### Math 241, Midterm 1 A Solutions May 4, 2009 Part I. 1

```Math 241, Midterm 1 A Solutions
May 4, 2009
Part I.
1. Circle T for true and F for false.
2xy
= −1 .
+ y2
2xy
b)
lim
= 0.
(x,y)→(0,0) x2 + y 2
a)
T
lim
(x,y)→(1,−1) x2
F
T
F
c) If f (x, y) = ln(y), then ∇f (x, y) = 1/y .
p
d) The domain of the function f (x, y) = z − x2 − y 2 consists of all points (x, y, z)
that lie above the cone z = x2 + y 2 .
p
dx + y dy
e) The differential of u = 2x + y 2 is du = p
.
2x + y 2
T
F
f) The set of points { (x, y, z) | x2 + y 2 = 1 } is a circle.
T
F
g) The graph of the surface y 2 + z 2 = 1 + x2 is an ellipsoid.
T
F
h) The maximum rate√of change of the function f (x, y, z) = tan(x + 2y + 3z) at the
point (−5, 1, 1) is 3 3 .
T
F
T
F
T
F
i) The critical point (0,0) of the function f (x, y) = x sin(y) is a saddle point.
T
F
j) The linear approximation of the function f (x, y) = sin(x − 3y) at (6,2) is
T
F
T (x, y) = x − 3y .
2. Fill in the blanks. Simplify all expressions.
a) An equation for the surface obtained by rotating the curve y = x2 + 1 about the y-axis is
y − 1 = x2 + z 2
This an elliptical paraboloid centered at (0,1,0)
b) Write the equation x2 + y 2 + z 2 = 2z in spherical coordinates:
c) If f (x, y, z) = cos(4x+3y +2z), then fxyz (x, y, z) =
d) The linear approximation of the function f (x, y, z) =
ρ2 = 2ρ cos(φ)
or ρ = 2 cos(φ)
24 sin(4x + 3y + 2z)
p
−
x2 + y 2 + z 2 at the point (3, 2, 6) is
xz + sin(x + y + z)
xy + sin(x + y + z)
.
.
f) If g(s, t) = s2 et , then the directional derivative of g at the point (2,0) in the direction of v = i + j is
√
∇g(2, 0) · √12 h1, 1i = 4 2
.
2
2
2
g) The critical points
√ of f (x, y) = x + y + x y + 4
are (0,0) and (± 2, −1). Classify each one as a
local max, local min, or saddle point.
The point
√(0,0)
( √2, −1)
(− 2, −1)
.
.
T (x, y, z) = 7 + 73 (x − 3) + 72 (y − 2) + 76 (z − 6) = 3x/7 + 2y/7 + 6z/7
e) If xyz = cos(x + y + z), then ∂z/∂y =
.
is classified as a
local min
Part II.
1. Find the volume of the rectangular box with largest volume if it lies in the first octant (x ≥ 0, y ≥ 0, z ≥ 0)
with three of its faces in the coordinate planes, one vertex at (0,0,0), and the opposite vertex in the plane
x + 2y + 3z = 6.
Let V be the volume of the box and (x, y, z) be the point at its vertex that is opposite to (0, 0, 0. We wish to
maximize V = xyz when x > 0, y > 0, z > 0, and x + 2y + 3z = 6. Since x = 6 − 2y − 3z the problem can be
reduced to two dimensions by substituting for x.
So, we will maximize V = (6 − 2y − 3z)yz on the open set y > 0 and z > 0. The maximum of V will be attained
at one of its critical points. To find the critical points simplify to V = 6yz − 2y 2 z − 3yz 2 so
Vy = 6z − 4yz − 3z 2 = z(6 − 4y − 3z)
Vz = 6y − 2y 2 − 6yz = 2y(3 − y − 3z) .
Since y and z are not 0, the critical point equations Vy = 0 and Vz = 0 are equivalent to the following system
of linear equations.
4y + 3z = 6
y + 3z = 3
Subtracting the second equation from the first we see that 3y = 3 so y = 1. From this it follows that z = 2/3,
and x = 6 − 2y − 3z = 2. The maximum volume is 2 · 1 · (2/3) = 4/3.
