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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 #161 : Polar

Convert $(-1, \frac{\pi}{4})$ to Cartesian coordinates.

Possible Answers:

$( -\sqrt2,-\sqrt2)$

$( \frac{\sqrt2}{2},\frac{\sqrt2}{2})$

$( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

$( \frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

$( -\frac{\sqrt2}{2},\frac{\sqrt2}{2})$

Correct answer:

$( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

Explanation:

Write the formulas to convert from polar to Cartesian.

$x=rcos(\theta)$

$y=rsin(\theta)$

The and $\theta$ values are known. Substitute both into each equation and solve for and .

$x=(-1)cos(\frac{\pi}{4}) = -\frac{\sqrt2}{2}$

$y=(-1)sin(\frac{\pi}{4}) = -\frac{\sqrt2}{2}$

The Cartesian coordinates are: $( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

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Example Question #324 : Parametric, Polar, And Vector

Convert $r=-4cos(\theta)$ to Cartesian coordinates and find the coordinates of the center.

Possible Answers:

(-2,0)

$(-\frac{1}{2},0)$

(-4,0)

$(-\frac{1}{4},0)$

$(-\frac{1}{8},0)$

Correct answer:

(-2,0)

Explanation:

Write the conversion formulas.

x^2+y^2=r^2

$x=rcos(\theta)$

Notice the r^2 term. If we multiplied by on both sides of the $r=-4cos(\theta)$ equation, we will get:

$r^2=-4rcos(\theta)$

r^2=-4x

Substitute this back into the first equation.

x^2+y^2=-4x

Add on both sides.

x^2+4x+y^2=0

Complete the square with the terms.

(x^2+4x+4)+y^2=4

This would then become:

(x+2)^2+y^2 = 4

This is a circle centered at (-2,0) with a radius of 4.

The answer is: (-2,0)

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Example Question #9 : Polar Calculations

Convert $\Big (3, \frac{\pi}{6} \Big)$ to Cartesian coordinates.

Possible Answers:

$\Bigg( \frac{\sqrt3}{2}, \frac{1}{2} \Bigg)$

$\Bigg (\frac{3 \sqrt3}{2} , \frac{3}{2} \Bigg)$

$\big(\sqrt3, 3 \big)$

$\Bigg(\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}\Bigg)$

Correct answer:

$\Bigg (\frac{3 \sqrt3}{2} , \frac{3}{2} \Bigg)$

Explanation:

When converting from polar to Cartesian coordinates, we must use the formulas

$x = r\cos \theta \; ;\; y=r\sin \theta$

the values r = 3 and $\theta = \frac{\pi}{6}$ are given, so we can calculate that

$x = 3\cdot \cos(\frac{\pi}{6}) = \frac{3\sqrt3}{2}$

and $y = 3\cdot \sin(\frac{\pi}{6}) = \frac{3}{2}$

So the Cartesian coordinate form is

$\Bigg( \frac{3\sqrt3}{2} , \frac{3}{2} \Bigg)$

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Example Question #10 : Polar Calculations

Determine the equation in polar coordinates of y=2x

Possible Answers:

$\theta=tan^{-1}(2)$

$\theta=tan^{-1}(\frac{1}{2})$

$\theta=tan(2)$

$\theta=2$

Correct answer:

$\theta=tan^{-1}(2)$

Explanation:

y=2x can be immediately transformed into polar form by:

$rsin(\theta)=2rcos(\theta)$ .

Dividing by ,

$sin(\theta)=2cos(\theta)$

Dividing both sides by $cos(\theta)$

$tan(\theta)=2$

$\theta=tan^{-1}(2)$

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Example Question #841 : Calculus Ii

Possible Answers:

$r=\sqrt{}2$

$\Theta=\frac{4}{\Pi}$

r=2

$\Theta=\frac{5\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

r=2

$\Theta=\frac{\Pi}{4}$

Correct answer:

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

Explanation:

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Example Question #12 : Polar Calculations

Determine how many points of intersection exist for the curves

$r=cos(2\Theta)$

and

$r = \frac{1}{2}$ .

