NAME: Calculus 3 Exam 1-September 18, 2012 Instructions ...

NAME:
Calculus 3
Exam 1-September 18, 2012
Instructions.
• SHOW WORK! and write clearly. Show all work, answers obtained apparently by divine inspiration
may get no credit.
• Try to write in the form of full sentences. I may penalize you for the failure to use full sentences. A sentence
must contain a verb. For example “5” or “x 2 +y 2 are NOT sentences. “x 2 +y 2 = 5” is a sentence; the symbol
“=” would be read “is equal,” and “is” is a verb. Failure to use an equal symbol where it should be used will
probably result in the loss of points.
• Do not use an equal symbol to connect quantities that are not equal, except when setting up equations.
Incorrect use of an equal symbol will result in loss of points.
• Use extra paper as necessary.
The Test.
1. (5 points) Find the center and radius of the sphere of equation
x 2 + y 2 + z 2 − 8x + 2y + 6z + 1 = 0.
Solution.
Completing the squares one gets
The center is at (4, −1, −3), the radius is 5.
(x − 4) 2 + (y + 1) 2 + (z + 3) 2 = 25.
2. (5 points) If u and v are vectors as shown in the picture, find u · v and u × v. Is u × v directed into the
page or out of it?
u · v = |u| |v| cos(30 ◦ ) = 30 √ 3.
This answers the question about u·v. Concerning
the cross product,
|u × v| = |u| |v| sin(30 ◦ ) = 30.
But u × v is a vector, so we are not
done yet. Since u, v, u × v have to
form a right system we can say that
u × v is a vector of length 30, pointing perpendicularly
out from the page.

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3. Find parametric equations of the following lines.
(a) (5 points) The line through the points (1, 0, −1) and (2, −3, 1).
Solution. A direction vector for this line is given by a = 〈2 − 1, −3 − 0, 1 − (−1)〉 = 〈1, −3, 2〉. From
this we can get the equations at once.Using the first point as base,
x = 1 + t, y = −3t, z = −1 + 2t.
(b) (10 points) The line through (2, −1, 0) and perpendicular to the plane of equation 5x + 2y − 3z = 6.
At what point does this line intersect the plane?
Solution. If the line is perpendicular to a plane, then a vector normal to the plane directs it. The
obvious vector normal to the plane in question is 〈5, 2, −3〉. Parametric equations are
x = 2 + 5t, y = −1 + 2t, z = −3t.
The line intersects the plane for t such that 5(2 + 5t) + 2(−1 + 2t) − 3(−3t) = 6. Solving for t we get
t = −1/19. Using this value for t we get the point (33/19, −21/19, 3/19).
4. (10 points) Find the equation of the plane through the points (2, 1, 1), (−1, −1, 10), and (1, 3, −4).
Solution.
Taking the point of coordinates (2, 1, 1, ) as the base point, the plane is parallel to the vectors
v = 〈−3, −2, 9〉 and w = 〈−1, 2, −5〉.
Taking their cross product to get a vector normal to the plane:
i j k
n =
−3 −2 9
= −8i − 24j − 8k.
∣ −1 2 −5 ∣
An acceptable, though sort of ugly, answer is −8(x − 2) − 24(y − 1) − 8(z − 1) = 0. We could multiply by
1 to get rid of the negative numbers. Or we can expand and rearrange to get 8x + 24y + 8z = 48 (also an
acceptable answer); dividing by 8,
x + 3y + z = 6.
If I were a student in this course, after finding the equation of the plane, I would check that it
works; that the given points satisfy it. If one of them didn’t, I would check my computations.
Of course, that’s me.
5. (5 points) Show that the points A(1, 1, 1), B(2, 6, −1), C(4, 0, 1), D(6, 10, −3) are coplanar, and find a fifth
point in the same plane.
Solution. One way of seeing the points are coplanar is to take one of them as a base, construct the three
vectors starting at that point and ending at the other points, and verify that their triple product is 0. Taking
A as base, the vectors are
−→
AB = 〈1, 5, −2〉,
−→
AC = 〈3, −1, 0〉, −→ AD = 〈5, 9, −4〉.
The triple product works out to
1 5 −2
3 −1 0
= 1((−1)(−4) − 0) − 5(3(−4) − 0) − 2((3)(9) − (−1)(5)) = 4 + 60 − 64 = 0.
∣ 5 9 −4 ∣
The triple product being 0, the points are coplanar. There are several ways of finding a fifth point. Here are
two ways.
Method 1. Any combination of the vectors −→ −→ −→
−→ −→ −→
AB, AC, AD with scalars (aAB + bAC + cAD where a, b, c
are constants) will give a vector that starts at the point A and ends at a point of the plane. Adding
the components of A to it will give a point of the plane. And we don’t even need to use all three of
the vectors. For example, 2 −→ AB = 〈2, 10, −4〉 lies entirely in the plane if we start it at A; adding the
components of A we get (3, 11, −3) as a fifth point in the plane.

