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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 #72 : Parametric, Polar, And Vector

Given x=2t-19 and y=4-8t , what is in terms of (rectangular form)?

Possible Answers:

y=4x-72

y=-4x-72

None of the above

y=-72x-4

y=72x-4

Correct answer:

y=-4x-72

Explanation:

Given x=2t-19 and y=4-8t , let's solve both equations for :

$x=2t-19\rightarrow t=\frac{x+19}{2}$

$y=4-8t\rightarrow t=-\frac{y-4}{8}$

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

$-\frac{y-4}{8}=\frac{x+19}{2}$

y-4=-4(x+19)

y=-4(x+19)+4

y=-4x-76+4

y=-4x-72

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Example Question #73 : Parametric, Polar, And Vector

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

Possible Answers:

$y=5\left(\pm \sqrt{\frac{x-3}{2}}\right)+8$

$y=5\sqrt{\frac{x+3}{2}}-8$

$y=5\sqrt{\frac{x+3}{2}}+8$

$y=5\sqrt{\frac{x-3}{2}}-8$

None of the above

Correct answer:

$y=5\left(\pm \sqrt{\frac{x-3}{2}}\right)+8$

Explanation:

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

$x=2t^{2}+3\rightarrow t=\pm \sqrt{\frac{x-3}{2}}$

$y=5t+8\rightarrow t=\frac{y-8}{5}$

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

$\frac{y-8}{5}=\pm \sqrt{\frac{x-3}{2}}$

$y-8=5\left(\pm\sqrt{\frac{x-3}{2}}\right)$

$y=5\left(\pm \sqrt{\frac{x-3}{2}}\right)+8$

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Example Question #74 : Parametric, Polar, And Vector

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

x=3t^2-6, y=9t+5, t=3

Possible Answers:

$\frac{32}{21}$

$\frac{1}{2}$

$\frac{1}{9}$

$\frac{21}{32}$

Correct answer:

$\frac{1}{2}$

Explanation:

The formula for dy/dx for parametric equations is given as:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ From the problem statement:

$\frac{dy}{dt}=9,\frac{dx}{dt}=6t$

If we plug in t=3, into the above equations:

$\frac{dy}{dx}=\frac{9}{6(3)}=\frac{9}{18}=\frac{1}{2}$

This is one of the answer choices.

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Example Question #81 : Parametric, Polar, And Vector

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

$x=\sqrt{t+5},y=5t-2, t=4$

Possible Answers:

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{6}{5}$

$\frac{1}{30}$

Correct answer:

Explanation:

The formula for dy/dx for parametric equations is given as:

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

From the problem statement:

$\frac{dy}{dt}=5,\frac{dx}{dt}=\frac{1}{2\sqrt{t+5}}$

If we plug in t=4, into the above equations:

$\frac{dy}{dx}=\frac{5}{\frac{1}{2\sqrt{9}}}=\frac{5}{\frac{1}{2(3)}}=\frac{5}{\frac{1}{6}}=30$

This is one of the answer choices.

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Example Question #201 : Ap Calculus Bc

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

$x=5\sec(t),y=3\tan(t), t=\frac{\pi}{6}$

Possible Answers:

$10\sqrt3$

$\frac{3}{10}$

$\frac{5}{6}$

$\frac{10}{3}$

$\frac{6}{5}$

Correct answer:

$\frac{6}{5}$

Explanation:

The formula for dy/dx for parametric equations is given as:

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

From the problem statement:

$\frac{dy}{dt}=3\sec^2(t),\frac{dx}{dt}=5\sec(t)\tan(t)$

If we plug these into the above equation we end up with:

$\frac{dy}{dx}=\frac{3\sec^2(t)}{5\sec(t)\tan(t)}=\frac{3\sec(t)}{5\tan(t)}=\frac{3\frac{1}{cos(t)}}{5\frac{sin(t)}{cos(t)}}$

$\frac{dy}{dx}=\frac{3}{5}\frac{1}{\cos(t)}\frac{\cos(t)}{\sin(t)}=\frac{3}{5}\frac{1}{\sin(t)}$

If we plug in our given value for t, we end up with:

$\frac{3}{5}\frac{1}{\sin(\frac{\pi}{6})}=\frac{3}{5}\frac{1}{\frac{1}{2}}=\frac{3}{5}*2=\frac{6}{5}$

This is one of the answer choices.

