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Calculus 2 : Polar

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

Example Questions

Example Question #791 : Calculus Ii

In which quadrant does the polar coordinate terminate?

(3,70^o)

Possible Answers:

III

Correct answer:

Explanation:

The coordinate (3,70^o) goes to the right units from the origin and is rotated 70^o counter-clockwise, terminating in

$Quadrant \,I$

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Example Question #792 : Calculus Ii

In which quadrant does the polar coordinate terminate?

(4,460^o)

Possible Answers:

III

Correct answer:

Explanation:

The coordinate (4,460^o) goes to the right units from the origin and is rotated 460^o counter-clockwise, terminating in

$Quadrant \,II$

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Example Question #793 : Calculus Ii

In which quadrant does the polar coordinate terminate?

(-2,45^o)

Possible Answers:

III

Correct answer:

III

Explanation:

The coordinate (-2,45^o) goes to the left units from the origin and is rotated 45^o counter-clockwise, terminating in

$Quadrant \,III$

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Example Question #794 : Calculus Ii

In which quadrant is the polar coordinate located?

(-3, 30^o)

Possible Answers:

III

Correct answer:

III

Explanation:

The polar coordinate

(-3, 30^o)

is graphed by moving units to the left of the origin and rotating 30^o counter-clockwise, resulting in

$Quadrant\,III$

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Example Question #801 : Calculus Ii

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 #802 : Calculus Ii

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 #803 : Calculus Ii

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 #804 : 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 III

I and II

II and IV

III and IV

I and IV

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:

$\sqrt3-2$

$\sqrt2/2$

None of the other answers.

$1+\sqrt3$

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 #1 : Derivatives Of Polar Form

Find the derivative of the following polar equation:

$r=3+8sin\theta$

Possible Answers:

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

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

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

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

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

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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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #791 : Calculus Ii

Example Question #792 : Calculus Ii

Example Question #793 : Calculus Ii

Example Question #794 : Calculus Ii

Example Question #801 : Calculus Ii

Example Question #802 : Calculus Ii

Example Question #803 : Calculus Ii

Example Question #804 : Calculus Ii

Example Question #1 : Derivatives Of Polar Form

Example Question #1 : Derivatives Of Polar Form

All Calculus 2 Resources

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