C Roettger, Spring 09
Math 265 Practice Exam 2 - Solutions
Problem 1 Consider the function
f (x, y) = y 2 ex
2 −4y
.
a) Find the gradient ∇f (x, y).
b) Find an equation for the tangent plane to the surface given by z = f (x, y)
in (x, y) = (2, 3).
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Solution. a) ∇f = yex −4y h2xy, 2 − 4yi.
b) Evaluate the gradient at (2, 3) to get ∇f (2, 3) = 3e−8 (12, −10). Then one
possible equation is
z = 9e−8 + 3e−8 [12(x − 2) − 10(y − 3)].
Problem 2
Consider the ellipsoids with equations F (x, y, z) = 8 and G(x, y, z) = 8,
where
F (x, y, z) = x2 + 4y 2 + (z − 2)2
G(x, y, z) = (x + 2)2 + 4y 2 + z 2
The point P0 = (−2, 1, 2) lies on the intersection of the two ellipsoids, which
is an ellipse. Find a vector equation for the tangent line to this ellipse in the
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point P0 .
Solution. First, we find the gradients of F and G at P0 .
∇F (x, y, z)
∇F (P0 )
∇G(x, y, z)
∇G(P0 )
=
=
=
=
(2x, 8y, 2(z − 2))
(−4, 8, 0)
(2(x + 2), 8y, 2z)
(0, 8, 4)
Then, we find an equation for the tangent plane to F = 8 at P0 :
−4(x + 2) + 8(y − 1) + 0(z − 2) = 0
which simplifies to −4x + 8y = 16. We do the same for the ellipsoid with
equation G = 8 at P0 :
0(x + 2) + 8(y − 1) + 4(z − 2) = 0
which simplifies to 8y + 4z = 16. The desired tangent line is the intersection
of these two planes. So we can use as direction vector
v = (−4, 8, 0) × (0, 8, 4) = (32, 16, −32).
and of course, it passes through the point P0 , so it is described by
r(t) = (−2, 1, 2) + t(32, 16, −32).
Problem 3 A gas particle in the plane is experiencing a pressure P =
x4 −x2 y 2 +y 4 in position (x, y). The particle is always moving in the direction
of lowest pressure - assume the particle moves in the direction where the
pressure decreases most rapidly.
a) Find a unit vector in the direction of the particle’s movement when it is
at point (2, 1).
b) Find all points (x, y) where the particle would move parallel to the x-axis
and sketch them (please include labeled axes with units).
Solution. a) The gradient of P is
∇P = h4x3 − 2xy 2 , −2x2 y + 4y 3 i.
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Evaluate this at (2, 1) to get ∇P (2, 1) = h28, −4i. Since the particle always
moves opposite to ∇P , the unit vector in its direction is
u=−
1
1
∇P (2, 1) = √ h−7, 1i.
||∇P (2, 1)||
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b) Since the movement is always parallel to the gradient, we need to find
points (x, y) where Py = 0. So the condition is
−2x2 y + 4y 3 = 0,
or after simplification: y = 0 or x2 √
= 2y 2 . This gives three straight lines
through the origin, y = 0, and x = ± 2y.
Problem 4 The quantities x, y are known to satisfy the equation F (x, y) = 3
with
F (x, y) = 3x + cos(π(x + y)).
In a neighborhood of x = 1, y = −1/2, this determines y = y(x) as a function
of x (so y(1) = −1/2).
a) Using partial derivatives of F , find y 0 = dy/dx at (x, y) = (1, −1/2).
b) Find an approximation to y(1.02) using differentials dy and dx = 0.02.
Round to four digits after the decimal point.
c) Now let x, y again be independent variables. Find an approximation to
F (2.04, −0.49) using differentials dx and dy at the point p0 = (2, −0.5).
Round to four digits after the decimal point.
Solution. a) The gradient is
∇F = h3 − π sin(π(x + y)), −π sin(π(x + y))i.
