Solutions to the posted sample exam.

Solutions to the posted sample exam.
1. Given vectors v =< 1, 2, 3 > and w =< −1, 2, 1 > answer the following:
(a) Find a unit vector perpendicular to both v, w.
√
√
√
Answer: v×w =< −4, −4, 4 > and this has length 4 3. Answer: ± <−4,−4,4>
= ± <−1,−1,1>
.
4 3
3
(b) Find the equation of the plane passing thru (1, 1, 2) and perpendicular to v.
Answer: Use the normal < 1, 2, 3 >. So (x − 1) + 2(y − 1) + 3(z − 2) = 0.
(c) Find the equation of a plane thru the origin and perpendicular to both v, w.
Answer: Since v and w are not parallel to each other, both cannot be perpendicular to the
same plane.
(d) What is the angle between v, w?
Answer: arccos( √146√6 ) = 0.8570719477 = 49.10660533◦ .
(e) What is the area of the triangle OP Q, where O is the origin, P is the point (1, 2, 3) and Q
is the point (−1, 2, 1)?
Answer:
√
4 3
2
= 3.464101616.
(f) Find the equation(s) of a line thru the origin and perpendicular to both v, w.
y
x
= −4
= z4 .
Answer: −4
2. Consider the parametric surface given by
r =< u + v, u2 , v 2 > .
What is the formula for the unit normal to this surface at a point u = a, v = b.
Using this or directly, find the equation of the tangent plane to the above surface at the point
P (2, 1, 1) (Hint: Find the values of u, v first.)
Answer:
So
ru ×rv
|ru ×rv |
evaluated at u = a, v = b.
<4ab,−2b,−2a>
√
.
2 4a2 b2 +a2 +b2
At P (2, 1, 1), we have u = 1 = v. A normal is < 4, −2, −2 > or < 2, −1, −1 >. We don’t need
the unit normal for the equation!
Equation 2(x − 2) − (y − 1) − (z − 1) = 0.
3. (a) The function f is defined as f (x, y, z) = xy + 2yz + 3xz. Given the substitution x =
∂f
u2 − v 2 , y = uv , z = u + v, calculate ∂f
∂u and ∂v .
Answer: ∂f
∂u = 2 (y + 3 z) u + (x + 2 z) v + 2 y + 3 x.
∂f = −2 (y + 3 z) v + (x + 2 z) u + 2 y + 3 x.
∂v
2
Also calculate ∂ f2 .
∂u
Answer: 2 (v + 3) u+(2 u + 2) v+(6 u+2 v)(1)+2 y+6 z. This simplifies to 6uv+18u+10v.
1
(b) Use Lagrange Multipliers to find largest value of the function x + 3y + 5z on the sphere
x2 + y 2 + z 2 = 4.
Answer: The equations to solve are:
1 = λ(2x), 3 = λ(2y), 5 = λ(2z), x2 + y 2 + z 2 − 4.
Solutions are:
√
2
x = 35
35, y =
6
35
√
35, z =
10
35
√
√
√
35, λ = 1/4 35, x + 3 y + 5 z = 2 35.
4. Calculate the derivative of the function f (x, y, z) = z 3 + x2 y + 3xyz, in the direction of the vector
< 3, 4, 5 > at the point P (1, 1, 2).
Also calculate directions of maximum and minimum derivative for the function at the same point
P.
Answer: ∇(f ) =< 2 xy + 3 yz, x2 + 3 xz, 3 z 2 + 3 xy >. At P (1, 1, 2) this gives < 8, 7, 15 >. So,
.
the answer is √127
50
√
√
The maximum derivative is in the direction < 8, 7, 15 >√with value 82 + 72 + 152 = 338. The
minimum is in the opposite direction with the value − 338.
5. Set up a double integral to find the surface area of the part of the paraboloid z = 9 − x2 − y 2
above the disk x2 + y 2 ≤ 1.
Also sketch the surface and finish the integration to find its surface area.
Answer:
Z Z q
4x2 + 4y 2 + 1 dA
D
where D is the unit disc. Using polar coordinates,
Z 2π Z 1
√
4r2 + 1r dr dθ.
θ=0 r=0
√
This evaluates to 2π( 5
5−1
).
12
6. Consider the triple integrals
I1 =
Z 1 Z 1−z Z 1−y−z
0
0
0
dx dy dz and I2 =
Z 1 Z 1−z Z 1−y−z
0
0
zdx dy dz
0
(a) First sketch the region of integration.
Answer: It is a tetrahedron with corners (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1).
(b) Explain a relation between the region of integration and the integral I1 .
