$ and y=3t+8 , what is in terms of (rectangular form)?

Possible Answers:

$y=3\sqrt{x-5}+8$

$y=3\sqrt{x+5}-8$

None of the above

$y=3(\pm\sqrt{x+5})+8$

$y=3\sqrt{x-5}-8$

Correct answer:

$y=3(\pm\sqrt{x+5})+8$

Explanation:

Given $x=t^{2}-5$ and y=3t+8 , et's solve both equations for :

$x=t^{2}-5 \rightarrow t=\pm\sqrt{x+5}$

$y=3t+8\rightarrow t=\frac{y-8}{3}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y-8}{3}=\pm\sqrt{x+5}$

$y-8=3(\pm\sqrt{x+5})$

$y=3(\pm\sqrt{x+5})+8$

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Example Question #603 : Calculus Ii

Given x=4+10t and y=2-t , what is in terms of (rectangular form)?

Possible Answers:

None of the above

$y=\frac{x+4}{10}-2$

$y=2-\frac{x-4}{10}$

$y=\frac{x-4}{10}-2$

$y=\frac{x+4}{10}+2$

Correct answer:

$y=2-\frac{x-4}{10}$

Explanation:

Given x=4+10t and y=2-t , let's solve both equations for :

$x=4+10t\rightarrow t=\frac{x-4}{10}$

$y=2-t\rightarrow t=2-y$

Since both equations equal , let's set them equal to each other and solve for :

$2-y=\frac{x-4}{10}$

$y=2-\frac{x-4}{10}$

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Example Question #91 : Parametric

Given x=7-9t and $y=3t^{2}$ , what is in terms of (rectangular form)?

Possible Answers:

$y=-3\left(\frac{7+x}{9}\right)^{2}$

$y=3\left(\frac{7-x}{9}\right)^{2}$

$y=-3\left(\frac{7-x}{9}\right)^{2}$

$y=3\left(\frac{7+x}{9}\right)^{2}$

None of the above

Correct answer:

$y=3\left(\frac{7-x}{9}\right)^{2}$

Explanation:

Given x=7-9t and $y=3t^{2}$ , let's solve both equations for :

$x=7-9t\rightarrow t=\frac{7-x}{9}$

$y=3t^{2}\rightarrow t=\sqrt{\frac{y}{3}}$

Since both equations equal , let's set them equal to each other and solve for :

$\sqrt{\frac{y}{3}}=\frac{7-x}{9}$

$\frac{y}{3}=\left(\frac{7-x}{9}\right)^{2}$

$y=3\left(\frac{7-x}{9}\right)^{2}$

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Example Question #604 : Calculus Ii

Given $x=6t^{2}-14$ and y=5t+1 , what is in terms of ?

Possible Answers:

$y=5\sqrt{\frac{x-14}{6}}-1$

None of the above

$y=5\sqrt{\frac{x-14}{6}}+1$

$y=5\left(\pm\sqrt{\frac{x+14}{6}}\right)+1$

$y=5\sqrt{\frac{x+14}{6}}-1$

Correct answer:

$y=5\left(\pm\sqrt{\frac{x+14}{6}}\right)+1$

Explanation:

Given $x=6t^{2}-14$ and y=5t+1 , let's solve both equations for :

$x=6t^{2}-14\rightarrow t=\pm\sqrt{\frac{x+14}{6}}$

$y=5t+1\rightarrow t=\frac{y-1}{5}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y-1}{5}=\pm\sqrt{\frac{x+14}{6}}$

$y-1=5\left(\pm\sqrt{\frac{x+14}{6}}\right)$

$y=5\left(\pm\sqrt{\frac{x+14}{6}}\right)+1$

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Example Question #93 : Parametric

Given x=2t-7 and $y=8-t^{2}$ , what is in terms of ?

Possible Answers:

$y=8+\left(\frac{x+7}{2}\right)^2$

$y=8-\left(\frac{x-7}{2}\right)^2$

None of the above

$y=8-\left(\frac{x+7}{2}\right)^2$

$y=8+\left(\frac{x-7}{2}\right)^2$

Correct answer:

$y=8-\left(\frac{x+7}{2}\right)^2$

Explanation:

Given x=2t-7 and $y=8-t^{2}$ , let's solve both equations for :

$x=2t-7\rightarrow t=\frac{x+7}{2}$

$y=8-t^{2}\rightarrow t=\sqrt{8-y}$

Since both equations equal , let's set them equal to each other and solve for :

$\sqrt{8-y}=\frac{x+7}{2}$

$8-y=\left(\frac{x+7}{2}\right)^2$

$y=8-\left(\frac{x+7}{2}\right)^2$

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Example Question #91 : Parametric, Polar, And Vector

Given the parametric equations

x(t)=t+t^2, y(t)=t^3

what is $\frac{dy}{dx}$ ?

Possible Answers:

$\frac{dy}{dx}=\frac{3t^2}{2t+1}$

$\frac{dy}{dx}=\frac{2t+1}{3t^2}$

$\frac{dy}{dx}=2t+1$

$\frac{dy}{dx}=3t^2$

Correct answer:

$\frac{dy}{dx}=\frac{3t^2}{2t+1}$

Explanation:

It is known that we can derive $\frac{dy}{dx}$ with the formula

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

So we just find $\frac{dy}{dt}, \frac{dx}{dt}$ :

In order to find these derivatives we will need to use the power rule which states,

$x^n \ dx =nx^{n-1}$ .