Then
y0 = −
3 − π sin(π(x + y))
Fx
=
Fy
π sin(π(x + y))
and after substituting x = 1, y = −1/2 we get y 0 (1) = (3 − π)/π.
b) With x0 = 1,
1 3−π
y(1.02) ≈ y(1) + dy = − +
· 0.02 ≈ −0.5009
2
π
c) First, F (p0 ) = 6. With p = (2.04, −0.49),
F (p) ≈ F (p0 ) + ∇F (p0 ).(p − p0 ) = 6 + (3 + π)(0.04) + π(0.01) ≈ 6.2771
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Problem 5 Consider the surface S given by F (x, y, z) = 9 with
F (x, y, z) = x2 + 2xz + y 2 + 2yz + 3z 2
a) Show that for (x, y, z) on the surface S, the gradient of F is never the zero
vector.
b) Find all points (x, y, z) on S where the tangent plane is horizontal.
Solution. a) First find
∇F (x, y, z) = 2hx + z, y + z, x + y + 3zi.
If x + z = 0 and y + z = 0 then obviously x = y = −z. Finally, from 0 = Fz
we get
0 = x + y + 3z = z
which means x = y = z = 0. But this is not a point on the surface S! nowhere else can ∇F be zero, according to our analysis.
b) These are points (x, y, z) on S where Fx = Fy = 0, so
x+z = 0
y+z = 0
or x = y = −z as before. Together with the condition F (x, y, z) = 9 this gives
us z = ±3. The two points with a horizontal tangent plane are (−3, −3.3),
(3, 3, −3).
Problem 6 Consider the function f (x, y) = x2 y − 6y 2 − 3x2 .
a) Using the First Derivative Test, find all candidates for local extrema of f
in the plane.
b) Determine for each of these whether it is a local maximum, a local minimum, or neither, using the Second Partials Test.
Solution. a) The gradient ∇f = h2xy − 6x, x2 − 12yi vanishes for x = 0
and y = 0 or y = 3, x = ±6. The candidates for local extrema are (0, 0),
(6, 3) and (−6, 3).
b) First calculate D = −12(2y − 6) − 4x2 = 72 − 24y − 4x2 . Then plug in the
three candidates to get
D
loc. max/min
point
(0, 0)
72 loc. max since fxx = −6 < 0,
(6, 3) −36
neither since D < 0,
(−6, 3) −36
neither since D < 0,
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Problem 7 Consider the function
f (x, y) = x2 y − y + 2
on the triangle T with vertices (0, −6), (0, 6), and (3, 0). Find the maximum
and minimum value of f (x, y) on T , and all points where they are attained.
Solution. We compute the gradient
∇f (x, y) = (2xy, x2 − 1).
This is the zero vector exactly when x = ±1 and y = 0, giving one critical
point (1, 0) in the interior of the domain. We note f (1, 0) = 2. Then we
compute maxima and minima on the edges of the triangle. For the edge on
the y-axis, f (0, y) = −y + 2 which has a maximum of 8 at (0, −6) and a
minimum of −4 at (0, 6). For the edge with equation y = 6 − 2x, we have
h(x) = f (x, 6 − 2x) = x2 (6 − 2x) − (6 − 2x) + 2
which has derivative
h0 (x) = −6x2 + 12x + 2.
This derivative equals zero for
√
√
144 + 48
= 6 ± 4 3.
2
For the edge with equation y = 2x − 6, we have
x=
12 ±
h(x) = f (x, 2x − 6) = −x2 (6 − 2x) − (2x − 6) + 2
which has derivative
h0 (x) = 6x2 − 12x − 2.
This is just the negative of the previous derivative, so has the same zeros. All
these x-values are outside of [0, 3], so the corresponding points are outside the
triangle T and will be discarded. We do need to include the endpoint (3, 0)
of the last two line segments, since endpoints always need to be included. So
these are the candidates for maxima and minima with the values of f (x, y)
there.
P0 (1, 0) (0, −6) (0, 6) (3, 0)
f (P0 )
2
8
−4
2
From the table, f (x, y) has the maximum value of 8 at (0, −6) and the
minimum value of −4 at (0, 6).
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