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MATH 105: PRACTICE PROBLEMS AND SOLUTIONS FOR CHAPTER 12: SPRING 2011
INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
Question 1: These problems deal with open sets. Open sets were covered in Spring 2010 but not that much in Spring 2011 or 2013, so do not worry about these problems if you’re in 105 in Spring 2011 or 2013; these are included for general interest.
(1) Let S = {(x, y, z) : 3x2 + 4y2 + 5z2 < 6}. Is S open? Solution: Yes: This is an ellipsoid where we do not include the boundary as we
have strictly less than 6. Given any point in the interior, we can find a sufficiently small radius.
(2) Let S = {(x, y) : x2 − y2 = 1}. Is S open? Solution: No: The set is the two branches of a hyperbola. These are one-
dimensional curve, and if we draw a ball about any point on either branch, most of the points in the ball will not be on the branch.
(3) Let S = {(x1, . . . , xn) : x21 + · · · + x2n < 1}. Is S open? Solution: Yes: This set is open, and in fact is just the n-dimensional sphere.
(4) Let S = {(x, y, z) : x2 + y2 ≤ z}. Is S open? Solution: No: This is a paraboloid where we include the boundary. If we take
any point (x, y, z) such that x2 + y2 − z, then we see that any ball centered at such a point hits both points inside and outside our region.
(5) Let S = {(x, y) : xy = 1}. Is S open? Solution: No: The reason is essentially the same as the reasoning in part (2). We
have a two-dimensional object in the plane; if we draw a ball about any point on either branch of the hyperbola, we’ll find many points not on the curve,
(6) Let S = {(x, y) : x2 + y2 > 1}. Is S open? Solution: This set is open. It is all points more than 1 unit from the origin.
Date: March 14, 2013. 1
2 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
Question 2: Compute the following limits (if they exist), or prove they do not. Remember log x means the logarithm of x base e.
(1) limx→1(x 4 − 2x3 + 3x2 + 4x − 5).
Solution: As x → 1, the expression is just 1 − 2 + 3 + 4 − 5 = 1, where we used the rules for limits of suns, differences and constant multiples.
(2) limx→2 sin(3x 2 − 12).
Solution: As f(x) = sin(3x2 − 12) is continuous (the sine function is continuous), the limit is just f(2), which is sin(12 − 12) = 0.
(3) limx→2 sin(3x2−12)
x−2 .
Solution: As we have 0/0, we must resort to other methods than simply substi-
tuting. Using L’Hopital’s rule, we find the limit is just limx→2 6x(cos(3x2−12)
1 , which is
12. This is because the denominator is always 1, and as x → 2 the numerator tends to 6 · 2 · cos 0, and cos 0 = 1.
(4) limx→0 log x x
. Solution: It is natural to want to use L’Hopital’s rule. Taking the derivatives, we
would find it equals limx→0 1/x 1 , which is undefined. Unfortunately, we don’t have 0/0
or ∞/∞. As x → 0, log x → −∞. Thus as x → 0 through positive values, we have a very large negative number divided by a small positive number, which is an extremely large negative number. Thus the limit tends to −∞ (or does not exist).
(5) limx→0 x
log x .
Solution: This is the reciprocal of the previous problem, and hence tends to 0.
(6) lim(x,y)→(0,0)(4xy cos(xy) + x 2 − y3).
Solution: As each function is continuous, the limit is obtained by substituting (0, 0) for (x, y); we may do this as we don’t have 0/0 or infinity anywhere. We find the limit equals 0 cos 0 + 0 − 0, which is 0.
(7) lim(x,y)→(0,0) x2y2−1 xy−1
.
Solution: As the limit of the numerator is -1 and the limit of the denominator is -1, we may use the limit of a quotient is the quotient of the limits, and hence the answer is −1/ − 1 or 1.
(8) lim(x,y)→(1,1) x2y2−1 xy−1
.
