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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 #285 : Parametric, Polar, And Vector

In which quadrant is the polar coordinate located?

(2, 45^o)

Possible Answers:

III

Correct answer:

Explanation:

The polar coordinate

(2, 45^o)

is graphed by moving units to the right of the origin and rotating 45^o counter-clockwise, resulting in

$Quadrant\,I$

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Example Question #286 : Parametric, Polar, And Vector

In which quadrant is the polar coordinate located?

(4, 100^o)

Possible Answers:

III

Correct answer:

Explanation:

The polar coordinate

(4, 100^o)

is graphed by moving units to the right of the origin and rotating 100^o counter-clockwise, resulting in

$Quadrant\,II$

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Example Question #287 : Parametric, Polar, And Vector

In which quadrant is the polar coordinate located?

(1,315^o)

Possible Answers:

III

Correct answer:

Explanation:

The polar coordinate

(1, 315^o)

is graphed by moving unit to the right of the origin and rotating 315^o counter-clockwise, resulting in

$Quadrant\,IV$

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Example Question #801 : Calculus Ii

Graph the following relationship in polar coordinates for $0\leq \theta\leq\pi/2$ :

$r(\theta)=2\sin{4\theta}$ ;

In which quadrants does the graph appear?

Possible Answers:

I and IV

III and IV

II and IV

I and III

I and II

Correct answer:

I and III

Explanation:

Looking at the graph of

$r(\theta)=2\sin{4\theta}$

with polar coordinates

$0\leq \theta\leq\pi/2$

Screen shot 2016 02 18 at 8.58.38 am

It is seen that the graph lies in quadrant one and three.

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Example Question #1 : Derivatives Of Polar Form

For the polar equation $r = cos(\theta) + sin(\theta),$ $0<\theta<2\pi$ , find $\frac{dy}{dx}$ when $\theta = \pi/3$ .

Possible Answers:

$\sqrt3-2$

$1+\sqrt3$

$\sqrt2/2$

None of the other answers.

Correct answer:

$\sqrt3-2$

Explanation:

When
$\theta = \pi/3, r = cos(\pi/3)+sin (\pi/3) = \frac{1+\sqrt3}{2}$ .

Taking the derivative of our given equation with respect to $\theta$ , we get

$\frac{dr}{d\theta} = -sin(\theta)+cos(\theta)$

To find $\frac{dy}{dx}$ , we use

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{dr}{d\theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta}cos(\theta)-rsin(\theta)}.$

Substituting our values of $r, \theta$ into this equation and simplifying carefully using algebra, we get the answer of $\sqrt{3}-2$ .

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Example Question #802 : Calculus Ii

Find the derivative of the following polar equation:

$r=3+8sin\theta$

Possible Answers:

$\frac{8cos\theta sin^2\theta+3cos^2\theta }{4cos^2\theta+3sin\theta-6sin^2\theta}$

$\frac{8cos\theta sin\theta+3sin\theta }{4sin^2\theta-3sin\theta-8cos^2\theta}$

$\frac{16sin^2\theta+3cos\theta }{8cos^2\theta-3cos\theta-8sin^2\theta}$

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

$\frac{8cos\theta sin\theta-6cos\theta }{4cos^2\theta+6sin\theta+3sin^2\theta}$

Correct answer:

$\frac{16cos\theta sin\theta+3cos\theta }{8cos^2\theta-3sin\theta-8sin^2\theta}$

Explanation:

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to $\theta$ . This gives us:

$\frac{dr}{d\theta}=8cos\theta$

Now that we know dr/d $\theta$ , we can plug this value into the equation for the derivative of an expression in polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }$

Simplifying the equation, we get our final answer for the derivative of r:

$\frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }$

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Example Question #1 : Derivatives Of Polar Form

Evaluate the area given the polar curve: $r=\frac{\theta}{2}$ from $[0,2\pi]$ .

Possible Answers:

$\frac{\pi^3}{3}$

$8\pi$

$\frac{\pi^3}{4}$

$\pi$

$\frac{\pi^3}{6}$

Correct answer:

$\frac{\pi^3}{3}$

Explanation:

Write the formula to find the area in between two polar equations.

$A=\frac{1}{2}\int_{a}^{b}[(\textup{Outer Radius}^2)-(\textup{Inner Radius})^2]\: d\theta$

The outer radius is $r=\frac{\theta}{2}$ .

