Exam 1 Math 251, Fall 2015 Section 520
Multiple choice problems are worth 10 points each. Work-out problems are
worth 15 each.
Section 1: Multiple choice problems. The default rule is that if no multiplechoice options are given, then (a)=6, (b)=7, (c)=8, (d)=9, and (e)=10. Please
give reasons for your answers, even though it’s multiple choice.
1. Find the radius of the sphere with center (6, 3, 7) and tangent to the plane
2x + 3y + 6z = 0.
• A vector perpendicular (normal) to the plane is n = h2, 3, 6i. The
scalar projection of the vector a = h6, 3, 7i from 0 to the center of
the sphere onto n is |h6, 3, 7i · h2, 3, 6i|/kh2, 3, 6ik = 63/7 = 9. You
bubble in (d).
2. Consider the points P = (3, 0, 3), Q = (−2, 1, 5), and R = (5, 2, 7). Let
v be the unit vector orthogonal to the plane through P , Q, and R with
positive y-component. If a, b, c, and d are single-digit integers (positive
or negative) so that
ha, b, ci
v= √ ,
d
find a + b + c + d.
• Let pq be the vector from P to Q, so pq = h−5, 1, 3i. Let pr be the
vector from P to R, so pr = h2, 2, 4i. Now pq
√ × pr = h0, 24, −12i is
normal to the given plane and has length 12 5, so v = √15 h0, 2, −1i.
That means a = 0, b = 2, c = −1, and d = 5, so a + b + c + d = 6.
Bubble in (a).
√
3. Let L be the length of the curve 2ti + et j + e−t k between t = 0 and
t = 2. Find the integer nearest L.
√
• To get L, integrate kr′ (t)k dt from 0 to 2. Now r′ (t) = h 2, et , −e−t i
which has length
p
p
2 + e2t + e−2t = (et )2 + 2 + (e−t )2 = et + e−t .
The integral works out to e2 − e−2 . Now e is a bit larger than 2.7,
so the nearest integer is 7. Bubble in (b).
4. A curve has the parametric equations x = 3t, y = 4t3/2 , z = −t2 . Its
curvature κ at the point corresponding to t = 1 can be written as
√
a b
κ= d
c
where a, b, c, d are positive integers, each less than 20. Find a + b − c − d.
1
• We have r′ = h3, 6t1/2 , −2ti, so kr′ k = 7 at t = 1. We have r′′ =
h0, 3t−1/2 , −2i, which at t = 1 is h0, 3, −2i. Now
κ=
kr′ × r′′ k
kr′ k3
√
which here works out to a numerator of kh−6, 6, 9ik = 3 17, and a
√ 3
denominator of kh3, 6, −2ik3 = 49 = 73 . So a = 3, b = 17, c = 7,
and d = 3. Bubble in (e).
5. The surface 2x2 − 8y 2 + z 2 = 8 is an
• (a) an elliptic cone
• (b) a hyperboloid of one sheet
• (c) a hyperboloid of two sheets
• (d) a hyperbolic paraboloid
• an elliptic paraboloid
It’s a hyperboloid of one sheet. Fixing y to any value and looking at what
happens in that cross section gives 2x2 + z 2 = A where A is positive. An
ellipse. All those cross sections are ellipses and the ones in the middle
are smaller while the ones further out are bigger. Fixing z = 0, say, gives
2x2 − 8y 2 = 8, which is a hyperbola. Everything points to hyperboloid of
one sheet.
6. Which of these formulas could have been the one to generate the graph
shown:
?
• (a) he−t cos(10t), e−t sin(10t), e−t i
• (b) ht, 1/(1 + t2 ), t2 i
• (c) hcos t, sin t, sin 5ti
• (d) hcos 4t, t, sin 4ti
• (e) ht, t2 , e−t i
The first one. This is straight from both our class session at the blackboard, and from the eHomework. The internal logic of it is that as t grows,
the ‘radius’ of the circular motion governed by the cos 10t and sin 10t goes
2
to 0, while the ‘height’ goes also to zero and at the same pace. At any
point along this curve, x2 + y 2 = z 2 which means the curve lies on a cone.
There was no need to remember this formula and its answer because the
information present in the problem points to the answer.
7. The contour plot of a function f of the form f (x, y) = Ax2 +By 2 is shown.
Which are the most reasonable candidates for A and B?
?
• (a) A = 3, B = −1
• (b) A = 3, B = 1
• (c) A = 1, B = −3
• (d) A = 1, B = 9
• (e) A = 9, B = −1.
