**Unformatted text preview: **MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS (1) For vectors A = 2i + 3j − k and B = −i + 3k, nd:
(a) A − 2B. (5 points per part) Solution: A − 2B = (2i + 3j − k) − 2 (−i + 3k)
= 4i + 3j − 7k.
(b) A · B.
Solution: A · B = (2i + 3j − k) · (−i + 3k)
= 2(−1) + 3(0) − 1(3)
= −5.
(c) The vector projection projB A of A onto B.
Solution: projB A =
=
=
= (A · B)
B
(B · B)
((2i + 3j − k) · (−i + 3k))
(−i + 3k)
((−i + 3k) · (−i + 3k))
−5
(−i + 3k)
10
3
1
i− j
2
2 (d) A × B.
Solution: A×B = i j k
2 3 −1
−1 0 3 = (9 − 0) i + (1 − 6) j + (0 + 3) k
= 9i − 5j + 3k,
which you should check is perpendicular to A and B, which this is. You
may think the computation of the j component of the cross-product is the
negative of what it should be, but if you look closely, it is correct.
(2) Find the line of intersection between the planes P1 dened by 2x + 3y − z = 0 and
P2 dened by x + 3y + z = 1. (10 points)
1 MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 2 The direction of the line is the cross-product of the two normal vectors
to the planes, Solution: A = i j k
2 3 −1
1 3 1 = (3 + 3) i + (−1 − 2) j + (6 − 3) k
= 6i − 3j + 3k,
which again you should check is perpendicular to the two normals. Then, you
nd a point on both planes. If x = 0, then 3y − z = 0 and 3y + z = 1. Subtract
the second equation from the rst, and you get −2z = −1, or z = 1/2. Add the
1
two equations, and you get 6y = 1, or y = 1/6, so 0, 6 , 1 should be on both
2
planes. Check that it satises both equations. Then, the vector equation of the
line of intersection, the line containing this point in the direction of A,is
1
1
j + k + t (6i − 3j + 3k) .
r(t) =
6
2
Its parametric equations are x = 6t, y = 1 − 3t, z = 1 + 3t.
6
2
(3) Find the arclength of the curve r(t) = et cos(t)i + et sin(t)j, t ∈ [0, 2π]. (10 points)
Solution: The arclength is given by (be careful to use the product rule)
2π r dt L =
0
2π (et cos(t) − et sin(t))2 + (et sin(t) + et cos(t))2 dt =
0
2π et = (cos(t) − sin(t))2 + (sin(t) + cos(t))2 dt 0
2π et = cos2 (t) − 2 cos(t) sin(t) + sin2 (t) + cos2 (t) + 2 cos(t) sin(t) + sin2 (t) dt 0
2π = √
et 2dt 0 √ t 2π
=
2e
0
√
2π
2 e −1 .
=
(4) (5 points per part) (a) If f (x, y) = exy , nd ∂f
∂x . Solution: ∂f
= yexy .
∂x
(b) If f (x, y) = (2x + 3y)2 , nd ∂f .
∂y
Solution: ∂f
∂y = 2 (2x + 3y) 3
= 6 (2x + 3y) . (5) Find the tangent plane to the surface S dened by x2 + y 2 −
(1, −1, 2). (10 points) z2
4 = 1 at the point MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 3 Here you use the fact that the surface is a level surface, so the gradient
at (1, −1, 2) is perpendicular to the surface there. Solution: N =
= x2 + y 2 − z2
4 1
2xi + 2yj − zk
2 (1,−1,2) (1,−1,2) = 2i − 2j − 1k.
So the equation of the tangent plane is 2(x − 1) − 2(y + 1) − (z − 2) = 0, or
2x − 2y − z = 2.
(6) Find the absolute maximum and the absolute minimum of the function f (x, y) =
2x2 − 4x + y 2 on the disk x2 + y 2 ≤ 9. (15 points)
Solution: You have to nd all critical points in the interior, and those on the
boundary.
Interior: Since
f = (4x − 4)i + 2yj, f = 0 only when x = 1 and y = 0.
f (1, 0) = 2 − 4 = −2.
2
2
Boundary: The boundary is the circle x +y = 9. Here it is best to use Lagrange
multipliers. You set f = (4x−4)i+2yj equal to λ g , where g is the constraint
equation, x2 + y 2 = 9. So, f = λ g
(4x − 4)i + 2yj = λ (2xi + 2yj) ,
or 4x − 4 = 2λx
2y = 2λy.
The second equation is easy to solve; either y = 0 or λ = 1. If y = 0, then
to satisfy the constraint, x = ±3, so we get two points (3, 0) and (−3, 0). If
λ = 1 we then look at the rst equation, which becomes 4x − 4 = 2x, or 2x = 4,
√
x = 2. If x = 2, then to√
satisfy the constraint equation, y = ± 5, which gives
√
us two more points, (2, 5) and (2, − 5). We only need to evaluate f on all
these points to nd the winner. f (1, 0) = −2 (the interior critical point)
f (3, 0) =
=
=
f (−3, 0) =
=
√
f (2, 5) =
=
√
f (2, − 5) = 2x2 − 4x + y 2 (3,0) 18 − 12 + 0
6
18 + 12 + 0
30
8−8+5
5
5. So, the maximum value is 30, at (−3, 0) and the minimum value is −2, at (1, 0). MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 1 (7) Evaluate 1 0 1 − y 2 dy dx. 4 (10 points) x Change the order of integration to solve this. Note that the region is
the triangle bounded by the lines x = 0, y = x, and y = 1. So Solution: 1 1 1 y 1 − y 2 dy dx =
0 1 − y 2 dx dy x 0 0
1 1 − y2x =
0
1 x=y
x=0 dy 1 − y 2 y dy. Set u = 1 − y 2 , =
0
0√ =
1 =
=
= 1
u − du
2
1√
udu 1
2 0
1 3/2
u
3
1
.