Solution: As we have 0/0, we must be careful. We cannot use L’Hopital’s rule as that is for one variable problems, and this has two. The easiest way to attack it is Thoreau it. Explicitly, notice that the numerator factors as (xy − 1)(xy + 1), as it is a difference of two squares. We can thus cancel the factor xy − 1 in the numerator and the denominator, and our problem is the same as evaluating lim(x,y)→(1,1)(xy +1), which is just 2. Note we could also have attacked the previous problem this way as well.
(9) lim(x,y)→(0,0) x4−x2y2+y4
x2+y2+x4y4 .
PRACTICE PROBLEMS 3
Solution: We have 0/0, so we have to be careful. If we use polar coordinates, we replace x with r cos θ, y with r sin θ, and then (x, y) → (0, 0) becomes r → 0 and θ does whatever it wants. Note that each term in the numerator is a multiple of r4, while the denominator is r2 + r8 cos4 θ sin4 θ. Specifically, we have
lim r→0
r4(cos4 θ − cos2 θ sin2 θ + sin4 θ) r2(1 + r6 cos4 θ sin4 θ)
= lim r→0
r2(cos4 θ − cos2 θ sin2 θ + sin4 θ) 1 + r6 cos4 θ sin4 θ
= 0.
The reason the limit is zero is that we can now use the quotient rule – the limit of a quotient is the quotient of the limits, as the denominator tends to 1 as r → 0. What is nice is that using polar coordinates allows us to check all possible paths of (x, y) tending to (0, 0).
(10) lim(x,y)→(0,0) x 2y3 cos
(
1 x2+y2
)
.
Solution: Remember that when we take limits, the point (x, y) is never (0, 0); thus the cosine term is always well defined, as we are never evaluating it at 1/0. The simplest way to determine the answer is to use the squeeze theorem. Note that for all choices of input, the absolute value of cosine is at most 1; however, as (x, y) → (0, 0) we have x2y3 rapidly tending to 0. Thus we are taking the limit of a product, one term tending to zero and the other at most 1 in absolute value. Thus the product tends to 0. Note that we cannot use the limit of a product is the product of the limits, as both
limits do not exist (the cosine piece fluctuates between -1 and 1); however, we do not need the limit of each piece to exist, only the limit of the product.
(11) lim(x,y)→(0,0) x2y3 cos
(
1
x2+y2
)
x2+y2 .
Solution: This problem is very similar to the previous. The only difference is that we now divide by x2 +y2. The cosine piece is still at most 1 in absolute value; we now analyze the x2y3/(x2 + y2) term. Using polar coordinates, we find this piece is just r5 cos2 θ sin3 θ/r2, which is just r3 cos2 θ sin3 θ. As (x, y) → (0, 0), r → 0 and hence this term tends to 0 as well. Thus, arguing similarly as the previous problem, we see that this limit is 0 as well.
(12) lim(x,y,z)→(0,0,0) x3+y3+z3
x2+y2+z2 .
Solution: This problem requires spherical coordinates, which are discussed in Sec- tion 1.4 (page 69). Sometimes physicists and mathematicians have different conven- tions for spherical coordinates. Using the book’s convention, we have
x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.
As (x, y, z) → (0, 0, 0) we have ρ → 0 and θ, φ vary however they want. Thus our limit becomes
lim ρ→0
ρ3 sin3 φ cos3 θ + ρ3 sin3 φ sin3 θ + ρ3 cos3 φ
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ + ρ2 cos2 φ .
4 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
A little algebra shows the denominator is just ρ2, while the numerator is a multiple of ρ3. We thus factor out a ρ2 and find our limit equals
lim ρ→0
ρ (
sin3 φ cos3 θ + sin3 φ sin3 θ + cos3 φ )
.
As the trig piece is at most 3 in absolute value (each term is at most 1) and ρ → 0, the product tends to 0 and thus the limit is 0.
PRACTICE PROBLEMS 5
Question 3: Plot the level sets of value c for each function below (do enough values of c so you can recognize the result).
