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A practice midterm for Calculus III. It contains 5 problems that cover topics such as vector functions, acceleration and velocity vectors, and normal planes. The problems require students to show their work and partial credit is given for incomplete work. The document also includes a formula sheet. The practice midterm is useful for students preparing for an exam in Calculus III.
Typology: Exams
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Name:
space.
Question Points Score
Total: 50
(a) The curve parametrized by 〈sin(2t), cos(3t), 1 + t
3 〉 never intersects the XY plane.
Solution: False. It intersects the XY plane at t = −1.
True False
(b) If the acceleration vector is perpendicular to the velocity vector, the object must
be going in a circle or helix.
Solution: False. Any motion with constant speed will have this property.
True False
(c) The graph of the function f (x, y) = x
2
2 is a hemisphere.
Solution: False.
True False
(d) For a vector function
r (t), we have
d(
r (t) ·
r (t))
dt
d
r (t)
dt
d
r (t)
dt
Solution: False. The derivative of a (dot) product must be computed by the
product rule.
True False
(e) If T , N , and B represent the unit tangent, normal, and binormal vectors, then
Solution: True. T, N, B form a frame just like the i, j, k.
True False
2 = x
2
2 have curvature
2 at the point (0, a)?
Solution: We can choose the parameterization 〈t,
a
2
2 〉. This will only trace
the y > 0 part of the curve, but that is sufficient since (0, a) is on this part.
We now compute the curvature at t = 0.
r(t) =
t,
a
2
2
r
′ (t) =
t
√
a
2
2
r
′′ (t) =
a
2
2
t
2
(a
2
2 )
3 / 2
a
2
(a
2
2 )
3 / 2
Therefore,
r
′ (0) = 〈 1 , 0 〉 = i
r
′′ (0) = 〈 0 , 1 /a〉 = (1/a)j.
Using
κ =
|r
′ × r
′′ |
|r
′ |
3
we get
κ =
|i × (1/a)j|
|i|
3
a
For κ = 2, we must have a = 1/2.
F (t) = 〈 0 , 16 cos(2t), 16 sin(2t)〉.
At t = 0, the object is at 〈 0 , 0 , 0 〉 and is travelling with velocity 〈 3 , 0 , − 4 〉.
(a) (5 points) How much distance does it travel between t = 0 and t = 10?
Solution: Using
F = m
a , we get that the acceleration is
a (t) = 〈 0 , 8 cos(2t), 8 sin(2t)〉.
Integrating, we get the velocity
v (t) = 〈 0 , 4 sin(2t), −4 cos(2t)〉 +
c.
Since
v (0) = 〈 3 , 0 , − 4 〉, we get
c = 〈 3 , 0 , 0 〉. Hence
v (t) = 〈 3 , 4 sin(2t), −4 cos(2t)〉.
At this point, we can compute the position, but we don’t need to. The speed is
v (t)| =
2
2 = 5.
Hence in 10 seconds, the object travels 50 units.
(b) (5 points) Write an equation of the normal plane to its motion at t = π.
Solution: We must calculate the position. Integrating
v (t), we get
r (t) = 〈 3 t, −2 cos(2t), −2 sin(2t)〉 +
c.
Since
r (0) = 〈 0 , 0 , 0 〉, we have
c = 〈 0 , 2 , 0 〉. Hence
r (t) = 〈 3 t, 2 − 2 cos(2t), −2 sin(2t)〉.
At t = π, we have
r (π) = 〈 3 π, 0 , 0 〉
r
′ (π) =
v (π) = 〈 3 , 0 , − 4 〉.
The normal plane is the plane through 〈 3 π, 0 , 0 〉 and perperdicular to 〈 3 , 0 , − 4 〉.
The equation is
3(x − 3 π) − 4 z = 0
that is: 3x − 4 z = 9π.
d
dx
x
n = nx
n− 1
d
dx
sin x = cos x
d
dx
cos x = − sin x
d
dx
tan x = sec
2 x
d
dx
cot x = − csc
2 x
d
dx
sec x = sec x tan x
d
dx
csc x = − csc x cot x
d
dx
e
x = e
x
d
dx
ln |x| =
1
x
d
dx
arcsin x =
1 √ 1 −x^2
d
dx
arccos x =
− 1 √ 1 −x^2
d
dx
arctan x =
1
1+x^2
(1) sin
2 x + cos
2 x = 1
(2) tan
2 x + 1 = sec
2 x
(3) 1 + cot
2 x = csc
2 x
(4) sin(x ± y) = sin x cos y ± cos x sin y
(5) cos(x ± y) = cos x cos y ∓ sin x sin y
(6) sin
2 x =
1 −cos 2x
2
(7) cos
2 x =
1+cos 2x
2
For a parametric space curve given by r(t)
(1) Curvature κ =
|r
′ (t) × r
′′ (t)|
|r
′ (t)|
3
(2) Tangent component of acceleration aT = |r
′ (t)|
r
′ (t) · r
′′ (t)
|r
′ (t)|
(3) Normal component of acceleration aN = κ|r
′ (t)|
|r
′ (t) × r
′′ (t)|
|r
′ (t)|
1
