Math 6C – Exam 2 – Fall 2002

* Solutions*

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1. Find a unit vector in the direction of ( 2, 5 ). The y–component equals what?

Just divide this vector by it’s length, and you have a unit vector in the same direction:

The y-component of this vector is thus

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2. A = (–3, 1) and B = (2, 5). The vector projection of A onto B has x–component equal to what?

The vector projection of A onto B is given by:

The x-component of this projection vector is thus

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3. A = (3, –2, 1) and B = (–1, 2, 5). Find a unit vector in the direction of B x A.

Its x–component equals what?

Now divide this vector by its length:

The x-component of this vector is thus

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4. a(t) = (p sin(pt), 5t⁴) , v(0) = (1,0), r(0) = (0,1).

Then what is the y-component of r(1) ?

Use the given initial condition to solve for the two constants of integration:

And now we integrate again to find the position as a function of time:

.

Use the given initial condition to solve for the two constants of integration:

.

And so we have

.

Hence, the y-component of r(1) is

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5. Find the unit tangent vector T for the plane curve r(t) = (sin (2t), e^3t). The absolute value of the x–component of T is what?

And the absolute value of the x-component of T is thus

.

6. Find the principal unit normal vector N for the plane curve r(t) = (sin (2t), e^3t). The absolute value of the x–component of N is what?

In problem #5 we found that

.

The principal unit normal vector for this curve is given by

.

Although we could (somewhat tediously) compute the principal normal via this formula, it’s easier to proceed as follows.

First note that this normal vector is simply a unit vector perpendicular to T for each t. There are two such perpendicular unit vectors – one is the “principal” one and the other its negative. (Since this problem asks for the absolute value of the principal normal vector’s x-component, we needn’t worry about which of these two vectors we find.)

Secondly notice that is perpendicular to , since their dot products is zero.
It follows that

,

and so the absolute value of the x–component of N is

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7. Find the principal unit binormal vector B for the plane curve r(t) = (sin (2t), e^3t). The absolute value of the x–component of B is what?

Simply realize that since this is a plane curve were T and N are always in the xy-plane, B must always be perpendicular to the xy-plane, so B = (0,0,1)! J

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8. Find the curvature, k, for r(t) = (sin (2t), e^3t) at t = p.

And so

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9. Find the radius of curvature (the radius of the osculating circle) of
at the point (0, 1).

Since the given parabolic curve is symmetric about the y-axis, and the given point is on the y-axis, it follows that the best-fit circle has its center on the y-axis. Suppose the center is (0,c). Then the equation of the osculating circle is

This circle passes through the point (0, 1), so we have that

hence

Thus the osculating circle is given by

The slope of the circle at (0, 1) is 0 (since the cosine has a horizontal tangent line there). Implicitly differentiating the above equation with respect to x, we get:

The first derivatives of the cosine and this circle are both 0 at x=0. The second derivative of the cosine is negative the cosine, so at x=0 is –1, which needs to also be the second derivative of the osculating circle at (0, 1).

At the point (0, 1), this becomes

,

and since this must equal –1, we have

Thus, the radius of curvature is

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10. Find length of the parametrized curve, from . (Just write down the integral that equals this arclength, but don’t solve it!)

, so

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11. Find cosine of the angle between the velocity and acceleration vectors of , at t=0.

,

hence

.

The cosine of the angle between two vectors is given by

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12. Find the velocity and acceleration for the space curve .

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13. Find the (smaller) angle (in radians) between velocity and acceleration vectors of the space curve

, at t=0.

Substitute t=0 into the results of #12:

Then

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14. Find the curvature of the space curve at t=0.

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15. Find the torsion of the space curve at t=0.

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