Multivariable Calculus Practice Midterm 2 Solutions
Prof. Fedorchuk
1.
2
(10 points) Let f (x, y, z) = xz + ey−x .
a. (4 pts) Compute the gradient ∇f .
b. (3 pts) Find the directional derivative
D
√ f (0, 0, 1).
0, 21 ,
3
2
c. (3 pts)
Find the unit vector pointing in the direction along which f (x, y, z)
increases most rapidly at the point (0, 0, 1).
Solution:
2
2
(a) We have ∇f = hz − 2xey−x , ey−x , xi and so ∇f (0, 0, 1) = h1, 1, 0i.
(b) Therefore
√
√
1 3
1 3
D
1 √3 f (0, 0, 1) = ∇f (0, 0, 1) · 0, ,
= h1, 1, 0i · 0, ,
= 1/2.
0, 2 , 2
2 2
2 2
(c) The direction of the largest increase of f (x, y, z) at (0, 0, 1) is ∇f (0, 0, 1) = h1, 1, 0i
and the unit vector in this direction is
1 1 √ , √ ,0 .
2 2
2.
(10 points)
a. (5 pts) Find an equation of the tangent plane at (1, 3) to the graph of
f (x, y) = xy 2 − xy + 3x3 y.
Solution: The equation of the tangent plane at (1, 3) is:
z − f (1, 3) = fx (1, 3)(x − 1) + fy (1, 3)(y − 3).
We now compute
f (1, 3) = 9 − 3 + 9 = 15
fx (1, 3) = (y 2 − y + 9x2 y)|(1,3) = 9 − 3 + 27 = 33
fy (1, 3) = (2xy − x + 3x3 )|(1,3) = 6 − 1 + 3 = 8.
Therefore, the equation of the tangent plane at (1, 3) is:
z − 15 = 33(x − 1) + 9(y − 3).
1
b. (5 pts)
surface
Find an equation of the tangent plane at the point (0, 3, −1) to the
g(x, y, z) = zex + ez+1 + xy + y = 3.
Solution: We compute
∇g(x, y, z) = hzex + y, x + 1, ex + ez+1 i.
The normal vector of the tangent plane at the point (0, 3, −1) is
∇g(0, 3, −1) = h2, 1, 2i,
so the tangent plane is
2x + (y − 3) + 2(z + 1) = 0.
3.
(10 points)
√
√
The surface contains curves r1 (t) = (t, t2 , t3 ) and r2 (u) = (1, 2 sin(u), 2 cos(u)). Find
the tangent plane to the surface at the point (1, 1, 1) and the line perpendicular to the
tangent plane and passing through the origin (0, 0, 0).
Solution: We are going to use the fact that the tangent plane to a surface at a point P
contains the tangent lines at P of all curves contained in the surface and passing through
P.
Here we have two curves contained in the surface and passing through P = (1, 1, 1).
The tangent line of the first curve at P (corresponding to t = 1) has direction vector
~v = r10 (1) = h1, 2, 3i and the tangent line of the second curve at P (corresponding to
u = π/4) has direction vector w
~ = r20 (π/4) = h0, 1, −1i. Since the normal vector of the
tangent plane to the surface at P is perpendicular to these two vectors, we can take the
normal vector of the tangent plane to be
~v × w
~ = h−5, 1, 1i.
Finally, the equation of the tangent plane at P = (1, 1, 1) is
−5(x − 1) + (y − 1) + (z − 1) = 0.
The line perpendicular to the tangent plane and passing through the origin (0, 0, 0) is