The elliptical contours are longer sideways than vertically, by a ratio of
about 3 to 1. Because we have ellipses, A and B will be positive. If we
want A ∗ 32 + B ∗ 02 to be on the same contour as A ∗ 02 + B ∗ 12 we must
have 9A = B. Taking A = 1 and B = 9 works. Bubble in (d).
8. Find, if it exists,
lim
(x,y)→(0,0)
4x sin(xy)
.
x2 + 2y 2
• (a) 0
• (b) 1
• (c) 2
• (d) 4
• (e) The limit does not exist.
The numerator has absolute
value less than 4x2 |y| because | sin u| ≤ |u| in
p
general. Now |y| ≤ x2 + y 2 , and of course x2 ≤ x2 + y 2 , so the whole
numerator has absolute value no greater than p
4(x2 + y 2 )3/2 . The entire
expression, then, has absolute value less than 5 x2 + y 2 . Since this goes
to 0, so does the original expression. The limit exists and it’s 0. Bubble
in (a).
3
Section 2: Write out your answers. Use words and sentences to explain your
logic. Here, you are wearing the hat of a student explaining your answer to
another fairly good student.
1. The polar coordinates r and θ of a point can be seen as functions of the
Cartesian coordinates x and y. Find ∂r/∂x, ∂r/∂y, ∂θ/∂x, and ∂θ/∂y,
evaluated at x = y = 1.
• One way to work this is head-on. Look at the picture relating polar
and Cartesian coordinates. A right triangle with vertices (0, 0), (x, 0),
and (x, y) has base of x, height of y, hypotenuse
of r, and angle of
p
radian measure θ at the origin. Sopr = x2 + y 2 and tan
pθ = y/x so
θ = arctan y/x. Now ∂r/∂x = x/ x2 + y 2 , ∂r/∂y = y/ x2 + y 2 ,
y ∂θ
1
=
·
− 2 .
∂x
1 + (y/x)2
x
You could leave this as it stood, but it does simplify to −y/(x2 + y 2 ).
Finally, ∂θ/∂y = x/(x2 + y 2 ).
• Another way to work it is by implicit differentiation. The pair of
equations linking Cartesian to polar in the other direction is x =
r cos θ, y = r sin θ. Differentiating both these with respect to x yields
1 = rx cos θ − θx (r sin θ),
0 = rx sin θ + θx (r cos θ).
Differentiating both with respect to y yields
0 = ry cos θ − θy (r cos θ),
1 = ry sin θ + θy cos θ.
Multiplying some equations by cos θ and others by sin θ allows to
cancel off the terms involving θx , highlighting rx , or the other way
around, highlighting θx and canceling terms involving rx . Likewise
with y partials. You get ∂r/∂x = cos θ, ∂r/∂y = sin θ, ∂θ/∂x =
− 1r sin θ, and ∂θ/∂y = 1r cos θ.
2. Consider this technical claim: “The normal vector at (6, 4, 8) to the surface
S with equation 2x2 − 8y 2 +√z 2 = 8 is perpendicular to both h2, 0, −3i and
h0, 1, 4i. When u = (3 ± 5)/2, every real number t yields a position
vector r(t) = h6, 4, 8i + tuh2, 0, −3i + th0, 1, 4i that lies on S, while if u is
anything else, only the r corresponding to t = 0 lies on S.”
(a) Which of the quadric surface types is S? ( Ellipsoid? Paraboloid?
etc.) It’s a hyperboloid of one sheet. Just as it was earlier in this
exam.
(b) Goofus has been asked to explain what the claim says (never mind
for the time being whether the claim is true or not, just, what is
it saying) in plain English, and has got off to a decent start but
4
then went wrong. He said “It means that the plane through (6, 4, 8)
and parallel to h2, 0, −3i and h0, 1, 4i intersects the surface S in an
ellipse”. The plane in question is the tangent plane to the surface.
It intersects along lines through (6, 4, 8) parallel to both choices
for
√
uh2, 0, −3i + h0, 1, 4i, the one √
corresponding to u = (3 + 5)/2 and
the one corresponding to (3 − 5)/2.
Edit Goofus’ explanation to correct the mistaken part of his attempted translation of the claim into less technical language.
(c) Is the technical claim true? Why or why not? If you can’t prove it
or give a counterexample (which would be best), for part credit, give
the best reason you can for whichever you think more likely, true
or false. Here’s the answer I’d give if short on time:
To see if the
√
claim is plausible,
I’d take t = 1 and u = (3 + 5)/2
√
√ and plug in
x = 6 + (3 + 5), y = 4 + 1, and z = 8 + (−3/2)(3 + 5) + 4 and see
if 2x2 − 8y 2 + z 2 = 8 came true. To prove, I’d do this with general t
and for both choices of u. My guess is it does, because such a thing
is possible as we’ve seen in class.
5