MATH 221.504
Exam 1 Solutions
October 8, 2015
1.
Let L be the line that is perpendicular to the plane given by 3(x − 2) − 5(y − 1) +
7(z + 1) = 0 and that passes through the point (2, 1, −1). What is the vector function
that defines L?
(A) r(t) = h−3 + 2t, 5 + t, −7 − ti
(B) r(t) = h3 − 2t, −5 − t, 7 + ti
(C) r(t) = h−2 + 3t, −1 − 5t, 1 + 7ti
(D) r(t) = h2 + 3t, 1 − 5t, −1 + 7ti
(E) None of the above
Solution: D From the equation of the plane, we see that n = h3, −5, 7i is a normal
vector to the plane. The line parallel to n that passes through (2, 1, −1) is given by
the answer in (D).
2.
Let f (x, y) = ln(x2 y + 3y 2 + 2). What is fyx ?
(A) −
(B)
x4 + 6x2 y + 18y 2 − 12
(x2 y + 3y 2 + 2)2
4x − 6xy 2
(x2 y + 3y 2 + 2)2
2x2 y 2 − 6y 3 − 4y
x2 y + 3y 2 + 2)2
2xy
(D) 2
x y + 3y 2 + 2
x2 + 6y
(E) 2
x y + 3y 2 + 2
(C) −
Solution: B We calculate fy and find
fy =
x2 + 6y
.
x2 y + 3y 2 + 2
We then differentiate with respect to x, and find
fyx = 4x − 6xy 2 (x2 y + 3y 2 + 2)2 .
1
3.
exy
Let g(x, y) = √
. What is the domain of g?
x 1−x−y
(A) {(x, y) | −∞ < x < ∞ and − ∞ < y < ∞}
(B) {(x, y) | x + y < 1 and x 6= 0 and y 6= 0}
(C) {(x, y) | x + y < 1 and x 6= 0}
(D) {(x, y) | x + y > 1 and x 6= 0}
(E) {(x, y) | x + y ≥ 1 and xy 6= 0}
Solution: C Since exy is defined for all values of x and y, the function g(x, y) is
undefined whenever the denominator is 0 or undefined. This occurs at the points in
(C).
4.
Let a = h2, −9, 1i and b = h6, 1, −3i. Consider the following statements:
I. a is parallel b.
II. a is perpendicular to b.
III. a is perpendicular to a × b.
Which of these statements are true?
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I and III only
Solution: D We check that a · b = 12 − 9 − 3 = 0, and so a and b are perpendicular.
Also for non-zero vectors a and b, we always have that a is perpendicular to a × b.
Therefore II and III are true, but I is not.
5.
Calculate the value of the limit
(A)
6x2 y 2 − 3y
.
(x,y)→(0,0) 4y − xy 2
lim
3
2
3
4
(C) −6
(B) −
(D) 3
(E) Does not exist
Solution: B We calculate
6x2 y 2 − 3y
y(6x2 y − 3)
6x2 y − 3
3
=
lim
=
lim
=− .
2
(x,y)→(0,0) 4y − xy
(x,y)→(0,0) y(4 − xy)
(x,y)→(0,0) 4 − xy
4
lim
2
6.
Suppose we are told that
lim
(x2 +y 2 ) = 10. Fill in the blanks below to complete
(x,y)→(−1,3)
the definition of the limit in this case.
For every ε > 0 there is some δ > 0 (depending on ε) so that whenever
,
it must be true that
.
Solution: By the definition of limit the first blank should be 0 < |(x, y) − (1, −3)| < δ
p
or 0 < (x + 1)2 + (y − 3)2 < δ . The second should be |x2 + y 2 − 10| < ε . Other
answers may have been possible, but these are the most straightforward ones.
7.
Let
f (x, y) = 4x2 y 3 − sin(πx + πy).
Find the equation of the tangent plane to the graph of z = f (x, y) above the point
(x, y) = (1, 1).
Solution: We need to find a point on the plane and a normal vector to the plane.
The point on the plane is
P = (1, 1, f (1, 1)) = (1, 1, 4).
We see that
fx = 8xy 3 − π cos(πx + πy),
fy = 12x2 y 2 − π cos(πx + πy),
and so a normal vector to the tangent plane is
n = hfx (1, 1), fy (1, 1), −1i = h8 − π, 12 − π, −1i.
