C Roettger, Fall 14
Math 265 Exam 2B - Solutions
Problem 1 Consider the function
f (x, y) =
x + 3y
x2 − y 2
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 P0 = (3, 2).
c) Approximate f (2.95, 2.02) using differentials at the point P0 .
Graph of f in blue, tangent plane red, intersection curves in green/yellow.
Solution. a) After we factor out 1/(x2 − y 2 )2 , we get
1
∇f =
((x2 − y 2 ) − 2x(x + 3y), 3(x2 − y 2 ) + 2y(x + 3y))
2
(x − y 2 )2
1
=
(−x2 − y 2 − 6xy, 3x2 + 3y 2 + 2xy).
2
2
2
(x − y )
1
(−49, 57). The value
b) Evaluate the gradient at (3, 2) to get ∇f (3, 2) = 25
of f (x, y) at P0 is 59 = 1.8. Then one possible equation is
z=
9 49
57
− (x − 3) + (y − 2).
5 25
25
(1)
This could also be expressed as
49x 57y
78
−
+z =
25
25
25
or, avoiding fractions after we multiply by 25, as
49x − 57y + 25z = 78.
c) We use the equation (1) and put in (x, y) = (2.95, 2.02). The result is
f (2.95, 2.02) ≈ 1.8 −
49 · (−0.05) 57 · 0.02
+
= 1.9436
25
25
The exact answer is 1.949 (accurate to three decimals).
Problem 2 A cylinder with equation x2 + y 2 = 25 intersects the plane given
by 3x + 2y − 9z = 0 in an ellipse. Find a vector equation for the tangent line
to that ellipse in the point P0 = (4, 3, 2).
Cylinder in blue, plane yellow, intersection curve red, tangent line green
Solution. The cylinder is described by F (x, y, z) = 25, with
F (x, y, z) = x2 + y 2 .
The plane is given by G(x, y, z) = 0, with
G(x, y, z) = 3x + 2y − 9z
So the gradients are
∇F = (2x, 2y, 0) ,
∇G = (3, 2, −9)
and at P0 , ∇F (P0 ) = (8, 6, 0). As direction vector for the desired tangent
line, we can use
v = (8, 6, 0) × (3, 2, −9) = (54, −72, 2)
and of course, it passes through the point P0 , so it is described by
r(t) = (4, 3, 2) + t(54, −72, 2).
Problem 3 A metal particle in position (x, y) in the plane is subject to an
electric potential U = x2 + 4xy + 2y 2 . The particle is always experiencing an
electric force in the direction of lowest electric potential – assume the electric
force is in the direction where the potential U decreases most rapidly.
a) Find a unit vector in the direction of the electric force at the point (2, 1).
b) Find all points (x, y) where the electric force has direction parallel to the
y-axis, and sketch them (please include labeled axes with units).
Solution. a) The gradient of U is
∇U = (2x + 4y, 4x + 4y).
Evaluate this at (2, 1) to get ∇U (2, 1) = (8, 12). Since the particle always
moves opposite to ∇U , the unit vector in its direction is
u=−
1
1
∇U (2, 1) = − √ (2, 3).
||∇U (2, 1)||
13
b) Since the movement is always parallel to the gradient, we need to find
points (x, y) where Ux = 0. So the condition is
2x + 4y = 0,
or after simplification: y = −x/2. This gives one straight line through the
origin.
Plot of −∇U in black, line of points for part b) in red. Arrows scaled to
same length.
Problem 4 Consider the surface S given by F (x, y, z) = 0 with
F (x, y, z) = (x2 + 2y 2 − 1)(z − 2)
a) Find ONE point on the surface S where the gradient of F is 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) = 2x(z − 2), 4y(z − 2), x2 + 2y 2 − 1 .
If z 6= 2 then x = y = 0, but then Fz 6= 0. So the only possibility is z = 2.
To ensure Fz = 0, we only need
x2 + 2y 2 − 1 = 0
This means (x, y, z) is a point on the surface S!. Geometrically, this is an
ellipse which is the intersection of the cylinder x2 + 2y 2 − 1 = 0 with the
plalne z = 2.
b) These are points (x, y, z) on S where Fx = Fy = 0, so
2x(z − 2) = 0
4y(z − 2) = 0
If z 6= 2, then this forces x = y = 0 and the point (x, y, z) is not on the
surface. If z = 2, then the conditions Fx = Fy = 0 are satisfied, and the
point (x, y, z) is on the surface. But we still need to require x2 + 2y 2 − 1 6= 0,
so we don’t have one of the points on the ellipse from part a), where the
gradient ∇F = 0 (because for these points, the tangent plane to S does not
exist).
Problem 5 Consider the function
f (x, y) = 48xy − 32x3 − 24y 2
on the square R with vertices (0, 0), (0, 1), (1, 0), and (1, 1). Find the maximum and minimum value of f (x, y) on R, and all points where they are
attained.
Solution. We compute the gradient
∇f (x, y) = (48y − 96x2 , 48x − 48y).
This is the zero vector exactly when x = y and
48(y − 2x2 ) = 0,
giving two critical points (0, 0) and (1/2, 1/2). We note f (1/2, 1/2) = 2.
Then we compute maxima and minima on the edges of the square. For the
edge on the x-axis, f (x, 0) = −32x3 which has a minimum of −32 at (1, 0)
and a maximum of 0 at (0, 0). For the edge on the y-axis, f (0, y) = −24y 2
which has a maximum of 0 at (0, 0) and a minimum of −24 at (0, 1). For the
edge with equation x = 1, we have
f (1, y) = 48y − 32 − 24y 2
which has derivative
df (1, y)
= 48 − 48y
dy
.
So this is increasing, the minimum is again at (0, 1) and the maximum at
(1, 1) with value −8. For the edge with equation y = 1, we have
h(x) = f (x, 1) = 48x − 32x3 − 24
which has derivative
h0 (x) = 48 − 96x2 ,
√
√
√
equal to zero for x = 1/ 2. At (1/ 2, 1), f (x, y) has the value 8 2 − 24.
We list all relevant points with the values of f (x, y).
√
P0 (0, 0) (1, 0) (1, 1) (0, 1) (1/
√ 2, 1) (1/2, 1/2)
f (P0 )
0
−32
−8
−24 8 2 − 24
2
From the table, f (x, y) has the maximum value of 2 at (1/2, 1/2) and the
minimum value of −32 at (1, 0).
Plot of f (x, y) in blue, over a square slightly larger than the unit square.
Red curves are the values of f (x, y) on top of the edge of the unit square.
You can see that the maximum really is in the center of the unit square.