FIRST PRACTICE MIDTERM
The following five questions are worth 20 points each, although many have
multiple parts. Show ALL your work in your answers to each question. You
have 50 minutes for the examination. Calculators are NOT permitted.
1. Let P, Q, and R be points in R3 with coordinates P = (0, 2, 1), Q =
(2, 0, 1), R = (0, 0, −1).
1a. Give a parametrization (in terms of auxiliary variables t and u) for the
plane containing P , Q, and R.
We have
f (t, u) = P + t(Q − P ) + u(R − P )
= (0, 2, 1) + t(2, −2, 0) + u(0, −2, −2)
= (2t, 2 − 2t − 2u, 1 − 2u).
1b. Give an equation of the form ax + by + cz = d for the plane containing
P , Q, and R.
[(x, y, z) − P ] · [(Q − P ) × (R − P )]
(x, y − 2, z − 1) · [(2, −2, 0) × (0, −2, −2)]
(x, y − 2, z − 1) · [(4, 4, −4)]
4x + 4(y − 2) − 4(z − 1)
x+y−z
=
=
=
=
=
0
0
0
0
1
2. Draw the level sets f (x, y) = c for the given values of c, in the given
domains.
2a. f (x, y) = |x| + |y|, −2 ≤ x ≤ 2, −2 ≤ y ≤ 2, c = −1, 0, 1, 2, 3, 4.
The level set for c = −1 is empty; the level set for c = 0 is the origin. The
level set for c = 1, 2, 3, 4 is a square with vertices at (0, c), (0, −c), (c, 0) and
(−c, 0).
2b. f (x, y) = xy, −2 ≤ x ≤ 2, −2 ≤ y ≤ 2, c = −2, −1, 0, 1, 2.
The level set for c = 0 is the union of the x and y axes. The other level sets
are the graphs of y = xc for c = −2, −1, 1, 2.
3. Define vectors ~u, ~v in R3 by ~u = (−2, 1, 0), ~v = (0, 2, −1). Let θ be the
angle between the two vectors.
3a. Compute the lengths of ~u and ~v .
1
2
FIRST PRACTICE MIDTERM
The length of ~u =
p
(−2)2 + 12 + 02 =
√
5. The length of ~v is also
√
5.
3b. Compute cos θ.
We have cos θ =
~
u·~v
5
= 25 .
3c. Give a unit vector orthogonal to both ~u and ~v .
The cross product ~u × ~v will be orthogonal to ~u and ~v . We have ~u × ~v =
(−1, −2, −4). Normalized, this is √121 (−1, −2, −4).
4. Let f (x, y) = x3 + x2 y + xy 2 + y 3 .
4a. Find
∂f
∂x
and
∂f
∂y .
∂f
= 3x2 + 2xy + y 2
∂x
∂f
= 3y 2 + 2xy + x2
∂y
.
4b. Find the directional derivative of f at P = (1, 1) in the direction of
~v = (1, −1).
This is equal to
∂f
∂f
(P ),
(P )) · ~v
∂x
∂y
which is
(6, 6) · (1, −1) = 6 + −6 = 0.
4c. Give the linear approximation to f near the point (1, −1).
∂f
∂f
(1, −1)(x − 1) +
(1, −1)(y + 1)
∂x
∂y
= 0 + 2(x − 1) + 2(y + 1)
= 2x + 2y.
L(x, y) = f (1, −1) +
5. Let f : R2 → R3 be defined by f (x, y) = (x2 , xy, y 2 ), and let g : R3 → R2
be defined by g(t, u, v) = (t + 2u + v, t − 2u + v).
5a. Compute the derivative matrices Df and Dg.


2x 0
Df =  y x 
0 2y
1 2 1
Dg =
1 −2 1
5b. Compute g ◦ f : R2 → R2 .
Substituting t = x2 , u = xy, v = y 2 into the definition of g, we get
(g ◦ f )(x, y) = (x2 + 2xy + y 2 , x2 − 2xy + y 2 ).
FIRST PRACTICE MIDTERM
3
5c. Compute D(g ◦ f ) directly, using your answer to 5b.
Taking partial derivatives, we find
2x + 2y 2x + 2y
D(g ◦ f ) =
.
2x − 2y 2y − 2x
5d. Compute D(g ◦ f ) by the chain rule and your answer to 5a.
D(g◦f )(x, y) = Dg(f (x, y))Df =


2x 0
1 2 1 
2x + 2y 2x + 2y

y x =
.
1 −2 1
2x − 2y 2y − 2x
0 2y