Practice Midterm Exam Solutions
Calculus 3, Section 001
Name:
Answer the questions in the spaces provided on the question sheets.
If you run out of room for an answer, continue on the back of the
page. No calculators are permitted. Good luck, have fun!
Question Points Score
1
10
2
10
3
10
4
6
5
6
Total:
42
1
1. Let ~v = h3, 2, 5i and w
~ = h1, −3, −4i be two three-dimensional vectors.
(a) (2 points) Find |~v |.
Solution: This indicates the length of the vector ~v , which is
|~v | =
√
√
√
32 + 22 + 52 = 9 + 4 + 25 = 38
(b) (2 points) Find ~v · w.
~
Solution:
~v · w
~ = (3)(1) + (2)(−3) + (5)(−4) = −23
(c) (3 points) Find ~v × w.
~
Solution:
~i ~j
~k
~v × w
~ = 3 2
5
1 −3 −4
2
5
=
−3 −4
~ 3 5
i − 1 −4
~ 3 2
j + 1 −3
~
k
= h7, 17, −11i
(d) (3 points) What is the area of the parallelogram which has ~v and w
~ as
two of its sides?
Solution: The area of this parallelogram is the same as the length of
the cross product of the vectors, which is
|~v × w|
~ = |h7, 17, −11i| =
√
72 + 172 + 112 =
√
459
2. (a) (4 points) Find a parametric representation of the line of intersection of
the planes defined by the equations
x + 2y + z = 3
and
x−y−z =4
Solution: If we add the two equations, we get 2x + y = 7. If we choose
x to parameterize the line, i.e. setting x = t, we get y = 7 − 2t from this
equation. It remains to solve for z in terms of t, which we can do by
replacing x and y in the first equation by their expressions in terms of
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t:
(t) + 2(7 − 2t) + z = 3
so
z = 3t − 11
Thus one possible parameterization of the line is
~r(t) = ht, 7 − 2t, 3t − 11i
(b) (6 points) Find an equation for the plane containing the line ht + 1, 2t +
3, 3t + 2i and the point (1, 1, 0).
Solution: We need two directions in the plane, and from the line we
have the direction h1, 2, 3i as this is the direction vector of the plane. If
we draw a vector starting at the given point (1, 1, 0) to any point on the
line, this vector also lies in the plane, so in particular we can take the
vector from (1, 1, 0) to (1, 3, 2), giving h0, 2, 2i. We now have two directions
in the plane, so we can find their cross product ~n = h1, 2, 3i × h0, 2, 2i =
h−2, −2, 2i. We can then write down the equation of the plane – one way
to do it is as
−2(x − 1) − 2(y − 1) + 2z = 0
3. Consider the circular helix given by the vector function
√
~r(t) = h 3 t, sin(t), cos(t)i
0
00
−
−
(a) (3 points) Find the derivatives →
r (t) and →
r (t).
Solution: We have
√
0
→
−
r (t) = h 3 , cos(t), − sin(t)i
00
→
−
r (t) = h0, − sin(t), − cos(t)i
− 0 0
00
−
−
(b) (3 points) Find →
r (t) and →
r (t) × →
r (t).
Solution: We have
q√
√
√
0
→
−
r (t) = ( 3 )2 + (cos(t))2 + (− sin(t))2 = 3 + 1 = 4 = 2.
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and
→
→
−
→
−
−i
j
k
√
0
00
→
−
−
r (t) × →
r (t) = 3 cos(t) − sin(t) 0 − sin(t) − cos(t) √
√
− 3 cos(t) →
− cos(t) − sin(t) →
− 3 − sin(t) →
= i − j + k − sin(t) − cos(t) 0 − cos(t) 0 − sin(t) √
√
→
−
→
− √
→
− √
= − i + 3 cos(t) j − 3 sin(t) k =< −1, 3 cos(t), − 3 sin(t) > .
−
(c) (4 points) Let κ(t) be the curvature of →
r (t). Find κ(t).
Solution: We have
q
√
√
√
√
0
00
→
−
→
−
r (t) × r (t) = (−1)2 + ( 3 cos(t))2 + (− 3 sin(t))2 = 1 + 3 = 4 = 2
Hence
→
0
00
−
−
r (t) × →
r (t)
2
2
1
= 3 = =
κ(t) =
3
0
→
−
2
8
4
| r (t)|
4. Consider the ellipse in the xy plane defined by the equation 9x2 + 4y 2 = 36.
(a) (2 points) Find a parameterization of the ellipse.
Solution: We can write the ellipse as
x2 y 2
+
=1
4
9
and thus we have a parameterization by x = 2 cos t and y = 3 sin t.
(b) (4 points) At which points on the ellipse is the position vector perpendicular to the tangent vector?
Solution: We can write the parameterization as ~r(t) = h2 cos t, 3 sin ti.
Then the tangent vectors will be given by
r~0 (t) = h−2 sin t, 3 cos ti
So if the position vector is to be perpendicular to the tangent vector we
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will need ~r(t) · r~0 (t) = 0, which means
0 = h2 cos t, 3 sin ti · h−2 sin t, 3 cos ti
= h−4 cos t sin t, 9 cos t sin ti
So we must have cos t sin t = 0, which means that either sine or cosine of
3π
π
t is 0. This happens when t = 0, , π, or
2
2
5. (6 points) Prove that the parameterized curve
1
~r(t) = (2t − cos(t) + 3)~i + (sin2 (t) + 4t)~j + (− cos2 (t) + 2 cos(t) + 1)~k
2
lies in a plane.
Solution:
There are several ways to do this. We could use r(t) to find three points on
the plane: for example, P = r(0) = h2, 0, 1i, Q = r(/2) = h3 + π, 1 + 2π, 1/2i,
and R = ~r(π) = h4 + 2π, 4π, −1i . So two vectors in the plane are P~Q =
~ = h1 + π, −1 + 2π, −3/2i . Then P~Q × QR
~ = h−2 −
h1 + π, 1 + 2π, −1/2i and QR
2π, 1 + π, −2 − 2πi = −(1 + π)h2, −1, 2i is a vector perpendicular to both vectors
and therefore the plane. So using the point P as a point on the plane,
the equation of the plane is 2(x − 2) − y + 2(z − 1) = 0. Alternately, r0 (t) is
everywhere tangent to the curve and therefore also tangent to the plane.
This is r0 (t) = h2 + sin(t), 2 cos(t) sin(t) + 4, cos(t) sin(t) − sin(t)i . Then two vectors
parallel to the plane are r0 (0) = h2, 4, 0i and r0 (/2) = h3, 4, −1i , so that a normal
vector to the plane is r0 (0)r0 (/2) = h−4, 2, −4i , or ~n = h2, −1, 2i. Then, using
r(0) = h2, 0, 1i as a point on the plane, we get the same equation as before.
One form of the equation for the plane is thus
2x − y + 2z = 6
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