Exam 1 Math 251, Fall 2015 section 519
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. Consider the sphere S with center (1, 2, 3) and radius 5. Let D be the
intersection of S with the x-y plane and let (a, b) be the coordinates of
the center of D and r be its radius. Find a + b + r.
• The center O = (1, 2, 3) of the sphere has z-coordinate 3, so it is 3
units above the x-y plane. The intersection of this sphere with that
plane is a circle with center C = (1, 2, 0). If P is a point on that
circle, then the triangle P OC is a right triangle with the right angle
at C. The height is 3 and the hypotenuse is 5 so the other side, the
radius of D, is 4. That way, 32 + 42 = 52 , which is what has to
happen according to the Pythagorean theorem.
So a = 1, b = 2, and r = 4, and a + b + r = 7. The answer is 7, so
you mark bubble (b).
2. Let P = (p1 , p2 , p3 ) be the point at which the lines r1 (t) = h3, 3, 0i +
th3, −3, 2i and r2 (s) = h6, 0, 2i + sh−3, 3, 0i intersect. Find p1 + p2 + p3 .
• Finding intersections of lines in 3-space can be a chore, but here,
there’s a lucky break. The z-coordinate of the second line is always
2 because changing s has no effect in the third entry. That means
that the point of intersection has to have z = 2, which means that
t = 1 because that way, 2t + 0 = 2. Now the point on the first line
corresponding to t = 1 is (6, 0, 2). These coordinates add to 8, so the
answer is 8. You mark bubble (c).
3. A curve has the parametric equations x = 2et , y = 2t, z = e−t . Its
curvature κ at the point corresponding to t = 0 can be expressed as a/b
where a and b are positive integers with no common factor. Find b − a.
• The ‘velocity vector’ r′ has formula h2et , 2, −e−t i and the second
derivative r′′ (‘acceleration’) has formula h2et , 0, e−t i. At t = 0, these
two vectors are h2, 2, −1i and h2, 0, 1i, respectively. The formula for
curvature is
kr′ × r′′ k
.
κ=
kr′ k3
The cross product whose
√ length is the numerator works out to h2, −4, 4i
so the numerator is 36 = 6. The vector whose
√ cubed length is
the denominator is h2, 2, −1i, which has length 4 + 4 + 1 = 3. So
κ = 6/27 = 2/9, a = 2, and b = 9. Thus b − a = 7, and you bubble
in (b).
1
4. 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
• There is no way to have y = 0 here, because the x and z contributions
to the formula are nonnegative and the formula has to total to −8.
There are choices of (x, z) that work if we take y = 2, say, because
then we need 2x2 + z 2 = 24 and there’s a whole ellipse worth of
points. Likewise if we take y = −2, because y 2 is still 4.
This makes our surface a thing with two parts, and the only answer
consistent with that is ‘hyperboloid of two sheets’. You bubble in
(c).
5. 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
This comes straight from the set we studied in class. Its equation was the
one in (e). If you wanted to just guess, the two reasonable contenders were
(b) and (e). But taking t = 0 in (b) leads to a point (0, 1, 0) that appears
to not be on the curve shown. So you bubble in (e). (The others are not
reasonable because there’s no sine waves and no spiraling or circling going
on, and that’s commonly found in the graphs of things like a, c, or d.
6. The equation xyz + y 2 z 3 = 96 determines z as an implicit function of x
and y in some domain. (We cannot have y = 0!) Find ∂z/∂y evaluated
at x = 4, y = 3, (which forces z = 2).
2
•
•
•
•
•
(a) −7/15
(b) −2/3
(c) −3/4
(d) −1
(e) −3/13.
You differentiate both sides of the equation that defines z, that is, xyz +
y 2 z 3 = 96, with respect to y, keeping in mind that the right side is a
constant so its derivative is automatically zero, while the left side responds
both to changes in y, and to (they had better be offsetting) changes in z,
through its dependence on y. This gives
xz + xyzy + 2yz 3 + 3y 2 z 2 zy = 0.
Now plug in 4 for x, 3 for y, and 2 for z. That gives
8 + 12zy + 48 + 108zy = 0.
From this you can work out what zy = ∂z/∂y should be: 56 + 120zy = 0,
so zy = −56/120 = −7/15. The answer is −7/15. You bubble in (a).
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 contours appear to be hyperbolas (and they are). The basic example
of a saddle shape is z = x2 − y 2 has this kind of a contour map, and this
one is the same except instead of A = B = 1, we have different values of
A and B. The function responds more quickly to changes in x than to
changes in y, which means that A > B > 0. The 0-level contours, the
ones going through (0, 0) where f = 0, are lines of slope somewhere in the
vicinity of ±2, judging from the graph. The only choice consistent
√ with
these observations is A = 3, B = 1, where the slopes of y = ± 3x are
about ±1.7. You bubble in (b).