Applying the power rule we get the following.

$\frac{dy}{dt}=3t^2$

$\frac{dx}{dt}=2t+1$

so we have

$\frac{dy}{dx}=\frac{3t^2}{2t+1}$ .

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Example Question #92 : Parametric, Polar, And Vector

Given the parametric equations

$x(t)=\tan t, y(t)=e^t$

what is $\frac{dy}{dx}$ ?

Possible Answers:

$\frac{dy}{dx}=e^t$

$\frac{dy}{dx}=\frac{\sec^2 t}{e^t}$

$\frac{dy}{dx}=\frac{e^t}{\sec^2 t}$

$\frac{dy}{dx}=\sec^2 t$

Correct answer:

$\frac{dy}{dx}=\frac{e^t}{\sec^2 t}$

Explanation:

It is known that we can derive $\frac{dy}{dx}$ with the formula

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

So we just find $\frac{dy}{dt}, \frac{dx}{dt}$ :

To find the derivatives we will need to use trigonometric and exponential rules.

Trigonometric Rule for tangent: tan(t)dt=sec^2(t)

Rules of Exponentials: $e^u dx=e^u\cdot \frac{du}{dx}$

Thus, applying the above rules we get the following derivatives.

$\frac{dy}{dt}=e^t$

$\frac{dx}{dt}=\sec^2 t$

so we have

$\frac{dy}{dx}=\frac{e^t}{\sec^2 t}$ .

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Example Question #93 : Parametric, Polar, And Vector

Find the arc length of the curve:

$x=5t+3, y=4t-5, 0\leq t\leq2$

Possible Answers:

$\sqrt{41}$

$2\sqrt{41}$

Correct answer:

$2\sqrt{41}$

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

a=0, b =2

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[5t+3]=5$

$\frac{dy}{dt}=\frac{d}{dt}[4t-5]=4$

Applying these into the above formula results in:

$L=\int_{0}^{2}\sqrt{(5)^2+(4)^2}\,dt=\int_{0}^{2}\sqrt{25+16}\,dt=\int_{0}^{2}\sqrt{41}\,dt$

$L=\sqrt{41}\int_{0}^{2}dt=\sqrt{41}\,(2-0)=2\sqrt{41}$

This is one of the answer choices.

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Example Question #91 : Parametric, Polar, And Vector

Find the arc length of the curve (Round to three significant digits):

$x=\frac{1}{4}t^3,\,y=\frac{1}{10}t^2,\,0\leq t\leq 1$

Possible Answers:

0.306

0.272

0.422

0.409

0.460

Correct answer:

0.272

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

a=0, b =1

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[\frac{1}{4}t^3]=\frac{3}{4}t^2$

$\frac{dy}{dt}=\frac{d}{dt}[\frac{1}{10}t^2]=\frac{1}{5}t$

Applying these into the above formula results in:

$L=\int_{0}^{1}\sqrt{(\frac{3}{4}t^2)^2+(\frac{1}{5}t)^2}\,dt=\int_{0}^{1}\sqrt{\frac{9}{16}t^4+\frac{1}{25}t^2}\,dt$

Looking at this integral, it seems pretty intimidating at first, but we can do some algebraic manipulation to make it look like a simple u-substitution problem. First we notice that there is a t^2 term that is common within the two separate functions. If we factor it out then the integral becomes:

$\int_{0}^{1}\sqrt{t^2[\frac{9}{16}t^2+\frac{1}{25}]}\,dt=\int_{0}^{1}t\sqrt{\frac{9}{16}t^2+\frac{1}{25}}\,dt$

Now we can apply the rules of u-substitution:

$u=\frac{9}{16}t^2+\frac{1}{25},\,du=\frac{18}{16}t\,dt=\frac{9}{8}t\,dt$

Therefore:

$\frac{8}{9}du=t\,dt$

Now we must deal with the new bounds of our integral:

$u(0)=\frac{9}{16}(0)^2+\frac{1}{25}=\frac{1}{25}$

$u(1)=\frac{9}{16}(1)^2+\frac{1}{25}=\frac{241}{400}$

Our new integral becomes:

$\frac{8}{9}\int_{\frac{1}{25}}^{\frac{241}{400}}\sqrt{u}\,du=\frac{8}{9}[\frac{2}{3}u^{\frac{3}{2}}]_{\frac{1}{25}}^{\frac{241}{400}}=\frac{16}{27}[(\frac{241}{400})^\frac{3}{2}-(\frac{1}{25})^\frac{3}{2}]$

Plugging this result into a calculator results in:

0.2723945261

Because the problem statement requires us to round this to three significant figures, the final result is:

0.272

This is one of the answer choices.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #601 : Calculus Ii

Example Question #602 : Calculus Ii

Example Question #603 : Calculus Ii

Example Question #91 : Parametric

Example Question #604 : Calculus Ii

Example Question #93 : Parametric

Example Question #91 : Parametric, Polar, And Vector

Example Question #92 : Parametric, Polar, And Vector

Example Question #93 : Parametric, Polar, And Vector

Find the arc length of the curve:

Example Question #91 : Parametric, Polar, And Vector

Find the arc length of the curve (Round to three significant digits):

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