Answer: I1 is the volume of the region of integration.
(c) Explain a relation between the region of integration and the integral I2 .
Answer: I2 is the moment of the region about the xy plane. Also, II12 is the z-coordinate
of the center of mass, assuming uniform density.
Even thought the question does not ask it, you should be able to evaluate I1 = 1/6 and
I2 = 1/24, so the z-coordinate of the centroid is 1/4.
2
7. (a) Given the vector field
F =< sin(x) , cos(x) , yz >
calculate its curl and divergence.
Answer: curl F =< z, 0, − sin(x) > and div F = y + cos(x).
(b) Is F a conservative field? Why? If it is a conservative field, find a function f such that
∇f = F .
Answer: No, it is not conservative, since curl F 6=< 0, 0, 0 >.
(c) Given the function g(x, y, z) = xyz + 2x + 3y + 4z compute its gradient ∇g and its Laplacian
∇ · ∇g.
Answer: ∇(g) =< yz + 2, xz + 3, xy + 4 > and the Laplacian ∇ · ∇(g) = 0.
8. When is a vector field said to be conservative?
Under certain conditions, a conservative vector
R
field F has the property that the integral C F · dr is independent of the chosen path C joining
two given points.
Carefully explain what this means. Be sure to illustrate your explanation by taking a simple
example of a conservative field and suitable paths.
Be sure to describe the conditions also.
Answer: A vector field F is said to be conservative if F = ∇(f ) for some function f .
If the components of F have continuous partial derivatives in a simplyR connected open domain
D and F = ∇(f ), then for any path C from A to B, we can show that C F · dr = f (B) − f (A).
This condition is equivalent to the simpler condition that for all closed paths, the integral is zero.
Typically, the condition is stated in terms of simple paths, but that is not essential. The paths
are assumed to be piecewise smooth to avoid technical problems with integration itself.
For example, if we take F =< y, x > then FR = ∇(xy) and since it has continuous derivatives of
all order over the whole plane, the integral C y dx + x dy = 0 for all closed paths C. We could
take a path from (0, 0) to (1, 1) either by a direct line or by a two line path (0, 0) to (1, 0) to
(1, 1).
The
integral for the first choice uses parametrization < t, t > as t goes from 0 to 1 and we get
R1
t
dt
+ t dt = 1. This is also equal to the difference of values of xy between the two points.
0
For the second choice of paths,
we useR parametrizations < t, 0 > and < 1, t > as t goes from 0 to
R1
1 respectively. Then we get 0 0 dt + 01 1 dt = 0 + 1 = 1.
In two dimensions, a necessary condition for a field < P, Q > to be conservative is Qx − Py = 0.
It is sufficient in a simply connected domain when the condition on partial derivatives holds.
, x > we see that it satisfies the condition, but its integral on a unit
For the field F =< x2−y
+y 2 x2 +y 2
circle is non zero. It fails the continuity condition at the point (0, 0).
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9. Use Green’s Theorem to evaluate the line integral
Z
((y − x2 sin(x))dx + (x2 − y 2 cos(y))dy
C
where C the positively oriented boundary of the region bounded by y = 4 − x2 , y = 0.
Be sure to sketch the region, indicating the orientation of the curve C.
Answer: We have P = (y − x2 sin(x)) and Q = (x2 − y 2 cos(y)). Thus, Qx − Py = 2x − 1. We
integrate this over the region D as described. The integral is seen to be:
Z 4−x2
Z 2
x=−2
(2x − 1) dy dx = −
y=0
32
.
3
10. Show that the line integral
Z
(3x2 y 3 + 3)dx + (3x3 y 2 + 2)dy
C
is independent of the path C as long as the path goes form a given point P to a given point Q.
Answer: We check Qx − Py = 9x2 y 2 − 9x2 y 2 = 0. So, it should be a conservative field, as all
functions are infinitely differentiable.
To find the function f , we note that fx = 3x2 y 3 + 3 so f = x3 y 3 + 3x + h(y). Since, fy = 3x3 y 2 + 2
we see that 3x3 y 2 + h0 (y) = 3x3 y 2 + 2, so we take h(y) = 2y. Thus, f = x3 y 3 + 3x + 2y. The
desired integral on a path from P to Q will be f (Q) − f (P ).
Use this to find the value of the integral when C is some path from P (−1, 0) to Q(1, 3).
Answer: We calculate f (1, 3) − f (−1, 0) = 39.
Use your answer to find the work done by the force field < 3x2 y 3 + 3, 3x3 y 2 + 2 > in moving an
object from P (−1, 0) to Q(1, 3).
Answer: It is the same as the above answer!
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