Name:
MATH 208
Solutions
Exam # 1
This test is worth 100 points. Show all of your work to receive full and partial credit, making
sure that all work and answers are legible. Books, notes, formula sheets, etc. are not allowed.
(8 points)
1. To the right of each equation, write the letter of the corresponding graph.
1. z = x + 2y + 3
B
2. z = sin(y)
A
p
2
3. z = cos( x + y 2 )
2
4. z = x − y
(12 points)
2
C
D
2. Indicate, by writing TRUE or FALSE, whether each of the following statements is correct as written.
(a) There is exactly one vector perpendicular to a given vector ~v .
FALSE (In 2-space, there are two such vectors. In 3-space, there are infinitely many)
(b) The plane x + 2y − 5z has normal vector ~i + ~j − 5~k.
FALSE (The normal vectors to this plane are the scalar multiples of ~i + 2~j − 5~k)
(c) ~i · (~j × ~k) = (~i × ~j) · ~k.
TRUE (~i · (~j × ~k) = ~i · ~i = 1, and (~i × ~j) · ~k = ~k · ~k = 1)
(d) For any real number λ, (λ~v ) · w
~ = ~v · (λw).
~
TRUE (This is a property of the dot product)
(e) If fx (a, b) < 0, then the vales of f increase as we move in the positive x-direction from the point
(a, b).
FALSE (They decrease as we move in the positive x-direction from the point (a, b))
(f) If f (x, y) is a function such that fx (x, y) and fy (x, y) are both constant, then f is linear.
TRUE (If fx (x, y) and fy (x, y) are both constant, then f has the form: f (x, y) = mx + ny + d for
some constants m, n and d.)
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MATH 208
Solutions
(10 points)
3. Show the following limit does not exist.
3x2
(x,y)→(0,0) x2 + y 2
lim
3x2
3x2
Let f (x, y) = 2
.
Then
f
(x,
0)
=
=3
x + y2
x2
so
lim
f (x, 0) =
(x,0)→(0,0)
3x2
3x2
3
and f (x, 2x) = 2
= 2 = ,
2
x + (2x)
5x
5
so
lim
(x,2x)→(0,0)
lim
3=3
(x,0)→(0,0)
f (x, 2x) =
lim
(x,2x)→(0,0)
3
3
=
5
5
Thus f (x, y) → 3 as (x, y) → (0, 0) along the line y = 0,
and f (x, y) →
3
5
as (x, y) → (0, 0) along the line y = 2x. Since 3 6= 53 , lim(x,y)→(0,0)
3x2
does not
x2 + y 2
exist.
note: Any other two lines which yield different limits will also work.
(20 points)
4. Let ~v = 2~i + 8~j − 5~k and w
~ = 2~i + ~k.
(a) Prove that ~v and w
~ are not perpendicular.
~v · w
~ = 2(2) + 8(0) − 5(1) = 4 − 5 = −1 6= 0. Thus ~v and w
~ are not perpendicular.
(b) Find the
~
√ unit vector that
√points in√the same direction as w.
kwk
~ = 22 + 02 + 12 = 4 + 1 = 5, so
~
√2 ~i + √1 ~
~u = kw
k is the required vector.
wk
~ =
5
5
(c) Find the projection of ~v onto w,
~ (ie. ~vparallel ).
4
√
~vparallel = (~v · ~u)~u = ( 5 + 0 − √55 )~u = (− √15 )( √25~i +
√1 ~
k)
5
= − 52~i − 15 ~k
(d) Find ~vperpendicular .
~vperpendicular = ~v − ~vparallel = (2 − (− 52 ))~i + (8 − 0)~j + (−5 − (− 15 ))~k =
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12~
5 i
+ 8~j −
24 ~
5 k
MATH 208
Solutions
(26 points)
5. Consider the points: A = (2, 1, 0), B = (0, 1, 3), C = (1, 0, 1).
~ and AC
~ into components.
(a) Resolve the vectors AB
~ = (0 − 2)~i + (1 − 1)~j + (3 − 0)~k = −2~i + 3~k
AB
~
AC = (1 − 2)~i + (0 − 1)~j + (1 − 0)~k = −~i − ~j + ~k
~ × AC.
~
(b) Compute AB
~i
~j ~k ~ × AC
~ = −2 0 3 = (0 + 3)~i + (−3 + 2)~j + (2 − 0)~k
AB
−1 −1 1
= 3~i − ~j + 2~k
(c) Find the area of the triangle ABC. (Hint: Use the previous part.)
~
~
The area of the triangle is half of the area
p of the parallelogram
√ with sides√AB and AC. The area
2
2
2
~
~
of the√parallelogram is kAB × ACk = 3 + (−1) + 2 = 9 + 1 + 4 = 14, so the triangle has
14
area
.
2
~ and BC.
~
(d) Find a vector of length 1050 which is perpendicular to both AB
~ × AC
~ is perpendicular to both AB
~ and AC,
~ and has length
AB
√
14, so the following vector works:
1050 ~
1050
2100
~ = 3150
√ (AB × AC)
√ ~i − √ ~j + √ ~k)
14
14
14
14
(e) Find the equation of the plane containing A, B and C.
Using the normal vector 3~i − ~j + 2~k and the point (2, 1, 0) we have:
0 = 3(x − 2) + (−1)(y − 1) + 2(z − 0)
which simplifies to:
5 = 3x − y + 2z.
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MATH 208
Solutions
(24 points)
6. Let f (x, y) =
1
xy .
(a) Find ∇f (x, y).
∂f /∂x = y1 (−x2 ) = − x12 y and
∂f /∂y = x1 (−y 2 ) = − xy1 2 ,
so ∇f = − x12 y~i −
1 ~
xy 2 j.
(b) Compute f~u (1, 1) where ~u =
√1 ~i
2
−
√1 ~
j.
2
f~u (1, 1) = ∇f (1, 1) · ~u = (− 121(1) )~i − 121(1) )~j) · ~u
= (−~i − ~j) · ( √12~i − √12~j) = (−1) √12 + (−1)(− √12 ) = − √12 +
√1
2
=0
(c) Find the equation of the plane tangent to the graph of f (x, y) at the point (x, y) = (1, 1).
z
z
z
z
z
= f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
= f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1)
= 1 + (−1)(x − 1) + (−1)(y − 1)
=1−x+1−y+1
= −x − y + 3
(d) Use the tangent plane to approximate f (1.1, 0.9).
f (1.1, 0.9) ≈ 1 + (−1)(1.1 − 1) + (−1)(0.9 − 1) = 1 − (0.1) − (−0.1) = 1
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