(1) f(x, y) = sin(x + y). Solution: The level sets are where x + y is constant. If we want to find the level
set of value c, we must find all (x, y) such that sin(x + y) = c, or equivalently all (x, y) such that x + y = arcsin(c). Note that this is the equation of a line, namely y = −x + arcsin(c). Of course, not all c are permissible; as the sine of any quantity is between −1 and 1, the only values of c leading to non-empty level sets are when −1 ≤ c ≤ 1. Note that the level sets are periodic. For example, for c = 0 we have arcsin(0) = 0, ±π, ±2π, . . . ; for c =
√ 2/2 we have arcsin(
√ 2/2) = π/4, π/4 ±
2π, π/4 ± 4π as well as 3π/4, 3π/4 ± 2π, 3π/4 ± 4π, . . . . See Figure 1.
(2) f(x, y) = (x + y) sin(x + y). Solution: This problem is similar to the previous; we will still have the function
constant whenever x + y is constant.
(3) f(x, y) = x2 − 4y2. Solution: The level sets are hyperbolas. See Figure 2.
(4) f(x, y) = x2 + 4y. Solution: The level sets are parabolas. We have x2 + 4y = c, which implies
y = −x2 4 + c
4 . Thus the level set of value c is a downward pointing parabola with
y-intercept c/4. See Figure 3.
(5) f(x, y) = ecos x. Solution: Note there is no y-dependence in the function. We want ecos x = c, which
means cos x = log c, or x = arccos(log c). Of course, we need to be careful and see which values of c are permissible. As the exponential of any number is positive, the level set is empty if c ≤ 0. Further, we know that the cosine is always between -1 and 1, and thus we only have non-empty level sets for 1/e ≤ c ≤ e. The answer will be a series of parallel lines. Specifically, y is arbitrary, which gives us a line. The reason we have a series of parallel lines is that if we increase x by 2π we do not change the value of its cosine. See Figure 4.
6 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
-10 -5 0 5 10
-10
-5
0
5
10
-10
-5
0
5
10 -10
-5
0
5
10
-1.0
-0.5
0.0
0.5
1.0
Figure 1. Level sets and plot of f(x, y) = sin(x + y).
Question 4: Find the gradients of the following functions:
(1) f(x, y, z) = xy + yz + zx. Solution: As
grad(f) = ∇f = (
∂f
∂x , ∂f
∂y , ∂f
∂z
)
,
we have
∇f = (y + z, x + z, x + y) .
PRACTICE PROBLEMS 7
-10 -5 0 5 10
-10
-5
0
5
10
Figure 2. Level sets of f(x, y) = x2 − 4y2.
-10 -5 0 5 10
-10
-5
0
5
10
Figure 3. Level sets of f(x, y) = x2 + 4y.
(2) f(x, y) = x cos(y) + y cos(x). Solution: As f : R2 → R, now the gradient is
grad(f) = ∇f = (
∂f
∂x , ∂f
∂y
)
.
8 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
-10 -5 0 5 10
-10
-5
0
5
10
Figure 4. Level sets of f(x, y) = ecos x.
Differentiating our function yields
∇f = (cos y − y sin x, −x sin y + cos x).
(3) f(x1, . . . , xn) = x1x2 · · ·xn. Solution: As f : Rn → R, now the gradient is
grad(f) = ∇f = (
∂f
∂x1 , ∂f
∂x2 , . . . ,
∂f
∂xn
)
.
Differentiating yields
∇f = (x2x3 · · ·xn, x1x3 · · · xn, . . . , x1 · · · xn−1).
A particularly nice way of writing this is to note that
∂f
∂xi = x1 · · · xi−1xi+1 · · · xn =
x1 · · ·xn xi
= f(x1, . . . , xn)
xi .
Thus
∇f = f(x1, . . . , xn) (
1
x1 , . . . ,
1
xn
)
.
(4) f(x, y, z) = 1701x24601 log(1793x5y4).
PRACTICE PROBLEMS 9
Solution: We could differentiate directly, but it is much easier to Thoreau the problem first and simplify Note
f(x, y, z) = 1701x24601 (log 1793 + 5 log x + 4 log y)
= (1701 log 1793)x24601 + 8505x24601 log x + 6804x24601 log y.
