Math 241
January 13, 2012
Solutions to the white Quiz 1
There are four questions, two on the front and two on the back. Do all four.
1. Do both parts
a) Circle T for true and F for false.
i. If x is held constant, say x = 4, then the trace of the graph of y 2 = x + 2z 2 is an ellipse. T
F
If x = 4 the equation is y 2 − 2z 2 = 4 and the graph is a hyperbola.
ii. The domain of the function f (x, y) = ln(x + y − 1) consists of all points (x, y) that lie
above the line x + y = 1.
(x, y) must satisfy x + y − 1 > 0 or x + y > 1. Therefore (x, y) is above the line x + y = 1.
iii. The graph of the function f (x, y) =
centered at the point (0, 0, 0).
p
T
F
x2 + 16y 2 − 16 is the upper half of an ellipsoid T
F
The graph consists of points (x, y, z) such that z ≥ 0 and z 2 = x2 + 16y 2 − 16. That
is, x2 + 16y 2 = z 2 + 16. For a given value of z ≥ 0, the trace is a ellipse parallel to the
xy-plane. The size of the ellipse increases as z increases.
√
y
consists of all points (x, y) such that y ≥ 0 .
T
iv. The domain of f (x, y) = 2
x + y2
F
The domain consists of all (x, y) such that y ≥ 0 except the point (0, 0).
v. The function F (x, y) =
1 − xy
is continuous at all points (x, y).
1 + x2 + y 2
T
F
At all points (x, y) the denominator is not 0.
b) Fill in the blanks or circle the correct answer.
i. If z =
ex
∂z
=
, then
y + x2
∂y
ii. If v = y cos(xy) , then dv =
−ex
(y + x2 )2
.
−y 2 sin(xy) dx + cos(xy) − xy sin(xy) dy
iii. The graph of the function f (x, y) = y 2 + 1 is a
paraboloid
sphere
.
cylinder . (circle one)
iv. If z = x2 − xy + 3y 2 and (x, y) changes from (3, −1) to (2.96, −0.95), then dz =
−0.73
.
dz = (2x − y) dx + (−x + 6y) dy = (6 + 1) · (−0.04) + (−3 − 6) · (0.05) = −0.28 − 0.45 = −0.73
2.
6x3 y
does not exist. Explain why not in the space below.
(x,y)→(0,0) 2x4 + y 4
lim
The expression is 0 at (x, 0) and at (0, y), so its limiting value is 0 as (x, y) approaches (0, 0) along either of
the two coordinate axes.
6x4
6
If y = x, then the expression is
=
= 2, and its limiting value is 2 as (x, y) approaches (0, 0)
4
4
2x + x
2+1
along the line y = x. Since these values are different for different paths, the two dimensional limit does not
exist.
3. A portion of the graph of the function f (x, y) = | x y | is
shown on the right.
In the space below sketch several level curves for f . Label
each one with the value attained by f at points on the
curve.
Given k ≥ 0 the level curve is |xy| = k or xy = ±k implying that y = ±k/x. The level curves for k = 0 are
the two axes. The level curves for k = 1 and 2 are also displayed. Since they are z values, they are labelled as
such.
4. Find the equation of the tangent plane to the surface z =
the space below.
Let f (x, y) =
√
√
xy at the point (1, 1, 1). Show all of your work in
y
x
xy = (xy)1/2 . Since fx (x, y) = 12 (xy)−1/2 · y = √
and fy (x, y) = 21 (xy)−1/2 · x = √ ,
2 xy
2 xy
f (1, 1) = 1
fx (1, 1) =
1
,
2
and fy (1, 1) =
1
.
2
Therefore, the tangent plane is the graph of the function
L(x, y) = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1)
1
1
= 1 + (x − 1) + (y − 1)
2
2
1
1 1
1
=1+ x− + y−
2
2 2
2
1
1
= x+ y.
2
2
The tangent plane equation can also be expressed in the form z = 12 (x + y).