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Calculus 2 : Graphing Polar Form

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Example Question #11 : Graphing Polar Form

Describe the graph of $rcos\theta=-2$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

vertical line at x=-2

line passing through the origin and $\theta=-2$

horizontal line at y=-2

circle centered around the origin with a radius of r=-2

Correct answer:

vertical line at x=-2

Explanation:

Graphing polar equations is different that plotting cartesian equations. Instead of plotting an (x,y) coordinate, polar graphs consist of an $(r,\theta)$ coordinate where is the radial distance of a point from the origin and $\theta$ is the angle above the x-axis.

Using the identity $x=rcos\theta$ , we see the graph of $rcos\theta=-2$ will have the same shape as the graph x=-2 , or a vertical line at x=-2 .

Fig3

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Example Question #12 : Graphing Polar Form

Describe the graph of $rsin\theta=1$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

line passing through the origin and $\theta=1$

horizontal line at y=1

circle centered at the origin with a radius of r=1

vertical line at x=1

Correct answer:

horizontal line at y=1

Explanation:

Using the identity $y=rsin\theta$ , we see the graph of $rsin\theta=1$ will have the same shape as the graph y=1 , or a horizontal line at y=1 .

Fig4

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Example Question #13 : Graphing Polar Form

Describe the graph of $r=4cos\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

circle centered around (0,2) with a radius of

line passing through the origin and x=4

circle centered around (2,0) with a radius of

circle centered around (4,0) with a radius of

Correct answer:

circle centered around (2,0) with a radius of

Explanation:

Substituting values of $\theta$ (in radians) between and $2\pi$ into our expression, we find values of r. We then plot each ordered pair, $(r,\theta)$ , using the value as the radius and $\theta$ as the angle. We get the graph below, a circle centered around (2,0) with a radius of .

Fig5

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Example Question #14 : Graphing Polar Form

Describe the graph of $r=-6sin\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

circle centered around (-6,0) with a radius of

circle centered around (0,-3) with a radius of

circle centered around (0,-6) with a radius of

circle centered around (0,-3) with a radius of

Correct answer:

circle centered around (0,-3) with a radius of

Explanation:

Fig6

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Example Question #15 : Graphing Polar Form

Describe the graph of $r=5+5cos\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

an upright cardioid

a cardioid (heart shape) rotated $90^{\circ}$ left

a cardioid (heart shape) rotated $90^{\circ}$ right

an upside-down cardioid

Correct answer:

a cardioid (heart shape) rotated $90^{\circ}$ left

Explanation:

From our equation, we know the shape of our graph will be a cardioid because our equation is in the form $r=a\pm bcos\theta$ where a=b . Our cardioid is symmetric about the x-axis because our equation includes the cosine function The point of the cardioid is at the origin. The y-intercepts are at (0,a) and (0,-a) . The x-intercept is at (a+b,0) .

We could also substitute values of $\theta$ (in radians) between and $2\pi$ into our expression, to find values of r. We then plot each ordered pair, $(r,\theta)$ , using the value as the radius and $\theta$ as the angle.

We get the graph below, a cardioid (heart shape) rotated $90^{\circ}$ left.

Fig7

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Example Question #16 : Graphing Polar Form

Describe the graph of $r=3-3sin\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

an upside down cardioid (heart shape)

an upright cardioid (heart shape)

a cardioid (heart shape) rotated $90^{\circ}$ right

a cardioid (heart shape) rotated $90^{\circ}$ left

Correct answer:

an upright cardioid (heart shape)

Explanation:

From our equation, we know the shape of our graph will be a cardioid because our equation is in the form $r=a\pm bsin\theta$ where a=b . Our cardioid is symmetric about the y-axis because our equation includes the sine function. The point of the cardioid is at the origin. The x-intercepts are at (a,0) and (-a,0) . The y-intercept is at (0,a+b) .

We get the graph below, an upright cardioid (heart shape).

Fig8

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Example Question #11 : Graphing Polar Form

Describe the graph of $r=4+2sin\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

An upside-down limacon without a loop

An upright limacon without a loop

A limacon without a loop rotated $90^{\circ}$ left

A limacon without a loop rotated $90^{\circ}$ right

Correct answer:

An upside-down limacon without a loop

Explanation:

From our equation, we know the shape of our graph will be a limacon because our equation is in the form $r=a\pm bsin\theta$ where a>b . This limacon will have no loop because a/b>1 . Our limacon is symmetric about the y-axis because our equation includes the sine function. The x-intercepts are at (a,0) and (-a,0) . The y-intercept is at (0,a+b) .

We get the graph below, an upside-down limacon.

Fig9

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Example Question #18 : Graphing Polar Form

Describe the graph of $r=9-6cos\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

A limacon without a loop rotated $90^{\circ}$ right

A limacon without a loop rotated $90^{\circ}$ left

An upright limacon without a loop

An upside down limacon without a loop

Correct answer:

A limacon without a loop rotated $90^{\circ}$ right

Explanation:

From our equation, we know the shape of our graph will be a limacon because our equation is in the form $r=a\pm bcos\theta$ where a>b . This limacon will have no loop because a/b>1 . Our limacon is symmetric about the x-axis because our equation includes the cosine function. The y-intercepts are at (0,a) and (0,-a) . The x-intercept is at -(a+b,0) .

We get the graph below, an limacon turned $90^{\circ}$ right.

Fig10

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Example Question #19 : Graphing Polar Form

Describe the graph of $r=3+6cos\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

a limacon with a loop turned $90^{\circ}$ left

a limacon with a loop turned $90^{\circ}$ right

an upright limacon with a loop

an upside-down limacon with a loop

Correct answer:

a limacon with a loop turned $90^{\circ}$ left

Explanation:

From our equation, we know the shape of our graph will be a limacon because our equation is in the form $r=a\pm bcos\theta$ where a<b . This limacon will have a loop because a/b<1 . The length of the loop is b-a . Our limacon is symmetric about the x-axis because our equation includes the cosine function. The y-intercepts are at (0,a) and (0,-a) . The x-intercept is at (a+b,0) .

We get the graph below, a limacon with a loop turned $90^{\circ}$ left.

Fig11

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Example Question #20 : Graphing Polar Form

Describe the graph of $r=4-7sin\theta$ from $0\leq\theta\leq2\pi$ .

Possible Answers:

a limacon with a loop turned $90^{\circ}$ right

an upside-down limacon with a loop

a limacon with a loop turned $90^{\circ}$ left

an upright limacon with a loop

Correct answer:

an upright limacon with a loop

Explanation:

From our equation, we know the shape of our graph will be a limacon because our equation is in the form $r=a\pm bsin\theta$ where a<b . This limacon will have a loop because a/b<1 . The length of the loop is b-a . Our limacon is symmetric about the y-axis because our equation includes the sine function. The x-intercepts are at (a,0) and (-a,0) . The y-intercept is at -(0,a+b) .

We get the graph below, an upright limacon with a loop.

Fig12

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Calculus 2 : Graphing Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Graphing Polar Form

Example Question #12 : Graphing Polar Form

Example Question #13 : Graphing Polar Form

Example Question #14 : Graphing Polar Form

Example Question #15 : Graphing Polar Form

Example Question #16 : Graphing Polar Form

Example Question #11 : Graphing Polar Form

Example Question #18 : Graphing Polar Form

Example Question #19 : Graphing Polar Form

Example Question #20 : Graphing Polar Form

All Calculus 2 Resources

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