MATH 251.504
Examination 1 Solutions
February 28, 2011
1.
Let L be the line given by parametric equations x = 3 + 2t, y = −2 + 5t, and z = −1 − 4t.
What is the equation of the plane that contains the point (3, −1, 5) and is perpendicular to
L?
(A) 2(x − 3) + 5(y + 1) − 4(z − 5) = 0
(B) 2(x − 3) + 5(y + 2) − 4(z + 4) = 0
(C) 3(x − 3) − 2(y + 1) − (z − 5) = 0
(D) 3(x − 3) − 2(y + 2) − (z + 4) = 0
(E) 3(x − 3) − (y + 2) + 5(z + 4) = 0
Solution: A The vector n = h2, 5, −4i is parallel to L, and so will make a normal vector
to the plane. Since we want (3, −1, 5) to be on the plane, we see that (A) is the correct
answer.
2.
Let f (x, y) = 2 cos(xy 2 − 3y). What is fxy ?
(A) −2y 4 cos(xy 2 − 3y)
(B) −4y(2xy − 3) cos(xy 2 − 3y)
(C) −4x sin(xy 2 − 3y) − 2(2xy − 3)2 cos(xy 2 − 3y)
(D) −4y sin(xy 2 − 3y) − 2y 2 (2xy − 3) cos(xy 2 − 3y)
(E) −4x(2xy − 3) sin(xy 2 − 3y) − 2y 2 cos(xy 2 − 3y)
Solution: D We first calculate fx = −2 sin(xy 2 − 3y) · y 2 , and then find that
fxy = −4y sin(xy 2 − 3y) − 2 cos(xy 2 − 3y) · (2xy − 3)y 2 ,
which simplifies to (D).
3.
Consider the curve in 3-space defined by the vector function r(t) = h3 sin(t), 4 cos(t), 4ti.
What is the unit tangent vector T to this curve when t = 0?
(A) h0, −4, 4i
(B) h3, −4, 4i
3
4
(C)
, 0,
5
5
3
4
4
(D) √ , − √ , √
41
41 41
4
4
(E) 0, − √ , √
32 32
Solution: C We first calculate r0 (0): since r0 (t) = h3 cos(t), −4 sin(t), 4i, we see that
0 (t)
r0 (0) = h3, 0, 4i. Since T(t) = |rr0 (t)|
, we see that
T(0) =
h3, 0, 4i
1
= h3, 0, 4i,
|h3, 0, 4i|
5
which is (C).
1
4.
A curve in 3-space is defined by a differentiable vector function r(t). Let T, N, and B be
the unit tangent vector, the unit normal vector, and the binormal vector at some point P0
on the curve, with P0 corresponding to t = t0 . Consider the following statements:
I. N is perpendicular to B.
II. T is parallel to r0 (t0 ).
III. B is perpendicular to the osculating plane at P0 .
Which of these statements is true?
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II and III
5.
Solution: E All answers are true. All three vectors T, N, and B are pair-wise perpendicular, and so I. is true. The unit tangent vector T is a scalar multiple of r0 (t0 ), so II. is
true. Finally, from the definition of the osculating plane at P0 , it has B as a normal vector,
and so III. is true.
−−→
−−→
Let P = (3, 1, 1), Q = (1, 1, −1), and R = (1, 5, −1). Let a = QP and b = QR.
(a) Find a × b.
(b) Is the triangle 4P QR a right triangle? Why or why not?
Solution: (a) We first calculate that a = h2, 0, 2i and b = h0, 4, 0i. Therefore,
i j k 0 2
2 2 2 0
i − a × b = 2 0 2 = 0 0 j + 0 4 k = −8i + 8k .
4 0
0 4 0 (b) Yes the triangle 4P QR is a right triangle. Since
a · b = 2 · 0 + 0 · 4 + 2 · 0 = 0,
it follows that a ⊥ b, and so 4P QR has a right angle at Q.
6.
Suppose f (x, y) is a differentiable function such that
fx = 3yexy ,
Let
fy = 3xexy .
3u
g(t, u) = f tu2 ,
.
t
∂g
∂g
Use the multivariable Chain Rule to find
and
in terms of t and u.
∂t
∂u
Solution: Using the chain rule,
∂g
3u
2 3u
2
2 3u
= fx tu ,
· u + fx tu ,
· − 2
∂t
t
t
t
3u 3u3 2
3u
3
= 3
e · u + 3(tu2 )e3u − 2 .
t
t
2
It is not necessary to simplify this completely, but if you do, you find that
the second part of the problem,
∂g
3u
3u
3
· 2tu + fx tu2 ,
·
= fx tu2 ,
∂u
t
t
t
3
3u 3u3
3
e · 2tu + 3(tu2 )e3u
.
= 3
t
t
If you simplify this expression , you find that
7.
∂g
∂u
∂g
∂t
= 0 . For
3
= 27u2 e3u .
Let
D
E
8
r(t) = 8t − 6, 6t2 − 5, −
+2 .
t+1
(a) The graph of z = f (x, y) is what kind of quadric surface?
f (x, y) = 5 − x2 − 3y 2
and
(b) Let P0 = (2, 1, −2). Find the equation of the tangent plane to the graph of z = f (x, y)
at P0 .
(c) The curve defined by r(t) also contains P0 . Find a tangent vector to this curve at P0 .
(d) Is the tangent vector from (c) perpendicular to the tangent plane from (b)? Why or
why not?
Solution: (a) The graph is an elliptic paraboloid .
(b) We first find a normal vector n to the tangent plane to the surface z = 5 − x2 − 3y 2 at
P0 = (2, 1, −2). We know that we can take n = hfx (2, 1), fy (2, 1), −1i. Since
fx (x, y) = −2x,
fy (x, y) = −6y,
we see that n = h−4, −6, −1i. Therefore the equation of the tangent plane is
−4(x − 2) − 6(y − 1) − (z + 2) = 0 .
(c) We need to determine the value of t at which the curve defined by r(t) passes through
P0 . Since we would need to have 8t − 6 = 2 from the x-coordinates, we see that t = 1. (We
can also quickly verify that r(t) = h2, 1, −2i.) Therefore a tangent vector to the curve at
P0 will be r0 (1). We calculate that
8
r0 (t) = 8, 12t,
,
(t + 1)2
and so r0 (1) = h8, 12, 2i .
(d) Yes they are perpendicular. The vector r0 (1) is perpendicular to the tangent plane to
the surface z = f (x, y) at P0 if and only if r0 (1) is a normal vector to the plane. That is, if
and only if r0 (1) is parallel to n. We easily see that
r0 (1) = h8, 12, 2i = (−2) · h−4, −6, −1i = (−2) · n,
and so indeed r0 (1) k n.
3