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W13 - Examples

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Parametric curves

Parametric circles

The standard equation of a circle of radius centered at the point :

This equation says that the distance from a point on the circle to the center point equals . This fact defines the circle.

Parametric coordinates for the circle:

For example, the unit circle is parametrized by and .

Parametric lines

Parametric coordinate functions for a line:

Compare this to the graph of linear function:

Vertical lines cannot be described as the graph of a function. We must use .

Parametric lines can describe all lines equally well, including horizontal and vertical lines.

A vertical line is achieved by setting and .

A horizontal line is achieved by setting and .

A non-vertical line may be achieved by setting and , and .

Assuming that , the parametric coordinate functions describe a line satisfying:

and therefore the slope is and the -intercept is .

The point-slope construction of a line has a parametric analogue:

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Parametric ellipses

The general equation of an ellipse centered at with half-axes and is:

This equation represents a stretched unit circle:

  • by in the -axis
  • by in the -axis

Parametric coordinate functions for the general ellipse:

Parametric cycloids

The cycloid is the curve traced by a pen attached to the rim of a wheel as it rolls.

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It is easy to describe the cycloid parametrically. Consider the geometry of the situation:

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The center of the wheel is moving rightwards at a constant speed of , so its position is . The angle is revolving at the same constant rate of (in radians) because the radius is .

The triangle shown has base , so the coordinate is . The coordinate is .

So the coordinates of the point are given parametrically by:

If the circle has another radius, say , then the parametric formulas change to:

Calculus with parametric curves

Tangent to a cycloid

Find the tangent line (described parametrically) to the cycloid when .

Solution

Compute and .

Find :

Find :

Plug in :

Plug in :

Apply formula: :

Calculate at :

Simplify:

So:

This is the slope for our line.

Need the point for our line. Find at .

Plug into parametric formulas:

Point-slope formulation of tangent line:

Inserting our data:

Vertical and horizontal tangents of the circle

Consider the circle parametrized by and . Find the points where the tangent lines are vertical or horizontal.

Solution

For the points with vertical tangent line, we find where the moving point has (purely vertical motion):

The moving point is at when , and at when .

For the points with horizontal tangent line, we find where the moving point has (purely horizontal motion):

The moving point is at when , and at when .

Finding the point with specified slope

Consider the parametric curve given by . Find the point where the slope of the tangent line to this curve equals 5.

Solution

Compute the derivatives:

Therefore the slope of the tangent line, in terms of :

Set up equation:

Solve. Obtain .

Find the point:

Perimeter of a circle

The perimeter of the circle is easily found. We have , and therefore:

Integrate around the circle:

Perimeter of an asteroid

Find the perimeter length of the ‘asteroid’ given parametrically by for .

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Solution

Notice: Throughout this problem we use the parameter instead of . This does not mean we are using polar coordinates!

Compute the derivatives in :

Compute the infinitesimal arc element.

Plug into the arc element, simplify:

Bounds of integration?

Easiest to use . This covers one edge of the asteroid. Then multiply by 4 for the final answer.

On the interval , the factor is positive. So we can drop the absolute value and integrate directly.

Absolute values matter!

If we tried to integrate on the whole range , then really does change sign.

To perform integration properly with these absolute values, we’d need to convert to a piecewise function by adding appropriate minus signs.

Integrate the arc element:

Finally, multiply by 4 to get the total perimeter:

Speed, distance, displacement

The parametric curve describes the position of a moving particle ( measuring seconds). (a) What is the speed function?

Suppose the particle travels for seconds starting at . (b) What is the total distance traveled? (c) What is the total displacement?

Solution

(a) Compute derivatives:

Now compute the speed.

Find sum of squares:

Get the speed function:

(b) Distance traveled by using speed. Compute total distance traveled function:

Integrate.

Substitute and .

New bounds are and .

Calculate:

Insert for the answer.

The distance traveled up to is:

This is our final answer.

(c)

Displacement formula:

Pythagorean formula for distance between given points.

Compute starting and ending points.

For starting point, insert :

For ending point, insert :

Plug points into distance formula.

Insert and :

This is our final answer.

Surface of revolution - parametric circle

By revolving the unit upper semicircle about the -axis, we can compute the surface area of the unit sphere.

The parametrization of the unit upper semicircle is: .

The derivative is: .

Therefore, the arc element:

Now for we choose because we are revolving about the -axis.

Plugging all this into the integral formula and evaluating gives:

Notice: This method is a little easier than the method using the graph .

Surface of revolution - parametric curve

Set up the integral which computes the surface area of the surface generated by revolving about the -axis the curve for .

Solution

For revolution about the -axis, we set .

Then compute :

Therefore the desired integral is: