Math 213 – Calculus III Practice Exam 2 Solutions Fall 2002
1. Let
2
g(x, y, z) = e−(x+y) + z 2 (x + y).
(a) What is the instantaneous rate of change of g at the point (2, −2, 1) in the direction of
the origin?
We want the directional derivative of g at (2, −2, 1) in the direction of the origin. A vector in
this direction is −2~i + 2~j − ~k, and a unit vector in this direction is ~u = √19 (−2~i + 2~j − ~k) =
³ ´
2~ 2~ 1~
− 3 i + 3 j − 3 k . The gradient of g is
³ 2
´ ³ 2
´
grad g(x, y, z) = −2(x + y)e−(x+y) + z 2 ~i + −2(x + y)e−(x+y) + z 2 ~j + (2z(x + y)) ~k,
and in particular
grad g(2, −2, 1) = ~i + ~j.
Then the instantaneous rate of change of g in the direction ~u at the point (2, −2, 1) is
³ ´ µ 2 2~ 1 ~
¶
~ ~ ~
g~u (2, −2, 1) = grad g(2, −2, 1) · ~u = i + j · − i + j − k = 0.
3 3 3
(b) Suppose that a piece of fruit is sitting on a table in a room, and at each point (x, y, z)
in the space within the room, g(x, y, z) gives the strength of the odor of the fruit. Fur-
thermore, suppose that a certain bug always flies in the direction in which the fruit odor
increases fastest. Suppose also that the bug always flies with a speed of 2 feet/second.
What is the velocity vector of the bug when it is at the position (2, −2, 1)?
Since the bug flies in the direction in which the fruit odor increases fastest, it flies in the
direction of grad g. It always has a speed of 2, so the velocity vector at (2, −2, 1) is
grad g(2, −2, 1) 2
2 = √ (~i + ~j).
kgrad g(2, −2, 1)k 2
2. The path of a particle in space is given by the functions x(t) = 2t, y(t) = cos(t), and
z(t) = sin(t). Suppose the temperature in this space is given by a function H(x, y, z).
(a) Find dH
dt , the rate of change of the temperature at the particle’s position. (Since the
actual function H(x, y, z) is not given, your answer will be in terms of derivatives of H.)
dH ∂H dx ∂H dy ∂H dz
dt = ∂x dt + ∂y dt + ∂z dt = 2 ∂H ∂H ∂H
∂x − sin t ∂y + cos t ∂z
∂H ∂H ∂H dH
(b) Suppose we know that at all points, ∂x > 0, ∂y < 0 and ∂z > 0. At t = 0, is dt
positive, zero, or negative?
dH
At t = 0, dt = 2 ∂H
∂x +
∂H
∂z > 0.
1
, 3. Let
f (x, y) = x3 − xy + cos(π(x + y)).
(a) Find a vector normal to the level curve f (x, y) = 1 at the point where x = 1, y = 1.
The gradient of f is normal to the level curve at each point. We find
grad f (x, y) = (3x2 − y − π sin(π(x + y)))~i + (−x − π sin(π(x + y)))~j, and
grad f (1, 1) = 2~i − ~j.
(b) Find the equation of the line tangent to the level curve f (x, y) = 1 at the point where
x = 1, y = 1.
The line is
2(x − 1) − (y − 1) = 0, or 2x − y = 1.
(c) Find a vector normal to the graph z = f (x, y) at the point x = 1, y = 1.
The graph is the level surface g(x, y, z) = 0 of the function g(x, y, z) = f (x, y)−z. The gradi-
ent of g is normal to the level surface at each point. We have grad g(x, y, z) = grad f (x, y)−~k.
Now f (1, 1) = 1, so a vector normal to the graph at (1, 1, 1) is
grad g(1, 1, 1) = grad f (1, 1) − ~k = 2~i − ~j − ~k.
(d) Find the equation of the plane tangent to the graph z = f (x, y) at the point x = 1,
y = 1.
The plane is 2(x − 1) − (y − 1) − (z − 1) = 0, or 2x − y − z = 0.
4. Let
f (x, y) = (x − y)3 + 2xy + x2 − y.
(a) Find the linear approximation L(x, y) near the point (1, 2).
First get the numbers: f (1, 2) = −1 + 4 + 1 − 2 = 2,
fx (x, y) = 3(x − y)2 + 2y + 2x, fx (1, 2) = 3 + 4 + 2 = 9,
fy (x, y) = −3(x − y)2 + 2x − 1, fy (1, 2) = −3 + 2 − 1 = −2.
Then L(x, y) = f (1, 2) + fx (1, 2)(x − 1) + fy (1, 2)(y − 2) = 2 + 9(x − 1) − 2(y − 2).
(b) Find the quadratic approximation Q(x, y) near the point (1, 2).
We need some more numbers:
fxx (x, y) = 6(x − y) + 2, fxx (1, 2) = −6 + 2 = −4,
fxy (x, y) = −6(x − y) + 2, fxy (1, 2) = 6 + 2 = 8,
fyy (x, y) = 6(x − y), fxy (1, 2) = −6.
Then
fxx (1, 2) fyy (1, 2)
Q(x, y) = L(x, y) + (x − 1)2 + fxy (1, 2)(x − 1)(y − 2) + (y − 2)2
2 2
= 2 + 9(x − 1) − 2(y − 2) − 2(x − 1)2 + 8(x − 1)(y − 2) − 3(y − 2)2 .
