Calculus 2 : Graphing Polar Form

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1 : Graphing Polar Form

Which of the following substitutions will help solve the following integral?

\displaystyle \int\frac{\sqrt{25x^2-4}}{x}dx

 

 

 

 

Possible Answers:

\displaystyle x = \frac{2}{5}sec\theta

\displaystyle u = 25x^2 - 4

\displaystyle u = \frac{1}{x}

\displaystyle x = \frac{2}{5} tan\theta

\displaystyle x = \frac{2}{5}sin\theta

Correct answer:

\displaystyle x = \frac{2}{5}sec\theta

Explanation:

As we can see in this integral, there is no reverse chain-rule u-substitution possible. The logical step is to use a trigonometric substitution. If one recalls that trig substitutions of the type \displaystyle x^2-a^2 could be solved with the substitution \displaystyle \frac{2}{5}sec\theta, then the answer is easily seen. However, we can also use a right triangle:                                                       Screen_shot_2015-04-18_at_6.33.31_pm

And thus we have: 

\displaystyle sec \theta = \frac{5x}{2}

or:

 \displaystyle x = \frac{2}{5}sec\theta

Example Question #261 : Parametric, Polar, And Vector

Graph the equation \displaystyle r=cos(\theta ) where \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

R_sinx

R_cosx

R_cosx_1

Faker_cosx

R_cos2x

Correct answer:

R_cosx

Explanation:

At angle \displaystyle 0 the graph as a radius of \displaystyle 1. As it approaches \displaystyle \frac{\pi }{2}, the radius approaches \displaystyle 0.

As the graph approaches \displaystyle \pi, the radius approaches \displaystyle -1.

Because this is a negative radius, the curve is drawn in the opposite quadrant between \displaystyle \frac{3\pi }{2} and \displaystyle 2\pi.

Between \displaystyle \pi and \displaystyle \frac{3\pi }{2}, the radius approaches \displaystyle 0 from \displaystyle -1 and redraws the curve in the first quadrant.

Between \displaystyle \frac{3\pi }{2} and \displaystyle 2\pi, the graph redraws the curve in the fourth quadrant as the radius approaches \displaystyle 1 from \displaystyle 0.    

Example Question #8 : Functions, Graphs, And Limits

Draw the graph of \displaystyle r=sin\theta from \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

Faker_cosx

R_sinx_1

R_sinx

R_sin2x

R_cosx

Correct answer:

R_sinx

Explanation:

Between \displaystyle 0 and \displaystyle \frac{\pi }{2}, the radius approaches \displaystyle 1 from \displaystyle 0.

From \displaystyle \frac{\pi }{2} to \displaystyle \pi the radius goes from \displaystyle 1 to \displaystyle 0.

Between \displaystyle \pi and \displaystyle \frac{3\pi }{2}, the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches \displaystyle -1.

From \displaystyle \frac{3\pi }{2} and \displaystyle 2\pi, the curve is redrawn in the second quadrant as the radius approaches \displaystyle 0 from \displaystyle -1.   

Example Question #12 : Polar Form

Draw the graph of \displaystyle r=cos(2\theta ) from \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

R2_cos2x

R_sin2x

R_cos2x

R_sin2x

R_cosx_1

Correct answer:

R_cos2x

Explanation:

Because this function has a period of \displaystyle \pi, the x-intercepts of the graph \displaystyle y=cos(2x)  happen at a reference angle of \displaystyle \frac{\pi }{4} (angles halfway between the angles of the axes).  

Between \displaystyle 0 and \displaystyle \frac{\pi }{4} the radius approaches \displaystyle 0 from \displaystyle 1.

Between \displaystyle \frac{\pi }{4} and \displaystyle \frac{\pi }{2}, the radius approaches \displaystyle -1 from \displaystyle 0 and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From \displaystyle \frac{\pi }{2} to \displaystyle \frac{3\pi }{4} the radius approaches \displaystyle 0 from \displaystyle -1 , and is drawn in the fourth quadrant, the opposite quadrant. 

Between \displaystyle \frac{3\pi }{4} and \displaystyle \pi, the radius approaches \displaystyle 1 from \displaystyle 0.

From \displaystyle \pi and \displaystyle \frac{5\pi }{4}, the radius approaches \displaystyle 0 from \displaystyle 1.

Between \displaystyle \frac{5\pi }{4} and \displaystyle \frac{3\pi }{2}, the radius approaches \displaystyle -1 from \displaystyle 0. Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between \displaystyle \frac{3\pi }{2} and \displaystyle \frac{7\pi }{4} the radius approaches \displaystyle 0 from \displaystyle -1 and is draw in the second quadrant.

Finally between \displaystyle \frac{7\pi }{4} and \displaystyle 2\pi, the radius approaches \displaystyle 1 from \displaystyle 0.                  

Example Question #261 : Parametric, Polar, And Vector

Draw the graph of \displaystyle r=sin(2\theta ) where \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

R_sin2x

R_cos2x

Faker_cosx

R_sinx

R_sinx_1

Correct answer:

R_sin2x

Explanation:

Because this function has a period of \displaystyle \pi, the amplitude of the graph \displaystyle y=sin(2x)  appear at a reference angle of \displaystyle \frac{\pi }{4} (angles halfway between the angles of the axes).  