Thus the equation of the tangent plane is
(8 − π)(x − 1) + (12 − π)(y − 1) − (z − 4) = 0 .
8.
Suppose
F (x, y, z) = x3 ey+z − y sin(x − z) = 0
∂z
defines z implicitly as a function of x and y. Find
.
∂x
Solution: We use the implicit differentiation formula for partial derivatives that
∂z
Fx
=− .
∂x
Fz
After differentiating we find that
∂z
3x2 ey+z − y cos(x − z)
= − 3 y+z
.
∂x
xe
+ y cos(x − z)
3
9.
Let
r(t) = ht, t2 , 23 t3 i,
−∞ < t < ∞,
define a parametric curve C in R3 . Let P be the point on C obtained when t = 1.
(a) Find the unit tangent vector T to C at P .
(b) Find the curvature κ of C at P .
(c) Your friend tells you that the principal unit normal vector to C at P is
N = h √12 , 0, − √12 i. Explain why your friend is incorrect.
Solution: For (a), we use the fact that at t = 1, we have
T=
r0 (1)
.
|r0 (1)|
0
2
0
0
So
√ we calculate r (t) = h1, 2t, 2t i, and so r (1) = h1, 2, 2i. It follows that |r (1)| =
1 + 4 + 4 = 3. Thus
T = 13 h1, 2, 2i = h 31 , 23 , 23 i .
For (b) it is easiest to use the formula for the curvature that
κ=
|r0 (1) × r00 (1)|
.
|r0 (1)|3
We already know r0 (1), so we calculate r00 (t) = h0, 2, 4ti and find r00 (1) = h0, 2, 4i.
Thus
r0 (1) × r00 (1) = h1, 2, 2i × h0, 2, 4i
i j k
= 1 2 2 0 2 4 = (8 − 4)i − 4j + 2k = h4, −4, 2i.
From this |r0 (1) × r00 (1)| =
|r0 (1)| = 3,
√
16 + 16 + 4 =
κ=
√
36 = 6, and so since we already know
6
2
= .
3
3
9
For (c), we recall that among other things N is supposed to be perpendicular to T,
and thus it must be that N · T = 0. However, if we calculate
1
1
1 2 2
1
2
1
N · T = h √ , 0, − √ i · h , , i = √ − √ = − √ 6= 0.
3 3 3
2
2
3 2 3 2
3 2
Thus this N cannot be the princal unite normal vector, since it is not perpendicular
to T.
4
10.
Let
r(t) = hsin(2t), 2 sin2 (t), 2 cos(t)i,
−∞ < t < ∞,
define a parametric curve E in R3 .
(a) Show that all of the points of E sit on a sphere centered at the origin in R3 and
find the equation of the sphere. (Perhaps useful: sin(2t) = 2 sin(t) cos(t), cos(2t) =
1 − 2 sin2 (t).)
π
(b) Find the point where the tangent line to E at t = intersects the xy-plane.
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Solution: To show (a), we first recall that the equation of a sphere centered at the
origin is x2 + y 2 + z 2 = r2 , where r is the radius. If we substitute in the coordinate
functions of r(t), we find
x(t)2 + y(t)2 + z(t)2 = sin2 (2t) + 4 sin4 (t) + 4 cos2 (t)
= 4 sin2 (t) cos2 (t) + 4 sin4 (t) + 4 cos2 (t)
= 4 sin2 (t)(cos2 (t) + sin2 (t)) + 4 cos2 (t)
= 4 sin2 (t) · 1 + 4 cos2 (t)
= 4.
Therefore the equation of the sphere on which all of the points of C lie is
x2 + y 2 + z 2 = 4 .
√
For (b), we note that r( π4 ) = h1, 1, 2i, and so we first need to find the equation of
√
the tangent line at the point P = (1, 1, 2). We know that r0 ( π4 ) is a tangent vector
at P , and
r0 (t) = h2 cos(2t), 4 sin(t) cos(t), −2 sin(t)i,
√
so r0 ( π4 ) = h0, 2, − 2i. From this we see that the tangent line at P is defined by
x = 1,
y = 1 + 2t,
√
2−
√
2t
Now the question is asking when this line intersects the xy-plane. This occurs when
z = 0, and so when t = 1. When t = 1, we then see that the point of intersection is
(1, 3, 0) .
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