The inner radius is r=0 .

Substitute the givens and evaluate the integral.

$A=\frac{1}{2}\int_{0}^{2\pi}\left(\frac{\theta}{2}\right)^2\: d\theta=\frac{1}{2}\int_{0}^{2\pi}\frac{\theta^2}{4}\: d\theta=\frac{1}{8}\left[\frac{\theta^3}{3}\right]_{0}^{2\pi}$

$\frac{1}{8}\left[\frac{\theta^3}{3}\right]_{0}^{2\pi}= \frac{1}{8}\left[\frac{(2\pi)^3}{3}\right] = \frac{\pi^3}{3}$

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Example Question #2 : Derivatives Of Polar Form

Find the derivative $\frac{dy}{dx}$ of the polar function $r(\theta)=3-4sin(\theta)$ .

Possible Answers:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=-4cos(\theta)$

$\frac{dy}{dx}=4cos(\theta)$

$\frac{dy}{dx}=\frac{-4cos(\theta)cos(\theta)+(3-4sin(\theta))sin(\theta)}{-4cos(\theta)sin(\theta)-(3-4sin(\theta))cos(\theta)}$

Correct answer:

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

Explanation:

The derivative of a polar function is found using the formula

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$

The only unknown piece is $r'(\theta)$ . Recall that the derivative of a constant is zero, and that

$\frac{d}{d\theta}sin(\theta)=cos(\theta)$ , so

$r'(\theta)=-4cos(\theta)$

Substiting $r'(\theta)\ and\ r(\theta)$ this into the derivative formula, we find

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos(\theta)cos(\theta)-(3-4sin(\theta))sin(\theta)}$

$\frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}$

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Example Question #803 : Calculus Ii

Find the first derivative of the polar function

$r(\theta)=sin(3\theta)$ .

Possible Answers:

$\frac{dy}{dx}=3cos(3\theta)$

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

$\frac{dy}{dx}=\frac{3sin(3\theta)cos(\theta)+cos(3\theta)sin(\theta)}{3sin(3\theta)sin(\theta)-cos(3\theta)cos(\theta)}$

$\frac{dy}{dx}=\frac{sin(3\theta)sin(\theta)+3cos(3\theta)cos(\theta)}{sin(3\theta)cos(\theta)-3cos(3\theta)sin(\theta)}$

Correct answer:

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

Explanation:

In general, the dervative of a function in polar coordinates can be written as

$\frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}$ .

Therefore, we need to find $r'(\theta)$ , and then substitute $r(\theta)\ and\ r'(\theta)$ into the derivative formula.

To find $r'(\theta)$ , the chain rule,

$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$ , is necessary.

We also need to know that

$\frac{d}{dx}sin(x)=cos(x)$ .

Therefore,

$r'(\theta)=3cos(3\theta)$ .

Substituting $r(\theta)\ and\ r'(\theta)$ into the derivative formula yields

$\frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}$

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Example Question #3 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=2\cos(\theta)-1$

Possible Answers:

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{4\cos(\theta)\sin(\theta)-\sin(\theta)}$

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

$\cos^2(\theta)-\sin^2(\theta)$

Correct answer:

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

Explanation:

The formula for the derivative of a polar function is

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{\frac{\mathrm{dr} }{\mathrm{d} \theta}\sin(\theta)+r\cos(\theta)}{\frac{\mathrm{dr} }{\mathrm{d} \theta}\cos(\theta)-r\sin(\theta)}$

First, we must find the derivative of the function given:

$\frac{\mathrm{dr} }{\mathrm{d} \theta}=-2\sin(\theta)$

Now, we plug in the derivative, as well as the original function, into the above formula to get

$\frac{-2sin^2(\theta)+2\cos^2(\theta)-\cos(\theta)}{-4\cos(\theta)\sin(\theta)+\sin(\theta)}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #285 : Parametric, Polar, And Vector

Example Question #286 : Parametric, Polar, And Vector

Example Question #287 : Parametric, Polar, And Vector

Example Question #801 : Calculus Ii

Example Question #1 : Derivatives Of Polar Form

Example Question #802 : Calculus Ii

Example Question #1 : Derivatives Of Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #803 : Calculus Ii

Example Question #3 : Derivatives Of Polar Form

All Calculus 2 Resources

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