Between \displaystyle 0 and \displaystyle \frac{\pi }{4} the radius approaches 1 from 0.

Between \displaystyle \frac{\pi }{4} and \displaystyle \frac{\pi }{2}, the radius approaches 0 from 1.

From \displaystyle \frac{\pi }{2} to \displaystyle \frac{3\pi }{4} the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between \displaystyle \frac{3\pi }{4} and \displaystyle \pi, the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From \displaystyle \pi and \displaystyle \frac{5\pi }{4}, the radius approaches 1 from 0. Between \displaystyle \frac{5\pi }{4} and \displaystyle \frac{3\pi }{2}, the radius approaches 0 from 1.

Then between \displaystyle \frac{3\pi }{2} and \displaystyle \frac{7\pi }{4} the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between \displaystyle \frac{7\pi }{4} and \displaystyle 2\pi, the curve is drawn in the second quadrant.                  

Example Question #1 : Graphing Polar Form

Graph \displaystyle r^2=cos(2\theta ) where \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

R2_sin2x

R_cosx

R_cos2x

R2_cos2x

R_sin2x

Correct answer:

R2_cos2x

Explanation:

Taking the graph of \displaystyle y=cos(2x), we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from \displaystyle 0 to \displaystyle \frac{\pi }{4}\displaystyle \frac{3\pi }{4} to \displaystyle \frac{5\pi }{4}, and \displaystyle \frac{7\pi }{4} to \displaystyle 2\pi.

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of \displaystyle \pm 1.

To draw the graph, the radius is 1 at \displaystyle 0 and traces to 0 at \displaystyle \frac{\pi }{4}. As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From \displaystyle \frac{3\pi }{4} to \displaystyle \pi, the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in \displaystyle \pi to \displaystyle 2\pi.    

Example Question #261 : Parametric, Polar, And Vector

Draw the curve of \displaystyle r^2=sin(2\theta ) from \displaystyle 0\leq \theta \leq2\pi.

Possible Answers:

R_sinx_1

R_sin2x

R2_cos2x

R2_sin2x

R_sinx

Correct answer:

R2_sin2x

Explanation:

Taking the graph of \displaystyle y=sin(2x), we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from \displaystyle 0 to \displaystyle \frac{\pi }{2} and \displaystyle \pi to \displaystyle \frac{3\pi }{2}

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of \displaystyle \pm 1.

To draw the graph, the radius is 0 at \displaystyle 0 and traces to 1 at \displaystyle \frac{\pi }{4}. As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From \displaystyle \frac{\pi }4{} to \displaystyle \frac{\pi }{2}, the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in \displaystyle \pi to \displaystyle 2\pi.    

Example Question #781 : Calculus Ii

What are the parameters by which one can describe the position of a point A in a polar coordinates plane?

Possible Answers:

By distance y from horizontal axis and distance x from vertical axis to point A.

By a position vector to a point A and an angle between horizontal axis and said vector (counter-clockwise positive) .

By a position vector to a point and an angle between vertical axis and said vector (counter-clockwise positive).

By a position vector to a point and an angle between vertical axis and said vector (clockwise positive).

By a position vector to a point A and an angle between horizontal axis and said vector (clockwise positive).

Correct answer:

By a position vector to a point A and an angle between horizontal axis and said vector (counter-clockwise positive) .

Explanation:

A point in polar coordinates is described by a position vector to a point A and an angle between horizontal axis and said vector. A convention for a positive angle is counter-clockwise.

Note, that in polar coordinates, position vector may also be of negative value, meaning pointing in the opposite direction.

Example Question #1 : Graphing Polar Form

Describe the graph of  \displaystyle \theta=\pi/4.

Possible Answers:

Circle centered around the origin with a radius  \displaystyle r=-\pi/4

Straight line passing through the origin and  \displaystyle \theta=\pi/4

Straight line passing through the origin and  \displaystyle \theta=-\pi/4

Circle centered around the origin with a radius  \displaystyle r=\pi/4

Correct answer:

Straight line passing through the origin and  \displaystyle \theta=\pi/4

Explanation:

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

When the graph of an equation in the form \displaystyle \theta=\alpha, where \displaystyle \alpha is an angle, the angle of the graph is constant and independent of the radius. This creates a straight line \displaystyle \alpha radians above the x-axis passing through the origin.

In this problem,  \displaystyle \theta=\pi/4 is a straight line \displaystyle \pi/4 radians or  \displaystyle 45^{\circ} about the x-axis passing through the origin.

Fig1

Example Question #3 : Graphing Polar Form

Describe the graph of \displaystyle r=4.

Possible Answers:

Circle centered around the origin with a radius of \displaystyle 4.

Cardiod centered around \displaystyle r=2 with a radius of \displaystyle 4.

Straight line passing through the origin and \displaystyle r=4

Limacon with inner loop centered around \displaystyle r=2

Correct answer:

Circle centered around the origin with a radius of \displaystyle 4.

Explanation:

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

When the graph of an equation in the form \displaystyle r=\lambda, where \displaystyle \lambda is a constant, the graph is a circle centered around the origin with a radius of \displaystyle \lambda.

In this problem, \displaystyle r=4 is a circle centered around the origin with a radius of \displaystyle 4.

Fig2

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