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Example Question #81 : Parametric, Polar, And Vector

Convert the following parametric equation into rectangular form:

$x=t+5, y=\sec^2(t)$

Possible Answers:

$y=\sec^2(x)$

$y=\sec^2(x-5)+C$

$y=\sec(x-5)$

$y=\sec^2(x-5)$

Correct answer:

$y=\sec^2(x-5)$

Explanation:

To convert from parametric to rectangular form, we must eliminate the parameter by finding t in terms of x or y:

t=x-5

Now, plug t into the equation for y:

$y=\sec^2(x-5)$

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Example Question #83 : Parametric, Polar, And Vector

Given x=5t+12 and y=9-3t , what is in terms of ?

Possible Answers:

$y=9+\frac{3}{5}(x+12)$

$y=9-\frac{3}{5}(x+12)$

$y=9-\frac{3}{5}(x-12)$

$y=9+\frac{3}{5}(x-12)$

None of the above

Correct answer:

$y=9-\frac{3}{5}(x-12)$

Explanation:

Given x=5t+12 and y=9-3t , let's solve both equations for :

$x=5t+12\rightarrow t=\frac{x-12}{5}$

$y=9-3t\rightarrow t=\frac{9-y}{3}$

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

$\frac{9-y}{3}=\frac{x-12}{5}$

$9-y=3(\frac{x-12}{5})$

$y=9-\frac{3}{5}(x-12)$

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Example Question #84 : Parametric, Polar, And Vector

What is a way of parameterizing the circle x^2+y^2=16 ? (Assume $0\leq t\leq2\pi$ ) There are many ways of doing this.

Possible Answers:

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

$x(t)=4\cos t$ , $y(t)=4\sin t$

$x(t)=4\cos t$ , $y(t)=16\sin t$

$x(t)=16\cos t$ , $y(t)=16\sin t$

Correct answer:

$x(t)=4\cos t$ , $y(t)=4\sin t$

Explanation:

To see why the parametric equations

$x(t)=4\cos t$ , $y(t)=4\sin t$

describe the circle

x^2+y^2=16

we do the following manipulation:

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

which means

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

x(t)^2+y(t)^2=16

which describes the circle of radius .

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Example Question #85 : Parametric, Polar, And Vector

What is not a way of parameterizing the ellipse

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

(Assume $0\leq t\leq 2\pi$ )

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The reason that the parametric equations

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

don't describe the ellipse

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

is because if do some algebraic manipulation, we get

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

which describes a different ellipse than

$\frac{x^2}{4}+\frac{y^2}{9}=1$ .

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Example Question #86 : Parametric, Polar, And Vector

Which of the following set of parametric equations parametrizes a section of parabola

y=1-x^2 ? (Assume $0\leq t\leq 2\pi$ )

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

We can see that the parametric equations

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

describe a section of the parabola

y=1-x^2

because if do some manipulations of the parametric equations, we get

$x(t)^2+y(t)=\cos^2 t+\sin^2 t=1$

$\Rightarrow x(t)^2+y(t)=1$

So then we get

y(t)=1-x(t)^2

which describes part of the parabola

y=1-x^2 .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #72 : Parametric, Polar, And Vector

Example Question #73 : Parametric, Polar, And Vector

Example Question #74 : Parametric, Polar, And Vector

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

Example Question #81 : Parametric, Polar, And Vector

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

Example Question #201 : Ap Calculus Bc

Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

Example Question #81 : Parametric, Polar, And Vector

Example Question #83 : Parametric, Polar, And Vector

Example Question #84 : Parametric, Polar, And Vector

Example Question #85 : Parametric, Polar, And Vector

Example Question #86 : Parametric, Polar, And Vector

All Calculus 2 Resources

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