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Math 2263 Name (Print):
Spring 2009 Student ID:
Midterm 1 Section Number:
February 19, 2009 Teaching Assistant:
Time Limit: 50 minutes Signature:
This exam contains 8 pages (including this cover page) and 7 problems. Check to see if any pages
are missing. Enter all requested information on the top of this page, and put your initials on the
top of every page, in case the pages become separated. Calculators may be used. Please turn off
cell phones.
Do not give numerical approximations to quantities such as sin 5, ฯ, or
- However, you should
simplify cos
ฯ
2
= 0, e
0 = 1, and so on.
The following rules apply:
- Show your work, in a reasonably neat and coherent way, in the space provided. All
answers must be justified by valid mathematical reasoning. To receive full credit on
a problem, you must show enough work so that your solution can be followed by someone
without a calculator.
- Mysterious or unsupported answers will not receive full credit. Your work should
be mathematically correct and carefully and legibly written.
- A correct answer, unsupported by calculations, explanation, or algebraic work
will receive no credit; an incorrect answer supported by substantially correct calculations
and explanations might still receive partial credit.
1 15 pts
2 15 pts
3 15 pts
4 15 pts
5 10 pts
6 15 pts
7 15 pts
TOTAL 100 pts
- (15 points) Find an equation for the plane passing through the two points ใx, y, zใ = ใ 4 , โ 2 , 2 ใ
and ใ 1 , 0 , โ 2 ใ so that the vector
i +
j +
k is tangent to the plane.
- (15 points) The lines given parametrically by
(x, y, z) = (7 + 2t, โ 1 โ t, โ 2 t), โโ < t < โ
and
(x, y, z) = (4 โ s, โ1 + 2s, 2 + 2s), โโ < s < โ
intersect at the point ใx, y, zใ = ใ 3 , 1 , 4 ใ. Find an equation for the plane which contains both
lines.
- (15 points) For the function f (x, y) = e
2 x+y sin y, find the second partial derivatives
f xx
, f xy
and f yy
- (15 points) The point ใx, y, zใ = ใ 1 , โ 1 , 2 ใ lies on the surface S:
x
2
2 โ 2 xy โ y
2 = 6.
Find the equation of the tangent plane to the surface S at ใ 1 , โ 1 , 2 ใ. Write it in the form
ax + by + cz = d.
- (15 points) (a) Find the gradient of the function f (x, y, z) = e
x ln(xy + z
2 ) at the point
ใx, y, zใ = ใโ 1 , 1 , 2 ใ.
(b) Find the directional derivative of f at the point ใx, y, zใ = ใโ 1 , 1 , 2 ใ in the direction of
the unit vector
~u =
i โ 2
j +
k
Math 2263 Spring 2009 Midterm 1, WITH SOLUTIONS- Page 2 of 2 February 19, 2009
- (10 points) Suppose z = f (x, y) is a function with first partial derivatives f x
(1, 2) = 3 and
fy(1, 2) = 5. If x and y are both functions of t: x = g(t) = โ 1 โ 2 t and y = h(t) = 6 + 4t, find
the derivative at t = โ1 :
dz
dt
d
dt
f
g(t), h(t)
SOLUTION: The chain rule says that
dz
dt
= fx
g(t), h(t)
dx
dt
g(t), h(t)
dy
dt
Compute that when t = โ 1 , x(โ1) = 1 and y(โ1) = 2, so in the chain rule, both fx and fy
are evaluated at (1, 2). We get:
dz
dt
- (15 points) The point ใx, y, zใ = ใ 1 , โ 1 , 2 ใ lies on the surface S:
x
2
2 โ 2 xy โ y
2 = 6.
Find the equation of the tangent plane to the surface S at ใ 1 , โ 1 , 2 ใ. Write it in the form
ax + by + cz = d.
SOLUTION: Compute the gradient of g(x, y, z) = x
2
2 โ 2 xy โ y
2 :
โg = (2x โ 2 y)
i +
(โ 2 x โ 2 y)
j + 2z
k. Then
โg(1, โ 1 , 2) = 4
i + 4
k is a normal vector to the surface S given by
g(x, y, z) = 6 at ใ 1 , โ 1 , 2 ใ. The equation of the tangent plane to S at ใ 1 , โ 1 , 2 ใ is
โg ยท
ใx, y, zใ โ ใ 1 , โ 1 , 2 ใ
= 4(x โ 1) โ 0(y + 1) + 4(z โ 2) = 0, or x + z = 3.
- (15 points) (a)Find the gradient of the function f (x, y, z) = e
x ln(xy + z
2 ) at the point
ใx, y, zใ = ใโ 1 , 1 , 2 ใ.
(15 points) SOLUTION: f x
= e
x ln(xy + z
2 ) +
ye
x
xy+z
2
; f y
xe
x
xy+z
2
; and f z
2 ze
x
xy+z
2
. So the
partial derivatives of f at ใx, y, zใ = ใโ 1 , 1 , 2 ใ are fx =
ln 3
e
1
3 e
, fy = โ
1
3 e
and fz =
4
3 e
. The
gradient of f is
โf (โ 1 , 1 , 2) =
[
ln 3
e
3 e
]
i โ
3 e
j +
3 e
k.
(b) Find the directional derivative of the function f (x, y, z) = e
x ln(xy + z
2 ) at the point
ใx, y, zใ = ใโ 1 , 1 , 2 ใ in the direction of the unit vector
~u =
i โ 2
j +
k
SOLUTION: Since ~u is a unit vector, we know that D ~u
f (โ 1 , 1 , 2) = ~u ยท
โf (โ 1 , 1 , 2) =
1
3
i โ 2
j +
k
[
ln 3
e
1
3 e
]
i โ
1
3 e
j +
4
3 e
k
1
3
2[
ln 3
e
1
3 e
] โ 2
1
3 e
4
3 e
2
3 e
ln 3 +
4
9 e
