1. Find parametric AND symmetric equations for the line of intersection of the planes
2x + y − 3z = 12 and 3x − y − z = 6.
Solution. To solve this problem it is necessary to find a point of intersection, and the
direction vector for the line.
A point of intersection can be found by setting z to whatever you desire (say, z = 0)
and solving the resulting system of 2 equations and 2 unknowns. Setting z = 0 yields
2x + y = 12, 3x − y = 6 which has the solution (18/5, 24/5). Thus the point (18/5,
24/5, 0) is in the intersection of the planes and is on the line we are after.
The direction vector can be found either by finding another point (by setting z = 1
for example) or by taking the cross product of the planes’ normal vectors. The cross
product is orthogonal to both of the two original vectors. Recall that a plane consists
of all vectors orthogonal to the normal vector, so the cross product is guaranteed to be
inside BOTH planes. This results in < 2, 1, −3 > × < 3, −1, −1 >=< −4, −7, −5 >.
Thus this is the direction.
Parametric equations: x = 18/5 − 4t, y = 24/5 − 7t, z = −5t
Symmetric equations;
x−18/5
−4
=
y−24/5
−7
=
z
−5
Note: There are many different possible ways the answer can be written, corresponding to different points being chosen and different scalings of the direction vector. For
example, I could have chosen < 4, 7, 5 > instead as the direction, or (6, 3, 9) as the
point.
2. For the function F (x, y) = ln(x2 + y 2 ), find the equation of the tangent plane at the
input (1, 0). What is the direction of fastest increase from this point, as a unit vector?
Solution. To find the tangent plane, we need the point of tangency and the gradient
, 2y > so ∇F (1, 0) =< 2, 0 >.
at that point. F (1, 0) = ln(1) = 0, and ∇F =< x22x
+y 2 x2 +y 2
Thus the tangent plane’s equation is T (x, y) = 0 + 2(x − 1) + 0(y − 0) = 2x − 2.
The direction of fastest increase is the direction of the gradient, < 2, 0 >. This direction as a unit vector is < 1, 0 > .
3. For the vector-valued function r(t) =< 2t − 1, t2 , 12 − 2t2 >, find the velocity, acceleration, and curvature of the function when t = 3
Solution. The velocity function is r’(t) =< 2, 2t, −4t > which evaluates at t = 1 to
< 2, 6, −12 >.
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The acceleration function is r”(t) =< 0, 2, −4 >, and is actually this vector at all
values of t.
There are two ways to find the curvature. The easiest is to use a formula you can find
. Using this formula you can
near the bottom of page 599 in your book: κ = ||r’×r”||
||r’||3
calculate:
κ=
||<2,6,−12>×<0,2,−4>||
||<2,6,−12>||3
√
=
80
184( 3/2)
= 0.00358
The other method available is the one I presented in class. You need to calculate T(t),
0
(t)
the unit tangent vector, then κ = Tr’(t)
and you can calculate the curvature by plugging
in t = 1.
T(t) =
r’(t)
||r0 (t)||
=
√ 1
4+20t2
< 2, 2t, −4t >
Calculating the derivative of T(t) component-wise is a little tedious, but possible. The
result is
T’(t) =< −40t(4 + 20t2 )−3/2 , 2(4 + 20t2 )−1/2 − 40t2 (4 + 20t2 )−3/2 , −4(4 + 20t2 )−1/2 +
80t2 (4 + 20t2 )−3/2 >
At this point you √
can plug in t = 1 and calculate the magnitude of the vector, then
divide by r’(1) = 184 to get the same result, κ = 0.00358.
Note. Because of the length and difficulty of the curvature calculation, I gave points
on the problem mostly for the calculation of the velocity and acceleration, and whether
an attempt was made at finding the curvature. If an attempt was made using a correct
method, only 2 points out of 25 were removed for making some calculation error.
Note. The most common error in the curvature calculation was plugging in t = 1 too
early. You must leave both parts of T(t) general in terms of t because the derivative
depends on t. You are not allowed to just change some of the instances of t to constants
before taking a derivative with respect to t - that will change the result!