There is no z dependence, so ∂f ∂z
= 0. To find ∂f ∂x , it is best not to completely expand.
It does help a bit to expand the logarithm term, but not to multiply everything out (though of course it is not wrong to do so). We thus have
∇f = (
1701 · 24601x24600 (log 1793 + 5 log x + 4 log y) + 1701x24601 · 5
x , 1701x24601 ·
4
y , 0
)
.
(5) f(x, y) = sin(x2 + y2). Solution: Using the chain rule, we have
∇f = (
cos(x2 + y2) · 2x, cos(x2 + y2) · 2y )
= 2 cos(x2 + y2) (x, y) ;
of course there is no need to simplify, but by pulling out these pieces we see the gra- dient is in the direction (x, y), which is hidden at first.
10 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
Question 5: Determine which functions below are differentiable. To be differ- entiable the tangent plane is supposed to do an excellent job approximating the function. A sufficient condition to ensure the function is differentiable is that the partial derivatives all exist and are continuous. This concept was covered more in Spring 2010 than later years, so if you are taking this in Spring 2011 or later do not worry as much about this problem.
We constantly use the result that if the partial derivatives exist and are continuous then
the function is differentiable. Recall grad(f) = ∇f = (
∂f ∂x1
, . . . , ∂f ∂xn
)
.
(1) f(x, y, z) = (xyz)4/3. Solution: Taking the derivatives, we find
∇f = (
4
3 x1/3(yz)4/3,
4
3 y1/3(xz)4/3,
4
3 z1/3(xy)4/3
)
.
Note the partial derivatives exist and are continuous, thus the function is differentiable.
(2) f(x, y) = (xy)2/3. Solution: This is a slight modification of the problem from class, where we had
(xy)1/3. A similar calculation (using the definition of the derivative) gives
∂f
∂x (0, 0) = 0,
∂f
∂y (0, 0) = 0.
If x 6= 0 we have ∂f ∂x
= 2 3 x−1/3y2/3, and if y 6= 0 we have ∂f
∂y = 2
3 x2/3y−1/3. Thus the
partial derivatives are not continuous, and we cannot just use our theorem above. It is possible that our function could be differentiable even though the partial derivatives are not continuous. We must go to the definition of the derivative. What is the tangent plane at (0, 0)? It is
z = f(0, 0) + ∂f
∂x (0, 0)(x − 0) +
∂f
∂y (0, 0)(y − 0) = 0.
Thus the function is differentiable if
lim (x,y)→(0,0)
(xy)2/3 − 0 ||(x, y) − (0, 0)||
exists and equals zero. This limit does not exist. To see this, let’s investigate several paths. Note the denominator is
√
x2 + y2. If we take the path x = 0 we get 0, which we also see along the path y = 0 or y = x or even y = mx. One might then be led to think the limit exists. If we try x = r cos θ and y = r sin θ, we find
lim r→0
(r2 cos θ sin θ)2/3
r = lim
r→0 r1/3(cos θ sin θ)2/3 = 0.
Thus, it does seem as if the limit is zero; unfortunately, there is a slight technical error in what we’ve done here and in class. It is a very subtle point, something I am not going to hold you responsible for on exams, but for completeness I will mention it here. Technically, we are not considering all
PRACTICE PROBLEMS 11
paths when we use polar coordinates; we are only checking along paths (x, y) where x2 + y2 = r → 0. For this problem, consider the path y = x1/6. Then
(xy)2/3 = (x · x1/6)2/3 = (x7/6)2/3 = x14/18; the limit will not exist. For exams, I will not give you a problem such as this, but I want you to
be aware of them.
(3) f(x1, . . . , xn) = (x1x2 · · ·xn)2. Solution: The partial derivatives are computed using the power rule (or the chain
rule). We have
∂f
∂x1 = 2(x1 · · · xn)2−1
∂(x1x2 · · · xn) ∂x1
= 2(x1 · · · xn) · x2 · · · xn.