2. Find the absolute maximum and minimum values of the function f (x, y) = xy 2 on the set
D = { (x, y) | x ≥ 0 , y ≥ 0 , x2 + y 2 ≤ 3 } .
The region D is shown on the right. Note that f
is non-negative on D and f (x, y) = 0 at all points
on the x and y axes. Therefore, the absolute
minimum of f on D is 0.
Since fx = y 2 and fy = 2xy, all of the points
on the x axis are critical points, and there are no
more. Consequently, the absolute maximum of f
must occur on the circular part of its boundary
where y 2 = 3 − x2 and
z = f (x, y) = x(3 − x2 ) .
2
Thus z = 3x −√
x3 and dz/dx = 3 − 3x2 = 3(1 −
√x ) implying that the maximum value of z is attained when
x = 1 and y = 2. The maximum value is f (1, 2) = 2.
3. Do both parts.
a) If ρ =
p
x2 + y 2 + z 2 , then show that
2
∂2ρ ∂2ρ ∂2ρ
+ 2+ 2 = .
∂x2
∂y
∂z
ρ
(Student solution)
Start with ρ2 = x2 + y 2 + z 2 and differentiate with respect to x to obtain, 2ρρx = 2x. Therefore, ρx =
x
.
ρ
Differentiate again with respect to x to obtain
ρxx =
ρ − xρx
ρ − x · (x/ρ)
1 x2
=
= − 3.
2
2
ρ
ρ
ρ ρ
A similar calculation will show that
ρyy =
1 y2
−
ρ ρ3
and
ρzz =
1 z2
−
.
ρ ρ3
Therefore,
3 x2 + y 2 + z 2
−
ρ
ρ3
2
3 ρ
3 1
2
= − 3 = − = .
ρ ρ
ρ ρ
ρ
ρxx + ρyy + ρzz =
b) If z = cos(xy) + y cos(x) , where x = u2 + v and y = u − v 2 , use the Chain Rule to find ∂z/∂u and ∂z/∂v.
According to the Chain Rule,
∂z
∂z ∂x ∂z ∂y
=
+
. Therefore,
∂u
∂x ∂u ∂y ∂u
∂z
= (−y sin(xy) − y sin(x)) · 2u + (−x sin(xy) + cos(x)) · 1 .
∂u
Similarly,
∂z
∂z ∂x ∂z ∂y
=
+
, so
∂v
∂x ∂v
∂y ∂v
∂z
= (−y sin(xy) − y sin(x)) · 1 + (−x sin(xy) + cos(x)) · (−2v) .
∂v
4. The surface on the right is the graph of the
function
f (x, y) = sin(πx) sin(πy) .
The point on the surface is
P = (0.2, 1.2, f (0.2, 1.2)) .
a) In the space below the surface make a
rough sketch of a contour map of this
function for 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 .
b) Based upon the surface, at the point P ,
(circle i, ii, iii, or iv)
i. fx (P ) > 0 and fy (P ) > 0.
ii. fx (P ) > 0 and fy (P ) < 0.
iii. fx (P ) < 0 and fy (P ) > 0.
iv. fx (P ) < 0 and fy (P ) < 0.
c) If a bug is at the point P , in what direction (from
the point (0.2, 1.2) in the xy-plane) should it move in
order to go up the surface at the fastest rate? Show
your work in the space below.
Draw the contour plot on these axes.
Let z = f (x, y). Then z increases the fastest when
the bug moves in the direction of the gradient of f at
the point (0.2, 1.2). Since
∇f (x, y) = hπ cos(πx) sin(πy), π sin(πx) cos(πy)i
the bug should move in the direction of the vector
∇f (0.2, 1.2) = hπ cos(0.2π) sin(1.2π), π sin(0.2π) cos(1.2π)i
≈ h−1.494, −1.494i .
d) What will be the rate that the bug moves upward if it walks from the point P in the direction you found
in part c)? Show your work below.
When the bug walks in the direction of the gradient the rate of change of z is
p
|∇f (0.2, 1.2)| ≈ 1.4942 + 1.4942 ≈ 2.113 .
```