Answers to Sample Questions for Exam 1
Math 2374, Spring 2002
rogness@math.umn.edu

Most of these are just sketches of the work to get the correct answers.  If you've done the problem correctly, you'll probably recognize the work I've done.  If you can't make heads or tails of what I'm doing, it probably means you need to go back and relearn the material in question.  In some cases I don't even do the problem, but rather describe how it can be done.  Warning: I wrote these answers pretty quickly and I haven't checked my work, so it's possible that there's a mistake in here.  If you think you've spotted one, email me at the address above.  Thanks!

1. Find the equation for the following piece of a single cone.  Note that the vertex is at (0,0,200), and the bottom is on the circle [Graphics:Images/index_gr_1.gif]in the xy-plane.

The picture shows a piece of the single cone [Graphics:Images/index_gr_2.gif].  Alternatively, we could say its a piece of the lower half of the double cone [Graphics:Images/index_gr_3.gif].  In either case we must say that we only want [Graphics:Images/index_gr_4.gif], or we'll get too much of the cone(s).

2.  (a) Find the equation for a plane containing the points (1,2,3), (2,4,9), and (-1,3,-4).

[Graphics:Images/index_gr_5.gif]
[Graphics:Images/index_gr_6.gif]
[Graphics:Images/index_gr_7.gif]
[Graphics:Images/index_gr_8.gif]

So, using the point (1,2,3), the cartesian equation for the plane is [Graphics:Images/index_gr_9.gif].  Your answer may look different and be correct; you should multiply them out to see if they're actually the same.

The way the problem is phrased, you could also give a parametric equation for the plane.  One possibility is [Graphics:Images/index_gr_10.gif].

(b)  Find the parametric equation for the plane [Graphics:Images/index_gr_11.gif].

The easiest way to do this is probably just find three points A, B, and C on the plane by any method you choose, then find the vectors [Graphics:Images/index_gr_12.gif] and [Graphics:Images/index_gr_13.gif], which you then use in your parametrization as in the answer for (a) above.  (Hint: three possible points are where the plane intersects the coordinate axes; these points are (0,0,3), (0,-3,0), and (9/2,0,0).

(c)  Find the cartesian equation for the plane [Graphics:Images/index_gr_14.gif].  

[Graphics:Images/index_gr_15.gif]
[Graphics:Images/index_gr_16.gif]

So the equation for the plane is [Graphics:Images/index_gr_17.gif].

(d) Which, if any, of the following vectors is normal to the plane [Graphics:Images/index_gr_18.gif] used in part (c)?
        (0,1,1), (2,2,2), (-3,-3,2), (0,1,0).

One method: a normal vector to the plane must be perpendicular to the two vectors used in the parametrization, (-1,1,0) and (2,4,9).  So you can use the dot product to determine which vectors are perpendicular to both of these.

Another method: From the Cartesian equation for the plane, we know a normal vector is (9,9,-6).  Any vector normal to the plane must be a multiple of this one, which means only (-3,-3,2) works.

3.  Suppose we have the following linear transformations:

[Graphics:Images/index_gr_19.gif]

[Graphics:Images/index_gr_20.gif]

(a) Find T(S([Graphics:Images/index_gr_21.gif])), if possible.  If it is not possible, explain why.

(b) Find S(T(
[Graphics:Images/index_gr_22.gif])), if possible.  If it is not possible, explain why.

T(S(v)) is not possible; S has three outputs, and T takes two inputs.

S(T(v)) is possible; T takes two inputs, and gives three outputs, while S takes three inputs and returns three as well.  You can work out what S(T(v)) is on your own; ask your TA if you're not sure if your answer is correct.

4.  (a) using the limit defintiion of partial derivatives, find the value of [Graphics:Images/index_gr_23.gif](0,0) for the function

[Graphics:Images/index_gr_24.gif]

(b) Using any valid method, find a general formula for ∂f/∂y at any point away from [Graphics:Images/index_gr_25.gif]

[Graphics:Images/index_gr_26.gif]
[Graphics:Images/index_gr_27.gif]
[Graphics:Images/index_gr_28.gif]
[Graphics:Images/index_gr_29.gif]
[Graphics:Images/index_gr_30.gif]
[Graphics:Images/index_gr_31.gif]

5.  (a) Plot the level curves for c=0,1,4,9 of the function

[Graphics:Images/index_gr_32.gif]

These are, in order, a point (the origin), and then ellipses which are increasingly bigger.

[Graphics:Images/index_gr_33.gif]

(b) Plot the level surfaces for c=1,4 of the function

[Graphics:Images/index_gr_34.gif]

These are spheres of radius 1 and 2, respectively.

6. Given a three dimensional vector [Graphics:Images/index_gr_35.gif], show that T([Graphics:Images/index_gr_36.gif])=([Graphics:Images/index_gr_37.gif]·[Graphics:Images/index_gr_38.gif])[Graphics:Images/index_gr_39.gif] is a linear transformation.  (Hint: write the right hand side using matrix notation and use the fact that k[Graphics:Images/index_gr_40.gif]=[Graphics:Images/index_gr_41.gif]k for a scalar k.)  [This is problem 22 in section 2.3]

Let [Graphics:Images/index_gr_42.gif] and [Graphics:Images/index_gr_43.gif]  Then ([Graphics:Images/index_gr_44.gif]·[Graphics:Images/index_gr_45.gif])[Graphics:Images/index_gr_46.gif]which, if you multiply the vector by the scalar, is a vector with somewhat messy entries.  But it is a linear transformation, and you should be able to find the matrix representing it.

7. Parametrize the intersection of the cylinder [Graphics:Images/index_gr_47.gif] and the plane [Graphics:Images/index_gr_48.gif].

[Graphics:Images/index_gr_49.gif]

[Graphics:Images/index_gr_50.gif]

[Graphics:Images/index_gr_51.gif]

8.  Give a parametrization for the line segment from the point (1,0,0) to the point (2,3,-4).  How would you change this to give a parametrization for the entire line containing these two points?

[Graphics:Images/index_gr_52.gif]
[Graphics:Images/index_gr_53.gif]
[Graphics:Images/index_gr_54.gif]

[Graphics:Images/index_gr_55.gif]

[Graphics:Images/index_gr_56.gif]

If we let t range over all real numbers, instead of just 0 to 1, then we would get the entire line.

9.  Working in [Graphics:Images/index_gr_57.gif], consider the points A=(0,0,1,0), B=(1,0,2,0), and C=(3,2,0,-1).

(a) Find the vectors [Graphics:Images/index_gr_58.gif] and [Graphics:Images/index_gr_59.gif].

[Graphics:Images/index_gr_60.gif]
[Graphics:Images/index_gr_61.gif]
[Graphics:Images/index_gr_62.gif]

(b) Find the angle between [Graphics:Images/index_gr_63.gif] and [Graphics:Images/index_gr_64.gif].

[Graphics:Images/index_gr_65.gif]
[Graphics:Images/index_gr_66.gif]
[Graphics:Images/index_gr_67.gif]
[Graphics:Images/index_gr_68.gif]
[Graphics:Images/index_gr_69.gif]

This is in radians; in degrees we have

[Graphics:Images/index_gr_70.gif]
[Graphics:Images/index_gr_71.gif]

(c) Find the area of the parallelogram spanned by [Graphics:Images/index_gr_72.gif] and [Graphics:Images/index_gr_73.gif].

[Graphics:Images/index_gr_74.gif]
[Graphics:Images/index_gr_75.gif]

Incidentally, finding the area of a four-dimensional parallelogram was an exam question last semester!


Converted by Mathematica      February 19, 2002