Note ∂f ∂x1
exists and is continuous; the other partial derivatives are calculated similarly, and also seen to be continuous. Thus the function is differentiable.
(4) f(x, y, z) = 1701x24601 log(1793x5y4). Solution: As the logarithm is only defined for positive inputs, we must have x > 0
and y 6= 0. Note the partial derivatives exist and are continuous, and thus the function is differentiable. If we needed to compute the derivatives, it might be worthwhile to Thoreau the logarithm term, and note
log(1793x5y4) = log 1793 + 5 log x + 4 log y.
(5) f(x, y) = sin(x2 + y2). Solution: This function is clearly differentiable. We have
∇f = (
cos(x2 + y2) · 2x, cos(x2 + y2) · 2y )
= 2 cos(x2 + y2) (x, y) ;
the partial derivatives exist and are continuous.
(6) f(x, y, z) = x3 cos(x) + y3 cos(y). Solution: This function is differentiable; the partial derivatives are
∇f = (3x2 cos x − x3 sin x, 3y2 cos y − y3 sin y), and these functions are continuous.
(7) f(x, y) = xy cos(1/y). Solution: If f is not defined at the origin, then it is clearly differentiable at every
point in its domain, as its partials exist and are continuous away from the origin. What if the function is defined at the origin? What should its value be? Using the squeeze theorem, we see that we may define the function at the origin (or, in fact, any time y = 0) to be 0, and with such a definition our function is continuous. Thus our definition is
f(x, y) =
{
xy cos(1/y) if y 6= 0 0 if y = 0.
12 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
If y 6= 0, the partial derivatives are ∂f
∂x = y cos(1/y),
∂f
∂y = x cos(1/y) +
x sin(1/y)
y ,
while if y = 0 the partial derivatives are
∂f
∂x = 0,
∂f
∂y = 0.
The partial derivatives exist but are not continuous (while ∂f ∂x
is continuous when
y = 0, ∂f ∂y
is not). This does not prove that the function is not differentiable, though;
it just means that we cannot appeal to our theorem that says the partial derivatives being continuous implies the function is differentiable. The tangent plane at (0, 0) is just z = 0. To be differentiable, we would need
lim (x,y)→(0,0)
f(x, y) − 0 ||(x, y) − (0, 0)||
to exist and equal 0. This limit does exist, as it is either involves terms such as xy cos(1/y)√
x2+y2 or 0√
x2+y2 . In both cases, we do get 0 as (x, y) → (0, 0). Thus our function
is differentiable, even though the partial derivatives are not continuous. In other words, we have found an example of a function which is differentiable, but whose partial derivatives, while existing, are not continuous.
PRACTICE PROBLEMS 13
Question 6: Find the tangent plane approximation to f(x, y) = exy + 2 sin(x + y) cos(x−y) at the point (x0, y0). Use the tangent plane plane to estimate f(−.01, .02) by choosing (x0, y0) appropriately.
Solution: The tangent plane is given by
z = f(x0, y0) + ∂f
∂x (x0, y0)(x − x0) +
∂f
∂y (y − y0).
To estimate the value at (−.01, .02), we take (x0, y0) = (0, 0). We have f(0, 0) = 1 and ∂f
∂x = yexy + 2 cos(x + y) cos(x − y) − 2 sin(x + y) sin(x − y)
∂f
∂y = xexy + 2 cos(x + y) cos(x − y) + 2 sin(x + y) sin(x + y),
which implies ∂f
∂x (0, 0) = 2,
∂f
∂y (0, 0) = 2.
Therefore the tangent plane approximation to f(−.01, .02) is
f(0, 0) + ∂f
∂x (0, 0)(−.01 − 0) + ∂f
∂y (0, 0)(.02 − 0) = 1 + 2(−.01) + 2(.02) = 1.02,
which is quite close to f(−.01, .02), which is approximately 1.0198.
Note one can avoid the product rule by recalling trig identities; in particular, as 2 sin(x + y) cos(x − y) = sin 2x + sin 2y, we can simplify the derivative.
14 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
Question 7 : Parametrize the following curves, and find the tangent line approximation at the given point. Note: we haven’t done too much with parametrization yet in 2013; I’m shifting the order of the material a bit. A curve in 3-dimensional space is parametrized by a
curve c if c(t) = (x(t), y(t), z(t)) traces out the curve.
(1) A circle of radius 5 centered at (2, 3) going counter-clockwise starting at the point (7, 3); find the tangent line at the point (7, 3). Solution: The answer is c(t) = (2, 3) + (5 cos t, 5 sin t), or c(t) = (2 + 5 cos t, 3 +
5 sin t). Note that when t = 0 we have c(0) = (7, 3). The tangent line goes through the point (7, 3) and is in the direction c′(0). As c′(t) = (−5 sin t, 5 cos t), we have c′(0) = (0, 5), the tangent line is just
(x, y) = c(0) + tc′(0) = (7, 3) + t(0, 5) = (7, 3 + 5t).
To check that our curve c(t) correctly parametrizes our circle, we must show
(x(t) − 2)2 + (y(t) − 3)2 = 52, which follows from straightforward algebra.
(2) A circle of radius 5 centered at (2, 3) going counter-clockwise starting at the point (5, 7); find the tangent line at the point (5, 7).
Solution: This is similar to the previous problem; the difference is now we want to start at the point (5, 7). Note that (5, 7) − (2, 3) = (3, 4); as
√ 32 + 42 = 5, we do
see that the point (5, 7) is on the circle of radius 5 centered at (2, 3). Let’s consider our curve from the previous part: c(t) = (2 + 5 cos t, 3 + 5 sin t). This traces out the right path, but it starts at the wrong point. Thus, consider
c(t) = (2 + 5 cos(t + t0), 3 + 5 sin(t + t0)) .
If we take t = 0, we now have c(0) = (2 + 5 cos t0, 3 + 5 sin t0); we just need to take cos t0 = 3/5 and sin t0 = 4/5, which means tan t0 = 4/3 or t0 = arctan(4/3). The tangent line is
(x, y) = c(0) + c′(0)t.
As c′(t) = (−5 sin(t + t0), 5 cos(t + t0)), we have c′(0) = (−5 sin t0, 5 cos t0) = (−4, 3),
leading to a tangent line of
(x, y) = (5, 7) + (−4, 3)t = (5 − 4t, 7 + 3t).
(3) The curve y = ex from x = 1 to x = 10; find the tangent line at the point (2, e2). Solution: We may parametrize the curve by c(t) = (t, et) for 1 ≤ t ≤ 10. The point
(2, e2) corresponds to t = 2. We have c(2) = (2, e2) and c′(t) = (1, et) so c′(2) = (1, e2) and hence the tangent line is
(x, y) = c(2) + c′(2)t = (2, e2) + (1, e2)t = (2 + t, e2 + e2t).
PRACTICE PROBLEMS 15
Question 8: Consider the parametrized curve c(t) = (cos t, 2 sin t). What path does this trace out in the plane? Is it periodic (i.e., does it repeat where it is), and if so, what is the period? Does a particle whose position is given by c(t) move at constant speed? If not, when is it moving fastest?
Solution: This path traces out an ellipse. Note that x(t) = cos t and y(t) = 2 sin t. Thus
4x(t)2 + y(t)2 = 4 cos2 t + 4 sin2 t = 4.
It is periodic; every time t increases by 2π we return to the starting point. As
c′(t) = (− sin t, 2 cos t), we have
||c′(t)|| = √
sin2 t + 4 cos2 t = √ 1 + 3 cos2 t.
Note the speed is not constant; the particle is traveling fastest when cos2 t = 1, or at time 0, π, 2π, . . . .
Question 9: Find the derivative of A(x, y, z) = (x3y+y2z+ex)(sin(x)+cos(y)−z+10). Solution: There are two ways to do this. The fist is to multiply everything out and then
take the derivative directly. The problem with that is the first factor has three terms and the second has four. Multiplying everything out would give 12 terms. It is thus better to use the product rule. Let the first factor be f(x, y, z) and the second be g(x, y, z). Then
(DA)(x, y, z) = (Df)(x, y, z)g(x) + f(x)(Dg)(x, y, z).
We thus find
(Df)(x, y, z) =
(
∂f
∂x , ∂f
∂y , ∂f
∂z
)
= (
3x2y + ex, x3 + 2yz, y2 )
,
while
(Dg)(x, y, z) =
(
∂g
∂x , ∂g
∂y , ∂g
∂z
)
= (cos(x), − sin(y), −1) . Putting all the pieces together, we find
(Dh)(x, y, z) =
(
∂h
∂x , ∂h
∂y , ∂h
∂z
)
= (
3x2y + ex, x3 + 2yz, y2 )
(sin(x) + cos(y) − z + 10) + (x3y + y2z + ex) (cos(x), − sin(y), −1) .
Question 10 : Let g(x, y, z) = (xy, yz, xz) and f(u, v, w) = u2 + v2. Set A(x, y, z) = f(g(x, y, z)). Compute DA.
Solution: Setting g(x, y, z) = (g1(x, y, z), g2(x, y, z), g3(x, y, z)), we have
(Dg)(x, y, z) =
∂g1 ∂x
∂g1 ∂y
∂g1 ∂z
∂g2 ∂x
∂g2 ∂y
∂g2 ∂z
∂g3 ∂x
∂g3 ∂y
∂g3 ∂z
=
y x 0 0 z y z 0 x
.
16 INSTRUCTOR: STEVEN MILLER (SJM1@WILLIAMS.EDU)
Similarly we find
(Df)(u, v, w) =
(
∂f
∂u , ∂f
∂v , ∂f
∂w
)
= (2u, 2v, 0),
which implies (Df)(g(x, y, z)) = (Df)(xy, yz, xz) = (2xy, 2yz, 0).
The chain rule says (DA)(x, y, z) = (Df)(g(x, y, z))(Dg)(x, y, z), so
(DA)(x, y, z) = (2xy, 2yz, 0)
y x 0 0 z y z 0 x
= (2xy2, 2x2y + 2yz2, 2y2z).
Question 11 : Let g(x, y, z) = xy2z3. Compute the directional derivative of g at (1, 1, 1) in the direction −→v , where −→v is a unit vector in the direction (3, 4, 12). In what direction is g increasing fastest?
Solution: The length of −→v is ||−→v || = √ 32 + 42 + 122 =
√ 9 + 16 + 144 =
√ 169 = 13;
thus a unit vector in this direction is −→u = (3/13, 4/13, 12/13). The gradient of g is simply (∇g)(x, y, z) = (y2z3, 2xyz3, 3xy2z2),
so (∇g)(1, 1, 1) = (1, 2, 3).
The directional derivative is thus
(∇g)(1, 1, 1)·(3/13, 4/13, 12/13) = (1, 2, 3)·(3/13, 4/13, 12/13) = 1 · 3 + 2 · 4 + 3 · 12 13
= 47
13 .
Our function g is increasing fastest in the direction of its gradient, i.e., in the direction (1, 2, 3).
Question 12 : Let g(x, y, z) = ex cos πy + z cos πx. If possible, find the tangent plane to the level set of value 1 for g(x, y, z) at (x, y, z) = (1, 1, 1). If possible, find the tangent plane to the level set of value 0 for g(x, y, z) at (x, y, z) = (0, 1, 1).
Solution: It is not possible to find the tangent plane in the first case, as the specified point is not on the level set (specifically, g(1, 1, 1) = e(−1) + 1(−1) = −(e + 1) 6= 1. For the second, the point is on the level set as g(0, 1, 1) = 0. We have
∇g = (ex cos πy − πz sin πx, −πex sin πy, cos πx) which implies
(∇g)(0, 1, 1) = (−1, 0, −1). The equation of the tangent plane is
(∇g)(0, 1, 1) · (x − 0, y − 1, z − 1) = 0, or
(−1, 0, −1) · (x − 0, y − 1, z − 1) = 0, which simplifies to
z = 1 − x.