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Method 2. Find the equation of the plane; then use it to give arbitrary values to two of the variables and
solve for the third. To find the equation of the plane we take two of the vectors, for example −→ −→ AB, AC
and form their cross product to get a vector normal to the plane.
i j k
n =
1 5 −2
= −2i − 6j − 16k.
∣ 3 −1 0 ∣
From this the equation of the plane works out to −2(x−1)−6(y −1)−16(z −1) = 0 or x+3y +8z = 12.
Taking for example y = z = 0, we get x = 12. So a fifth point is (12, 0, 0) .
6. Consider the curve given parametrically by x = sin 2t, y = t, z = cos 2t.
(a) (10 points) Find the tangent, normal and binormal vectors (T(t), N(t), B(t)).
Solution. The position function for this curve is
r(t) = 〈sin 2t, t, cos 2t〉.
Thus r ′ (t) = 〈2 cos 2t, 1, −2 sin 2t〉, |r ′ (t)| = √ 4 cos 2 2t + 1 2 + 4 sin 2 2t = √ 5. Thus
Next, N(t) = (1/|T ′ (t)|)T ′ (t);
T(t) = 1
|r ′ (t)| r′ (t) = 1 √
5
〈2 cos 2t, 1, −2 sin 2t〉.
T ′ (t) = √ 1 〈−4 sin 2t, 0, −4 cos 2t〉, |T ′ (t)| = 1 √ 5
√16 sin 2 2t + 16 cos 2 2t = √ 4 .
5 5
Thus
Finally
B(t) = T(t) × N(t) =
∣
N(t) = 〈− sin 2t, 0, − cos 2t〉.
i j k
2√
5
cos 2t √5 1
− √ 2 5
sin 2t
− sin 2t 0 − cos 2t
∣ = − √ 1 cos 2ti + √ 2 j + √ 1 sin 2tk.
5 5 5
Thus (to keep the notation consistent)
B(t) = 〈− 1 √
5
cos 2t,
2
√
5
,
1
√
5
sin 2t〉.
(b) (5 points) Find an equation for the osculating plane at the point where t = π/4.
Solution. The osculating plane is normal to the binormal vector. At π/4 the binormal vector is
〈0, 2/ √ 5, 1/ √ 5〉. Since any vector parallel to the binormal will do, we can multiply by √ 5 and use
n = 〈0, 2, 1〉. The point on the curve corresponding to t = π/4 is (1, π/4, 0). Putting it all together, the
equation for the osculating plane works out to 2(y − π/4) + z = 0 or
2y + z = π 2 .
7. (10 points) A particle starts from the origin with velocity v 0 = i + 3j − 4k. Its acceleration as a function of
time is given by a(t) = −3t 2 i + 4tj + 4t 3 k. Determine its position function r(t).
Solution.
The velocity of the particle is given by
∫ ∫
v(t) = a(t) dt = (−3t 2 i + 4tj + 4t 3 k) dt = −t 3 i + 2t 2 j + t 4 k + c.
Setting t = 0, mv(0) = v 0 we see that c = v 0 = i + 3j − 4k so that
v(t) = (−t 3 + 1)i + (2t 2 + 3)j + (t 4 − 4)k.

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Now
∫
r(t) =
∫
v(t) dt =
[(−t 3 + 1)i + (2t 2 + 3)j + (t 4 − 4)k] dt = (− 1 4 t4 + t)i + ( 2 3 t3 + 3t)j + ( 1 5 t5 − 4t)k + c.
Setting t = 0 we should have r(0) = 0; this gives c = 0. The answer is
(− 1 4 t4 + t)i + ( 2 3 t3 + 3t)j + ( 1 5 t5 − 4t)k.
8. The position vector of a particle is given by
r(t) = 〈3t 2 , 3t 4 , 2t 6 〉.
Suppose t is measured in seconds and distance is measured in meters. The particle starts at time t = 0
(a) (5 points) Where is the particle at time t = 1 second?
Solution. At (3, 3, 2).
(b) (10 points) What distance has it travelled (measured along the trajectory!) at time t = 1?
Solution.
the distance will be given by
s(1) =
∫ 1
0
|r ′ (t)| dt =
∫ 1
0
√
36t2 + 144t 6 + 144t 1 0 dt.
Here is where the hint can be helpful. Or having done some of the textbook exercise, where this situation
happens several times. 36t 2 + 144t 6 + 144t 1 0 = (6t + 12t 5 ) 2 . So we have
The distance equal 5 meters.
s(1) =
∫ 1
0
∣ ∣∣∣
1
(6t + 12t 5 ) dt = 3t 2 + 2t 6 = 5.
(c) (10 points) At what time will it have travelled half the distance it travelled from time t = 0 to time
t = 1? In other words, in part b, you found a certain distance s. The question now is: For what value
of t is the distance s/2?
Solution. If t is that time, we should have
For t such that 3t 2 + 2t 6 = 2.5.
∫
5
t
∣ ∣∣∣
t
2 = s(t) = (6u + 12u 5 ) du = 3u 2 + 2u 6 = 3t 2 + 2t 6 .
0
0
0
Hint: If you are careful with the computations, don’t do anything foolish, any quantity that you have inside
a square root should be a perfect square. Look at it carefully.

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9. (10 points) At what point does the curve y = e x have maximum curvature. Justify that the curvature
is indeed maximum at that point. In case you forgot to add the formula for curvature of a curve given
by y = f(x) to your “cheat sheet,”, here it is:
κ(x) =
By the given formula, the curvature works out to
κ(x) =
|f ′′ (x)|
(1 + (f ′ (x)) 2 ) 3/2 .
e x
(1 + e 2x ) 3/2 .
Using Calculus techniques, we find the derivative and set it to 0. We get (after some simplifying)
κ ′ (x) = ex (1 − 2e 2x )
(1 + e 2x ) 5/2 .
There is only one zero, only one critical point, the value of x for which 2e 2x = 1; that is, x = (1/2) ln(1/2) =
− ln 2/2. Noticing that if x < (1/2) ln(1/2) we have κ ′ (x) > 0, and κ ′ (x) < 0 for x > (1/2) ln(1/2), proves
that we have a maximum. Answer: At x = − 1 ln 2.
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