3
8. Find, if it exists,
lim
(x,y)→(0,0)
4 sin(xy)
.
x2 + 2y 2
• (a) 0
• (b) 1
• (c) 2
• (d) 4
• (e) The limit does not exist.
2
)
and the limit of this is 4/3 as y
Along the line y = x, we have 4 sin(y
3y 2
goes to zero, because for small inputs, the sine of a number is very close
to the input itself. Since 4/3 is not available as an answer, you can be
tactical and bubble in (e). But a solid reason is available with a little
more thought: along the x-axis, the limit is 0 because the expression is 0
all the way in. With different limits along different paths, you can bubble
in (e) with extra heavy lead.
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 ∂ 2 r/∂x2 evaluated at x = y = 1.
The one basic formula for polar coordinates is that x = r cos θ, y = r sin θ.
What is r, here? It’s thepdistance from (x, y)pto the origin. In Cartesian
r = x2 + y 2 . Now taking the
coordinates, that’s just x2 + y 2 . So p
partial with respect to x once gives x/ x2 + y 2 , and taking it a second
time gives
p
2
1 x2 + y 2 − √ x2 2
y2
x +y
=
p
p
3
3.
x2 + y 2
x2 + y 2
Another way to go at this is to do everything with implicit differentiation
starting from x = r cos θ, y = r sin θ. Differentiating the first equation
with respect to x gives 1 = rx cos θ − rθx sin θ, and differentiating the
second with respect to x gives 0 = rx sin θ + rθx cos θ. That’s a system
of two equations in two unknowns, rx and θx . You can cancel the terms
involving θx by multiplying the first equation by cos θ and the second by
sin θ and adding. That gives cos θ = rx = ∂r/∂x. Halfway there! Now to
differentiate cos θ with respect to x, keeping in mind that θ depends on x
through the same pair of formulas used already above. We’ll need ∂θ/∂x =
θx . We almost have it already because the formula 0 = rx sin θ + rθx cos θ
involves θx and stuff we already know. Substituting in cos θ for rx gives
4
0 = cos θ sin θ + rθx cos θ. Canceling the factor of cos θ and solving for θx
gives θx = − 1r sin θ. The derivative of cos is − sin, so our answer is
f rac∂ 2 r∂x2 =
1
sin2 θ.
r
p
2
2
These
p are consistent because if we substitute sin θ = y/ x + y and
r = x2 + y 2 in the new answer we get the old one.
2. Consider this technical claim: “The surface S with equation x2 + 21 y 2 +
√ √
√
√
1 2
3 z = 1 has the property that if u = h1/ 2, 0, 3/ 2i and v = h0, 2, 0i,
then u · v = 0, kuk = kvk, and for all t, (cos t)u + (sin t)v is the position
vector of a point on S.”
(a) Which type of quadric surface is S? (Ellipsoid? Paraboloid? etc.)
It’s an ellipsoid.
(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 on the wrong track, writing that
“The surface is a paraboloid, [wrong, that should say ellipsoid ] and
the plane through the points with position vectors u, v and h1, 0, 0i
[no, h1, 0, 0i is not in the plane of the circle and he should instead use
−u, or −v (both on the circle) or 0, the center of the circle] intersects
S in an ellipse. [He should have said circle, because plain language
doesn’t beat about the bush and cautiously say ‘ellipse’ when the
more accurate ‘circle’ is available.] Correct him. Two of his three
errors have been flagged for you.
(c) Is the technical claim true? Prove or disprove, or, failing that, for
part credit make a reasoned case for whichever you think more likely.
It’s true. It’s a bit neat that by aiming your cross section plane just
so, you can get a circular cross section out of this lumpy ellipsoid.
The question wouldn’t have anything neat about it if the claim were
false, so it figures to be true. The serious
√ answer√ to this requires
some
computation.
Plug
in
x
=
cos(t)/
2, y = 2 sin t, and z =
p
3/2 cos t. Square these and add.
You
get
2 cos2 t + 2 sin2 t, so the
√
distance to the origin is always 2. That confirms it’s a circle. As to
being on the ellipsoid, well, that had equation x2 + y 2 /2 + z 2 /3 = 1.
Plugging in again for x, y, z gives
1
1
1
1
1 3
x2 + y 2 + z 2 = cos2 t + · 2 sin2 t + · cos2 t
2
3
2
2
3 2
1
1
2
2
2
= cos t + sin t + cos t = 1.
2
2
5