2
1. Let
2
g(x, y, z) = e−(x+y) + z 2 (x + y).
(a) What is the instantaneous rate of change of g at the point (2, −2, 1) in the direction of
the origin?
We want the directional derivative of g at (2, −2, 1) in the direction of the origin. A vector in
this direction is −2~i + 2~j − ~k, and a unit vector in this direction is ~u = √19 (−2~i + 2~j − ~k) =
³ ´
2~ 2~ 1~
− 3 i + 3 j − 3 k . The gradient of g is
³ 2
´ ³ 2
´
grad g(x, y, z) = −2(x + y)e−(x+y) + z 2 ~i + −2(x + y)e−(x+y) + z 2 ~j + (2z(x + y)) ~k,
and in particular
grad g(2, −2, 1) = ~i + ~j.
Then the instantaneous rate of change of g in the direction ~u at the point (2, −2, 1) is
³ ´ µ 2 2~ 1 ~
¶
~ ~ ~
g~u (2, −2, 1) = grad g(2, −2, 1) · ~u = i + j · − i + j − k = 0.
3 3 3
(b) Suppose that a piece of fruit is sitting on a table in a room, and at each point (x, y, z)
in the space within the room, g(x, y, z) gives the strength of the odor of the fruit. Fur-
thermore, suppose that a certain bug always flies in the direction in which the fruit odor
increases fastest. Suppose also that the bug always flies with a speed of 2 feet/second.
What is the velocity vector of the bug when it is at the position (2, −2, 1)?
Since the bug flies in the direction in which the fruit odor increases fastest, it flies in the
direction of grad g. It always has a speed of 2, so the velocity vector at (2, −2, 1) is
grad g(2, −2, 1) 2
2 = √ (~i + ~j).
kgrad g(2, −2, 1)k 2
2. The path of a particle in space is given by the functions x(t) = 2t, y(t) = cos(t), and
z(t) = sin(t). Suppose the temperature in this space is given by a function H(x, y, z).
(a) Find dH
dt , the rate of change of the temperature at the particle’s position. (Since the
actual function H(x, y, z) is not given, your answer will be in terms of derivatives of H.)
dH ∂H dx ∂H dy ∂H dz
dt = ∂x dt + ∂y dt + ∂z dt = 2 ∂H ∂H ∂H
∂x − sin t ∂y + cos t ∂z
∂H ∂H ∂H dH
(b) Suppose we know that at all points, ∂x > 0, ∂y < 0 and ∂z > 0. At t = 0, is dt
positive, zero, or negative?
dH
At t = 0, dt = 2 ∂H
∂x +
∂H
∂z > 0.
1
, 3. Let
f (x, y) = x3 − xy + cos(π(x + y)).
(a) Find a vector normal to the level curve f (x, y) = 1 at the point where x = 1, y = 1.
The gradient of f is normal to the level curve at each point. We find
grad f (x, y) = (3x2 − y − π sin(π(x + y)))~i + (−x − π sin(π(x + y)))~j, and
grad f (1, 1) = 2~i − ~j.
(b) Find the equation of the line tangent to the level curve f (x, y) = 1 at the point where
x = 1, y = 1.
The line is
2(x − 1) − (y − 1) = 0, or 2x − y = 1.
(c) Find a vector normal to the graph z = f (x, y) at the point x = 1, y = 1.
The graph is the level surface g(x, y, z) = 0 of the function g(x, y, z) = f (x, y)−z. The gradi-
ent of g is normal to the level surface at each point. We have grad g(x, y, z) = grad f (x, y)−~k.
Now f (1, 1) = 1, so a vector normal to the graph at (1, 1, 1) is
grad g(1, 1, 1) = grad f (1, 1) − ~k = 2~i − ~j − ~k.
(d) Find the equation of the plane tangent to the graph z = f (x, y) at the point x = 1,
y = 1.
The plane is 2(x − 1) − (y − 1) − (z − 1) = 0, or 2x − y − z = 0.
4. Let
f (x, y) = (x − y)3 + 2xy + x2 − y.
(a) Find the linear approximation L(x, y) near the point (1, 2).
First get the numbers: f (1, 2) = −1 + 4 + 1 − 2 = 2,
fx (x, y) = 3(x − y)2 + 2y + 2x, fx (1, 2) = 3 + 4 + 2 = 9,
fy (x, y) = −3(x − y)2 + 2x − 1, fy (1, 2) = −3 + 2 − 1 = −2.
Then L(x, y) = f (1, 2) + fx (1, 2)(x − 1) + fy (1, 2)(y − 2) = 2 + 9(x − 1) − 2(y − 2).
(b) Find the quadratic approximation Q(x, y) near the point (1, 2).
We need some more numbers:
fxx (x, y) = 6(x − y) + 2, fxx (1, 2) = −6 + 2 = −4,
fxy (x, y) = −6(x − y) + 2, fxy (1, 2) = 6 + 2 = 8,
fyy (x, y) = 6(x − y), fxy (1, 2) = −6.
Then
fxx (1, 2) fyy (1, 2)
Q(x, y) = L(x, y) + (x − 1)2 + fxy (1, 2)(x − 1)(y − 2) + (y − 2)2
2 2
= 2 + 9(x − 1) − 2(y − 2) − 2(x − 1)2 + 8(x − 1)(y − 2) − 3(y − 2)2 .
2
