Math 2263
Practice Midterm 1
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Show all work for each problem so that partial credit may be given. Answers without supporting work will receive no credit. Clearly indicate
your answers.
Make sure that you have 6 problems on your test. If any exam material
is missing, or if any questions are unclear, raise your hand or bring your
exam to the proctor.
Good luck!
1
1. The height of a hill is given by the function f (x, y) = x2 − y 2 − 2x +
2y + 2.
(a) Consider the graph z = f (x, y). What is the name of this quadric
surface?
Solution: We can rewrite the equation for the graph as z − 2 =
(x − 1)2 − (y − 1)2 , so this is a hyperbolic paraboloid.
(b) A raindrop falls at the point (1, 0) and rolls downhill in the direction of steepest descent. Give a unit vector u describing this direction.
Solution: The direction of steepest descent is the direction of −∇f (1, 0) =
h0, −2i. The unit vector in this direction is h0, −1i.
2
2. A function f (x, y) is harmonic if it satisfies the Laplace equation
∂ 2f
∂ 2f
+
= 0.
∂x2
∂y 2
Show that f (x, y) = x3 − 3xy 2 is harmonic.
Solution: To show that this function is harmonic we start by taking
the second order partial derivatives of f (x, y).
∂ 2f
= 6x
∂x2
∂ 2f
= −6x
∂y 2
So for this specific f we know
∂ 2f
∂ 2f
+
= 6x − 6x = 0.
∂x2
∂y 2
This shows that the function is harmonic.
3
3. Consider the surface x2 + y 2 + 2x − 4y − z = −7.
(a) Find parametric equations for the normal line to this surface at
the point (x, y) = (−2, 2).
Solution: The point on the surface is (−2, 2, 3). The gradient of the
function F (x, y, z) = x2 + y 2 + 2x − 4y − z will give us a direction
vector for this normal line.
∇F (−2, 2, 3) = h−2, 0, −1i
r(t) = h−2, 2, 3i + th−2, 0, −1i
In parametric form, this is
x = −2 − 2t,
y = 2,
z = 3 − t.
(b) Find an expression for the tangent plane at the pont (−2, 2). Write
the plane equation in the form x + Ay + Bz + C = 0.
Solution:
n · (r − r0 ) = 0
∇F (−2, 2, 3) · hx + 2, y − 2, z − 3i = 0
x + z/2 + 1/2 = 0
4
4. Let g(x, y) = xexy−2 .
(a) Find the tangent plane for g at the point (1, 2). Write the plane in
the form z = Ax + By + C.
Solution:
g(1, 2) = 1
gx (1, 2) = xyexy−2 + exy−2 |(1,2) = 3
gy (1, 2 = x2 exy−2 |(1,2) = 1
z = z0 + fx (x − x0 ) + fy (y − y0 )
z = 1 + 3(x − 1) + (y − 2)
z = 3x + y − 4
(b) Using a linear approximation, estimate the value of f at the point
(1.2, 1.9). (A calculator tells us that this value is approximately 1.587756.)
Solution:
f (x, y) ≈ 3x + y − 4
f (1.2, 1.9) ≈ 3(1.2) + 1.9 − 4 = 1.5
5
5. Suppose we have a function p = f (q, r, s), where q(x, y) = xy , r(x, y) =
exy , and s(x, y) = x − y. Suppose also that
∂f
∂f
∂f
(4, e, −3/2) = 2e,
(4, e, −3/2) = 4, and
(4, e, −3/2) = e.
∂q
∂r
∂s
(a) Find Df (x, y), the derivative matrix of f viewed as a function of
x and y, evaluated at the point (x, y) = (1/2, 2).
Solution:
Df (4, 3, −3/2) =
(b) Find
∂p
∂x
Solution:
when (x, y) = (1/2, 2).
∂f
∂x
= −7e
6
−7e 5e
6. (a) Suppose we know that the points (2, 3), (−1, 2), and (3, 1) are critical points of a function g(x, y). What can we conclude about the
partial derivatives of g?
Solution: gx = 0 and gy = 0 at each of these points
(b) Suppose we know that the second derivative matrix of g(x, y)
satisfies
4 2
−1 4
2
2
D g(2, 3) =
, D g(−1, 2) =
,
2 9
4 15
4 6
2
and D g(3, 1) =
.
6 9
For each critical point from part (a), determine whether the Second
Derivatives Test applies to that point. If it does apply, classify the
critical point as a local maximum, a local minimum, or a saddle
point.
Solution:
(2, 3): local minimum
(−1, 2): saddle point
(3, 1): Second Derivatives Test does not apply
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