x = −5t
y=t

z = t
2
4.
(10 points) The function f (x, y) = x2 + y 2 + xy + 9x has 1 critical point. Find it,
and identify it as a local minimum, local maximum, or a saddle point.
Solution: To find critical points, we have to solve the system
(
fx (x, y) = 2x + y + 9 = 0
fy (x, y) = 2y + x = 0
Solving, we obtain y = 3 and x = −6. Hence (x, y) = (−6, 3) is the only critical point.
Next we use the 2nd derivative test:
fxx (−6, 3) = 2
fyy (−6, 3) = 2
fxy (−6, 3) = 1.
The determinant at (−6, 3) is
D(−6, 3) = fxx (−6, 3)fyy (−6, 3) − (fxy (−6, 3))2 = 4 − 1 = 3 > 0.
Since fxx (−6, 3) > 0, the 2nd derivative test says that (−6, 3) is a local minimum.
5.
(10 points) Consider the function f (x, y) = x3 y − 3xy 2 . Find and classify all critical
points of f .
Solution: To find critical points, we have to solve the system
(
fx (x, y) = 3x2 y − 3y 2 = 3y(x2 − y) = 0
fy (x, y) = x3 − 6xy = x(x2 − 6y) = 0
The only solution of this system is (x, y) = (0, 0), hence (0, 0) is the only critical point.
Since fxx (0, 0) = fyy (0, 0) = fxy (0, 0), the 2nd derivative test is inconclusive.
However note that along y = x2 , the function is f (x, x2 ) = x5 − 3x5 = −2x5 . Since
−2x5 has neither maximum nor minimum at x = 0 (−2x5 is strictly decreasing because
(−2x5 )0 = −10x4 ≤ 0). We conclude that (0, 0) is a saddle point of f (x, y), i.e. (0, 0) is
neither local minimum nor local maximum.
6.
(10 points) Find the maximum and minimum values of f (x, y) = x − 2y subject to
2
the constraint x4 + y 2 = 2.
Solution: The method of Lagrange multipliers says that extrema of f (x, y) subject to the
2
constraint are found among the solutions of ∇f (x, y) = λ∇g(x, y), where g(x, y) = x4 +y 2 .
We rewrite this condition in the following form:


1 = fx = λgx = λ(x/2)


−2 = fy = λgy = λ(2y)
2


 x + y2 = 2
4
3
or


1 = λx/2


−2 = 2λy
2


 x + y2 = 2
4
This gives x = 2/λ and y = −1/λ. Plugging these into the constraint, we obtain:
(2/λ)2
+ (−1/λ)2 = 2/λ2 ,
4
which gives λ = 1 or λ = −1. It follows that the two critical points of f (x, y) subject to
the constraint are (2, −1) and (−2, 1).
We have f (2, −1) = 4 and f (−2, 1) = −4. Therefore, f (2, −1) = 4 is the maximum value
2
of f (x, y) subject to the constraint x4 + y 2 = 2; and f (−2, 1) = −4 is the minimum value
2
of f (x, y) subject to the constraint x4 + y 2 = 2.
2=
7.
(10 points) Evaluate each of the following iterated integrals.
a. (5 pts)
Z 1Z x
ex+2y dydx
0
Solution:
Z 1Z
x
e
0
x+2y
Z
1
dydx =
0
0
Z
0
1 x+2y x
e
dx
2
0
1
1 3x
(e − ex )dx
2
0 3x
ex 1
e3 e
e
e3 e
−
− + 1/3.
=
= ( − ) − (1/6 − 1/2) =
6
2
6
2
6
2
0
=
b. (5 pts)
Z
0
Solution: The region
16
Z
4
√
sin(y 3 )dydx
x
√
R = {(x, y)|0 ≤ x ≤ 16, x ≤ y ≤ 4}
is given to us in the form of an y-simple region. Since it is difficult to integrate sin(y 3 ),
we
form of an x-simple region. To do this, observe that
√ are going to rewrite R in the
x ≤ y ≤ 4 implies that x ≤ y 2 and that for any y in the interval [0, 4], x has to satisfy
only 0 ≤ x ≤ y 2 . Hence
R = {(x, y)|0 ≤ y ≤ 4, 0 ≤ x ≤ y 2 }.
4
Therefore
Z
16
4
√
x
0
We now compute
Z 4Z
Z
Z
3
Z
y2
sin(y )dydx =
0
y2
Z
3
sin(y )dxdy =
0
4
0
sin(y 3 )dxdy.
0
4
y 2 sin(y 3 )dy
0
Z
=
0
64
1
sin(u)du
3
1
1
= (− cos(64) − cos(0)) = (1 − cos(64)).
3
3
8.
(10 points) Compute the volume of the solid in the first octant bounded by
2x + 2y + z = 4.
Solution: We are integrating the function f (x, y) = 4 − 2x − 2y over the region
R = {(x, y) : x ≥ 0, y ≥ 0, 2x + 2y ≤ 4} = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x}.
The volume is thus
Z 2 Z 2−x
Z 2
2−x
(4 − 2x − 2y)dydx =
(4y − 2xy − y 2 ) dx
0
0
0
0
Z 2
2
2
3
2
=
(x − 4x + 4)dx = (x /3 − 2x + 4x) = 8/3.
0
0
5