C Roettger, Fall 14
Math 265 Exam 2A - Solutions
Problem 1 Consider the function
f (x, y) =
x + 2y
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 (2.98, 2.03) 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
((x2 − y 2 ) − 2x(x + 2y), 2(x2 − y 2 ) + 2y(x + 2y))
2
2
2
(x − y )
1
=
(−x2 − y 2 − 4xy, 2x2 + 2y 2 + 2xy).
2
(x − y 2 )2
∇f =
1
b) Evaluate the gradient at (3, 2) to get ∇f (3, 2) = 25
(−37, 38). The value
7
of f (x, y) at P0 is 5 = 1.4. Then one possible equation is
z=
7 37
38
− (x − 3) + (y − 2).
5 25
25
(1)
This could also be expressed as
37x 38y
14
−
+z =
25
25
5
or, avoiding fractions after we multiply by 25, as
37x − 38y + 25z = 70.
c) We use the equation (1) and put in (x, y) = (2.98, 2.03). The result is
f (2.98, 2.03) ≈ 1.4 −
37 · (−0.02) 38 · 0.03
+
= 1.4752
25
25
The exact answer is 1.4791 (accurate to three decimals).
Problem 2 A cylinder with equation x2 + y 2 = 25 intersects the plane given
by 2x + 3y − 9z = 0 in an ellipse. Find a vector equation for the tangent line
to that ellipse in the point P0 = (3, 4, 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) = 2x + 3y − 9z
So the gradients are
∇F = (2x, 2y, 0) ,
∇G = (2, 3, −9)
and at P0 , ∇F (P0 ) = (6, 8, 0). As direction vector for the desired tangent
line, we can use
v = (6, 8, 0) × (2, 3, −9) = (−72, 54, 2)
and of course, it passes through the point P0 , so it is described by
r(t) = (3, 4, 2) + t(−72, 54, 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) = 6x2 − 2x3 + 3y 2 + 6xy.
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 = (12x − 6x2 + 6y, 6y + 6x) vanishes for
y = −x and x = 0 or x = 1. The candidates for local extrema are (0, 0),
(1, −1).
b) First calculate fxx = 12−12x, fxy = 6, fyy = 6 and then D = 72(1−x)−36.
Then plug in the candidates to get
D
loc. max/min
point
(0, 0)
36 loc. min. since fxx = 36 > 0,
neither since D < 0.
(1, −1) −36
Contour plot of f , overlaid with plot of ∇f . The former shows a saddle
point at (1, −1) and a local extremum at (0, 0). Only by looking at the
arrows pointing away from (0, 0) can you tell that the extremum is actually
a local minimum. If water was raining down on the surface z = f (x, y), you
would get a little pond there.