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Tangent to a cycloid

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

Solution

(1) Compute and .

Find :

Find :

(2) Plug in :

Plug in :

(3) Apply formula: :

Calculate at :

Simplify:

So:

This is the slope for our line.

(4) Need the point for our line. Find at .

Plug into parametric formulas:

(5) 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

(1) 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 .

(2) 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

(1) Compute the derivatives:

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

(2) Set up equation:

Solve. Obtain .

(3) Find the point:

Perimeter of a circle

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

(2) Integrate around the circle:

Perimeter of an asteroid

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

Solution

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

Compute the derivatives in :

(2) Compute the infinitesimal arc element.

Plug into the arc element, simplify:

(3) 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.

(4) 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)

(1) Compute derivatives:

(2) Now compute the speed.

Find sum of squares:

Get the speed function:

(b) Distance traveled by using speed.

(1) Compute total distance traveled function:

(2) Integrate.

Substitute and .

New bounds are and .

Calculate:

(3) Insert for the answer.

The distance traveled up to is:

This is our final answer.

(c)

(1) Displacement formula:

Pythagorean formula for distance between given points.

(2) Compute starting and ending points.

For starting point, insert :

For ending point, insert :

(3) Plug points into distance formula.

Insert and :

This is our final answer.

Surface of revolution - parametric circle

(1) 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: .

(2) Therefore, the arc element:

(3) 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: