Solutions to Exam 1
1
We find the gradient:
∇f (x, y, z) = 2x i + 3z j + 3y k.
At the point in question, we have
∇f (−1, 1, 2) = −2 i + 6 j + 3 k.
This implies that the equation of the plane is
−2(x + 1) + 6(y − 1) + 3(z − 2) = 0.
To see where the plane hits the x-axis, we set y and z equal to zero and solve for x. This means we have
−2(x + 1) + 6(−1) + 3(−2) = 0 =⇒ −2x − 14 = 0 =⇒ x = −7.
The answer is thus (−7, 0, 0), G.
1
2
Define g(x, y) = x2 + y 2 − 1. Then by the method of Lagrange multipliers, the minimum must be a solution
to fx (x, y) = λgx (x, y), fy (x, y) = λgy (x, y) and g(x, y) = 0. We compute
fx (x, y) = 3
fy (x, y) = 4
gx (x, y) = 2x
gy (x, y) = 2y.
This means we must solve



2λx = 3


2λy = 4



 x2 + y 2 = 1.
Squaring the first two equations and adding them together yields 4λ2 (x2 + y 2 ) = 25. Plugging in x2 + y 2 = 1
gives 4λ2 = 25 =⇒ λ = ± 52 . Using the first two equations again, now implies
3
3
=±
2λ
5
2
4
y= =±
λ
5
x=
where we require x and y to have the same sign. Of the two possible values of (x, y), f is maximum when
both x and y are positive. Thus the greatest of the two values is
9 16
3 4
,
= +
= 5.
f
5 5
5
5
The answer is thus 5, F.
2
3
We sketch the region of integration R below:
y
(π 2 , π)
3
R
2
1
x
1
2
3
4
5
6
7
8
Switching the order of integration, we have
Z
0
π2
Z
π
√
x
sin(y)
dydx =
y2
Z
π
0
Z
Z
y2
0
sin(y)
dxdy
y2
π
=
sin(y)dy
π
= − cos(y)
0
0
= cos(y)
0
π
= 1 − (−1)
= 2.
The answer is thus 2, D.
3
9
10
4
We sketch the region in the xy plane below:
y
(4, 4)
4
3
2
1
x
1
2
3
4
We must check the interior for critical points, the three boundary line segments, and the three boundary
points. To check for critical points, we compute partial derivatives:
fx (x, y) = 2x − 4
fy (x, y) = 2y − 6.
We find that both fx and fy are zero only at the point (2, 3), which is outside of our region. We thus must
check the three boundary line segments. On y = 0, we have f (x, 0) = x2 − 4x, with f 0 (x, 0) = 2x − 4. The
only critical point when restricted to this segment is at x = 2. We then note that f (2, 0) = −4.
On x = 4, we have f (4, y) = y 2 − 6y, with f 0 (4, y) = 2y − 6. The only critical point on this segment is
at y = 3. We note that f (4, 3) = −9.
On x = y, we have f (x, x) = x2 − 4x + x2 − 6x = 2x2 − 10x, with f 0 (x, x) = 4x − 10. The only critical point when restricted to this segment is x = 25 . We note that f ( 52 , 52 ) = 2 ·
25
4
− 25 = − 25
2 . It remains to
check the three boundary points. We have
f (4, 4) = −8
f (0, 0) = 0
f (4, 0) = 0.
Since the minimum must occur at one the points we checked, the minimum must be
4
25
, G.
2
5
We sketch the region in the xy plane below:
y
3
(2, 2)
2
1
x
1
2
3
In this region, 0 ≤ x ≤ y, and 0 ≤ y ≤ 2. We then have that the volume of this region is
y
Z 2Z y
Z 2
2
3
2 (3x + 2xy)dxdy =
(x + x y) dy
0
0
0
Z
=
=
5
2y 3 dy
0
4 2
y 2 0
= 8.
The answer is thus 8, C.
0
2
6
Let f (L, R) be the amount of metal used for an oil car where the cylinder has radius R and length L. Then
we have
2
2
f (L, R) = |2 · 4πR
{z } + |2πRL
{z } = 8πR + 2πRL
end caps
cylinder
where the 2 in front of the volume of the spheres is due to the fact that the spherical caps require twice as
much metal used. Let V (L, R) denote the volume of such a container; we then have
V (L, R) =
4 3
2
πR + πR
| {z L} .
|3 {z }
cylinder
end caps
We must thus minimize f (L, R) under the constraint V (L, R) = 144π. By the theory of Lagrange multipliers,
this minimum must occur when fL (L, R) = λVL (L, R), fR (L, R) = λVR (L, R), and V (L, R) = 0. We compute
fL (L, R) = 16πR + 2πL
fR (L, R) = 2πR
VL (L, R) = 2πRL + 4πR2
VR (L, R) = πR2 .
This means we must solve:



16πR + 2πL = 2πλRL + 4πλR2


2πR = πλR2



 πR2 L + 4 πR3 = 144π.
3
The second equation implies Rλ = 2, i.e. λ =
2
R.
Plugging this into the first equation gives
16πR + 2πL = 4πL + 8πR =⇒ 8πR = 2πL =⇒ R =
L
.
4
Finally, plugging this value into the final equation gives:
πL3
πL3
πL3
+
= 144π =⇒
= 144π =⇒ L3 = 123 =⇒ L = 12.
16
3 · 16
12
The answer is thus 12, C.
6
7
Since f (1, 0) is easy to calculate (it is 1), we will approximate the value of f (.98, .3) by using the directional
derivative at (1, 0). Let u = (.98 − 1)i + (.3 − 0)j = −.02i + .3j be the vector from (1, 0) to (.98, .3). We now
compute the gradient:
∇f (x, y) = 4x3 ey i + x4 ey j.
This implies that
∇f (1, 0) = 4i + j.
The approximate change is thus u · ∇f (1, 0); this means that the approximate value is
f (.98, .3) ≈ f (1, 0) + u · ∇f (1, 0)
= 1 + (−.02i + .3j) · (4i + j)
= 1 + 4(−.02) + .3(1)
= 1 − .08 + .3
= 1.22.
The answer is thus 1.22, B.
7
8
We find first derivatives of f :
fx (x, y) = 3x2 − 12
fy (x, y) = −2y + 2.
To find the critical points, we must solve for fx (x, y) = 0 and fy (x, y) = 0 simultaneously. We see that
fy (x, y) = −2y + 2 = 0 only when y = 1, and fx (x, y) = 3x2 − 12 = 0 only when x = ±2. The only critical
points are thus (−2, 1) and (2, 1); to classify them, we use the second derivative test. We have
fxx (x, y) = 6x
fyy (x, y) = −2
fxy (x, y) = 0.
This
2
fxx (x, y)fyy (x, y) − (fxy (x, y)) = −12x.
2
At (2, 1), we have that fxx (2, 1)fyy (2, 1) − (fxy (2, 1)) = −12 < 0, implying that (2, 1) is a saddle point. At
2
(−2, 1) we have that fxx (−2, 1)fyy (−2, 1) − (fxy (−2, 1)) = 12 > 0, and fxx (−2, 1) = −12 < 0, implying that
(−2, 1) is a local maximum. The answer is that there is a saddle point at (2, 1) and a local max at (−2, 1), G.
8
9
By the chain rule, we know that
∂f
∂f ∂u ∂f ∂v
=
·
+
·
.
∂x
∂u ∂x
∂v ∂x
We then compute
∂f
∂u
Similarly, we compute
= 2u + v + sin(u). At (2, −1), this takes the value of
∂f
∂v
= u + ev−1 . At (2, −1), this takes the value of
these values—and those given for
∂u
∂x (2, −1)
and
∂v
∂x (2, −1)—into
9
∂f
∂u (2, −1)
= 0 + 1 + 0 = 1.
= 0 + e0 = 1. Plugging
the chain rule above gives
∂f
(2, −1) = 1 · (−2) + 1 · 3 = 1.
∂x
The answer is thus 1, E .
∂f
∂u (2, −1)
10
These are circles, and the equation of each level curve must thus be
(x − x0 )2 + (y − y0 )2 = c
for varying c, where (x0 , y0 ) is the center of the circle. Since these circles are centered at (0, −1), we have
that the equations must all be of the form
x2 + (y + 1)2 = c
for varying c. This implies that the function is x2 + (y + 1)2 , F.
10