Possible Answers:

Correct answer:

Explanation:

Solving the equations $r=cos(2\Theta )$ and $r=\frac{1}{2}$ yields $cos(2\Theta)=\frac{1}{2}$ .

Hence,

$2\Theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}.$

Therefore, the values of $\Theta$ between and $2\pi$ that satisfy both equations are:

$\Theta=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}.$

From this, it can be deduced that there are four points of intersection between the given curves:

$(\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6}),(\frac{1}{2},\frac{7\pi}{6}),(\frac{1}{2},\frac{11\pi}{6}).$

However, an identical graph to $r=\frac{1}{2}$ in polar coordinates is $r=-\frac{1}{2}$ , since these two equations describe the same circle with a radius $\frac{1}{2}$ units long. Therefore, the equations $r=cos(2\Theta)$ and $r=-\frac{1}{2}$ must also be solved to yield the remaining points of intersection:

$cos(2\Theta)=-\frac{1}{2}$ ,

$\Rightarrow 2\Theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}.$

$\Rightarrow \Theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.$

From this, it can be deduced that there are four other points of intersection between the given curves:

$(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{4\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3}).$

Hence, there are eight total points of intersection between the curves $r=cos(2\Theta)$ and $r=\frac{1}{2}$ .

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Example Question #162 : Polar

Convert $r=-3cos(\theta)$ to Cartesian coordinates

Possible Answers:

x+y=-3y

x^2+y^2=y

x^2+y^2=-3y

x^2+y^2=-3x

x+y=-3x

Correct answer:

x^2+y^2=-3x

Explanation:

we are given $r=-3cos(\theta)$ and we know that

$x=rcos(\theta )$

$y=rsin(\theta )$

r^2=x^2+y^2

we have r and 3cos(theta). Multiplying each side of the equation by r would give us

$r^2=-3rcos(\theta )$

substitute out the parts we know from the formulas above

x^2+y^2=-3x

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Example Question #161 : Polar

Convert the polar point $(-5, 5 \pi/6)$ into cartesian coordinates.

Possible Answers:

$(-5, 5 \pi/6)$

$( 5\sqrt3/4,5\sqrt3/4)$

$(5\sqrt3/2, -5/2)$

$(5/2, 5\sqrt3/2)$

Correct answer:

$(5\sqrt3/2, -5/2)$

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=-5 and $\theta=5 \pi/6$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=-5*cos(5\pi/6)=-5(-\sqrt3/2)=5\sqrt3/2$

$y=-5*sin(5\pi/6)=-5(1/2)=-5/2$

The cartesian point is $(5\sqrt3/2, -5/2)$ .

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Example Question #162 : Polar

Convert the polar point $(2, \pi/3)$ into cartesian coordinates.

Possible Answers:

$(\sqrt3,1)$

(1,1)

$(1,\sqrt3)$

$( 5\sqrt3/4,5\sqrt3/4)$

Correct answer:

$(1,\sqrt3)$

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=2 and $\theta= \pi/3$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=2*cos(\pi/3)=2(1/2)=1$

$y=2*sin(\pi/3)=2(\sqrt3/2)=\sqrt3$

The cartesian point is $(1,\sqrt3)$ .

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Example Question #841 : Calculus Ii

Convert the polar point $(-8, \pi)$ into cartesian coordinates.

Possible Answers:

$(-8, \pi)$

(8,0)

(8,1)

$(1,\sqrt3)$

Correct answer:

(8,0)

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=-8 and $\theta=\pi$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=-8*cos(\pi)=-8(-1)=8$

$y=-8*sin(\pi)=-8(0)=0$

The cartesian point is (8,0) .

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #161 : Polar

Example Question #324 : Parametric, Polar, And Vector

Example Question #9 : Polar Calculations

Example Question #10 : Polar Calculations

Example Question #841 : Calculus Ii

Example Question #12 : Polar Calculations

Example Question #162 : Polar

Example Question #161 : Polar

Example Question #162 : Polar

Example Question #841 : Calculus Ii

All Calculus 2 Resources

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