M408D Second Midterm Exam with Solution

Exam Given November 5, 2002

1. Planes:

a) Find the equation of the plane that goes through the point (2,1,0) and has normal vector (2,1,-2).

Let tex2html_wrap_inline48 . The equation of the plane is tex2html_wrap_inline50 , or 2 x + y -2 z =5.

b) Find the distance from the point (4,13,5) to the plane in part (a).

The distance is tex2html_wrap_inline54 .

c) Find the equation of the plane that goes through the three points (2,1,0), (3,1,4) and (0,0,0) [No, this isn't the same plane as part a)].

Since one of the points is the origin, the normal vector to this plane is tex2html_wrap_inline56 , so the plane is 4x - 8y - z =0.

d) Find the cosine of the angle between the planes of part (a) and (c).

This is tex2html_wrap_inline60 .

2. Lines

a) Find the equation, in symmetric form, for the line through the point (2,1,0) in the direction (2,1,-2).

The direction vector is d=(2,1,-2), the base point is tex2html_wrap_inline64 , so the line is


b) How far is the origin (0,0,0) from this line?

The vector from tex2html_wrap_inline68 to the origin is v=(-2,-1,0). The distance is tex2html_wrap_inline72

c) Find the equation, in symmetric form, for the line through the two points (2,1,0) and (3,-1,4).

The direction vector is d' = (3,-1,4)-(2,1,0) = (1,-2,4), so our line is


You could also write


d) Find the equation (in symmetric form) of the line through (2,1,0) that is perpendicular to the lines of parts (a) and (c).

The vector normal to d and d' is tex2html_wrap_inline84 . Rescale it to (0,2,1). Our line is


(If you don't rescale the direction vector, then you get an uglier, but completely equivalent, answer.)

3. Parametrized curves: Consider the parametrized curve


a) Compute the position, velocity, unit tangent vector and speed at time t=1.

tex2html_wrap_inline94 , so position = tex2html_wrap_inline96 , velocity = tex2html_wrap_inline98 , speed = |(1,2,2)| = 3, and unit tangent vector = tex2html_wrap_inline102 .

b) Compute the arc-length of the curve from t=0 to t=2.

Note that the speed is tex2html_wrap_inline108 , so the arclength is tex2html_wrap_inline110 .

4. Polar coordinates.

a) Sketch the curve tex2html_wrap_inline112 . Mark clearly the angles (if any) where the curve goes through the origin, and the angles where r is maximal.

There are two lobes, one to the right and one to the left. The curve goes through the origin at tex2html_wrap_inline116 and tex2html_wrap_inline118 . The farthest points are when tex2html_wrap_inline120 , so tex2html_wrap_inline122 , so tex2html_wrap_inline124 . That is, the positive and negative x directions.

b) Find (in polar coordinates!) the points where this curve intersects the circle r=1/2.

Note that r is never negative, so we don't have to worry about ``accidental'' intersections. Set tex2html_wrap_inline132 , so tex2html_wrap_inline134 , to tex2html_wrap_inline136 , so tex2html_wrap_inline138 . For tex2html_wrap_inline140 , these are the points tex2html_wrap_inline142 .

c) Write down a definite integral that gives the area, in the first quadrant, inside the curve tex2html_wrap_inline144 but outside the circle r=1/2.


Extra credit: Evaluate this integral. Expand it out and use the double-angle formula tex2html_wrap_inline150 to convert the integral to


5. Partial derivatives. Consider the function of two variables tex2html_wrap_inline154 .

a) Compute tex2html_wrap_inline156 and tex2html_wrap_inline158 and evaluate these at the point (1,1).

tex2html_wrap_inline160 , evaluated at (1,1) gives 5. tex2html_wrap_inline162 , evaluated at (1,1) gives 3.

b) Use these to estimate the value of F(1.01, 1) and the value of F(1, 1.01).

The change in F from (1,1) to (1.01,1) is roughly tex2html_wrap_inline170 , so tex2html_wrap_inline172 . The change in F from (1,1) to (1,1.01) is roughly tex2html_wrap_inline176 , so tex2html_wrap_inline178 .

c) Estimate the value of F(1.02, 1.01).

Add the effect of changing x to the effect of changing y: 4 + 5(0.02) + 3(0.01) = 4.13.

d) Compute the second-order partial derivatives tex2html_wrap_inline186tex2html_wrap_inline188 , and tex2html_wrap_inline190 .

In order, the answers are 6xy, 2, and tex2html_wrap_inline196 .

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Lorenzo Sadun

Tue Nov 5 10:26:20 CST 2002