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4. For the equation − x4 + y25 − z9 = 1, graph the cross sections corresponding to x = 2
and y = 10. What type of quadric surface is this?
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Solution. For x = 2: The equation in y and z becomes y25 −
√
√
hyperbola with axes 5 2 and 3 2. The graph is shown below.
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For y = 10: The equation in x and z becomes − x4 −
√
√
axes 2 3 and 3 3. The graph is shown below.
z2
9
z2
9
= 2. This is a
= −3 This is an ellipse with
Overall, the surface described by the original equation is a hyperboloid of two sheets.
This is basically a 3-dimensional hyperbola where cross sections cutting through the
hyperbolas look like ellipses. The hyperbola is aligned with the y-axis. The surface
begins at y = ±5 and increasing y increases the size of the ellipse cross section.
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Note. Some people graphed just a single ”cross section” corresponding to both x = 2
and y = 10. This gives you just two points. This does not really make sense from the
perspective of ”cross section” because we only used that term when describing a 2-D
slice in a 3-D object, which is the result from plugging in a value for only 1 variable.
Also notice that problem asked for ”cross sections” (plural), indicating x = 2 and
y = 10 should have been separate cross sections.
Note. On the same kind of note, the term ”quadric surface” refers to a 3-dimensional
object - namely, a surface existing in 3 dimensions described by an equation in x, y, and
z with no terms above degree 2. When the problem asks you ”What type of quadric
surface is this”, the only thing it can be referring to is the original equation. The two
cross sections above are 2-dimensional curves within a plane, in particular what we
would call ”conic sections” (the 2-D equivalent of quadric surfaces).
If someone did not know the exact name Hyperbola of Two Sheets for the name of the
quadric surface, I still accepted the answer as long as they either described the surface
completely or drew a picture of it.
BONUS: Find the following limit, or prove it does not exist. Explain your answer
completely.
x2 y − z 3
(x,y,z)→(0,0,0) x2 + y 2 + z 2
lim
Solution. The limit exists and is 0. The easiest way to show this is to convert the
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expression to spherical coordinates:
ρ2 sin2 (φ) cos2 (θ)ρ sin(φ) sin(θ) − ρ3 cos3 (φ)
ρ→0
ρ2
lim
ρ(sin3 (φ) cos2 (θ) sin(θ) − cos3 (φ))
ρ→0
1
lim
This expression is now a number tending to 0 multiplied by a quantity independent of
ρ which is bounded between −2 and 2, so the limit is 0.
Note. I’d like to reiterate here that no matter how many different approaches you
make to (0, 0, 0) and get 0, they do not ”prove” that the limit is actually 0 unless you
use some kind of reasoning like above.
Note. I saw a lot of people simply plug in (0, 0, 0) and get 00 , then declare the limit does
not exist because of division by 0. This represents to me a complete misunderstanding
of the concept of a limit. Practically every limit we take in calculus, including those we
take to define the derivative and integral (the most important calculus concepts) are
limits of the form 00 . This is an indeterminate form and the limit could still turn out to
be anything! If you made this mistake on the bonus problem and do not understand
what I am trying to say here, you MUST come talk to me about this! It is unacceptable
to be halfway through Calculus 3 and not understand what a limit is.
Note. Some people attempted to use some form of L’Hopital’s rule to simplify the
expression. I mentioned in class, and I’ll say so again. There is NO equivalent for
L’Hopital’s rule in 2 or more variable limits! Just to be sure, I did a search online and
I did turn up one article just published two years ago. You can check it out here if
you’re interested: http://arxiv.org/pdf/1209.0363.pdf. The article gives an algorithm
for using L’Hopital’s rule in 2 variables, but ONLY if the zeroes of the denominator
occur along a smooth curve through the point the limit is taken at. If the zero is
isolated, like in the Bonus Problem (only (0, 0, 0) is a zero of the denominator), there
is still no known L’Hopital’s rule analog. For problems such as this the paper only
gives some suggestions along the same lines as what we did in class.
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