# Solutions to Problem Set 14

Math 311-01/02

2-16-2018

[Tangent, normal, and binormal problems]

1. Sketch the graph of for .

Then find the unit tangent and unit normal vectors at .

Parametrize the curve as

Then

Since , the unit tangent vector is

Since this is a curve in the plane, I can find a vector normal to by switching the components of and negating one of them. There are two ways to do this:

The graph near shows that the unit normal must point to the left and downward. Hence, I use , which has length . The unit normal is

2. Sketch the graph of for .

Then find the unit tangent and unit normal vectors at .

Since , the unit tangent vector is

Since this is a curve in the plane, I can find a vector normal to by switching the components of and negating one of them. There are two ways to do this:

The graph near shows that the unit normal must point to the left and upward. Hence, I use , which has length 5. The unit normal is

3. Find parametric equations for the osculating circle to at .

I need the point on the curve, the curvature, the unit tangent, and the unit normal.

The point on the curve is .

There are two unit vectors perpendicular to :

At , the unit normal must point to the right. Hence, .

For the curvature, I'll use the parametric formula

First,

I can regard the velocity and acceleration vectors as 3-dimensional vectors by making the z-components 0:

Then

Therefore, the curvature is

The radius of the osculating circle is the reciprocal of the curvature: .

The center of the circle is

Hence, the circle is

4. Find the unit tangent, the unit normal, and the binormal to

Now

The unit tangent is

Hence, the unit tangent vector at is .

Next,

Hence, the unit normal is .

To find the binormal, I'll take the cross product.

5. Find the unit tangent, the unit normal, and the binormal to

Now

The unit tangent is

Hence, the unit tangent vector at is .

Since I only want the unit normal and binormal at , I may assume is positive and drop the absolute values:

Then

Hence, the unit normal is .

To find the binormal, I'll take the cross product.

When we are not sure, we are alive. - Graham Greene

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