3 1
0 (8) Find the x-coordinate x of the center of mass of the of the region bounded by the
lines y = 2 − x, x = 0, and y = 0, if the density ρ is given by ρ = x. (10 points)
Solution: x =
=
= =
= = R xdm
R dm
2 2−x
x (xdydx)
0 0
2 2−x
(xdydx)
0 0
2 2−x 2
x dydx
0 0
2
2−x
dx
0 yx|0
2
2 2−x dx
0 yx 0
2
0 (2 − x)xdx
2
2
0 (2 − x)x dx
2
x2 − 1 x3 0
3
2 3
1 4 2
3x − 4x 0
4
3 34
43
= 1.
= (9) Find the volume of the region bounded by the surfaces z = x2 +y 2 , and z = 2−x2 −y 2 .
(10 points) MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 5 I would do this in cylindrical coordinates. Note that the two surfaces
intersect when z = 1, along a circle of radius 1, center on the z -axis. Solution: dV V ol =
2π R
1 2−r2 = rdzdrdθ
0 0
1 r2 2r − 2r3 dr = 2π
0 1
r − r4
2 1 2 = 2π 0 = π.
(10) (7 points per part)
(a) Find the gradient of f (x, y, z) = xy − 2x + 3y + 4yz 2 .
Solution: f = (y − 2) i + x + 3 + 4z 2 j + (8yz) k. (b) Find the divergence of F = xyi + y 2 j + 3xzk.
Solution: div(F) =
·F
= y + 2y + 3x
= 3x + 3y.
(c) Let F = xyzi + x2 j − xyk. Compute the curl of F, × F. Solution: curl(F) =
= ×F
i
j
∂
∂x ∂
∂y
x2 k
∂
∂z xyz
−xy
= (−x − 0) i + (xy + y) j + (2x − xz) k
= −xi + (xy + y) j + (2x − xz) k.
(11) Determine whether the vector eld F = (x2 + xy 2 )i + (x2 y + y 2 )j is conservative. If
it is conservative, nd a potential function f (x, y) so that f = F. (10 points)
Solution: Since this vector eld is dened everywhere, all we have to do is check
the mixed partials condition ∂P = ∂Q . But
∂y
∂x ∂(x2 + xy 2 )
∂y
2y + y2)
∂(x
∂x = 2xy, and
= 2xy, so the condition is satised and the vector eld is conservative (you should
have expected this, since then you have to nd the potential function, and we
want to see whether you can do that). The potential function f (x, y) is then MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 6 obtained by nding P dx (with the "constant of integration" depending on y )
and re-dierentiating to compare to Q f (x, y) =
=
= P dx
(x2 + xy 2 )dx
1 3 1 2 2
x + x y + k(y),
3
2 so, since ∂f
∂y
2
2
(x y + y ) = x2 y + k (y),
Q = 1
1
so k (y) = y 2 , k(y) = 1 y 3 , and a potential function f (x, y) = 3 x3 + 1 x2 y 2 + 3 y 3 .
3
2
A +C is optional.
(12) Find the line integral C x2 sin(x) + y 2 dx + (x + ey ) dy , where C is the circle of
radius 1, centered at the origin, traversed in the positive direction (counterclockwise). (10 points) This has to be a Green's theorem problem. Recall the trick that
ydA = yArea, where y is the y -coordinate of the center of gravity, which
R
in this case is at the origin, so y = 0. Solution: x2 sin(x) + y 2 dx + (x + ey ) dy =
C (1 − 2y) dA
R = Area − 2yArea
= π.
2
2
(13) Find the surface integral
S f dS of the function f (x, y, z) = x +y over the sphere
x2 + y 2 + z 2 = 1. (10 points)
Solution: This is not a ux problem, so you can't use any of the theorems. You
have to do that directly. Parametrize the sphere by r(u, v) = cos(u) sin(v)i +
sin(u) sin(v)j + cos(v)k. This comes from spherical coordinates, with ρ = 1,
u = θ and v = φ. From that you can see that you cover the whole sphere taking
u ∈ [0, 2π] and, for each u, v ∈ [0, π]. Then, it will take a while to compute
∂r
∂r
∂r
∂r
∂u × ∂v directly, but in the end it comes out that ∂u × ∂v = sin(v). You
can see this by again considering spherical coordinates and remembering how
we got dV , but you should probably work out that cross product. Then,
2π π x2 + y 2 dS =
S cos2 (u) sin2 (v) + sin2 (u) sin2 (v) sin(v)dvdu
0 0
2π π sin3 (v)dvdu =
0 0
π 1 − cos2 (v) sin(v)dv = 2π
0 You can do the u integral with no problem since what you get out of the v integral will have no u-dependence. Then set t = cos(v), so dt = − sin(t)dt. MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 7 Also, when v = 0, t = 1 and when v = π , t = −1, so
−1 x2 + y 2 dS = 2π
S 1 − t2 (−dt)
1 = 2π 1 1
t − t3
3 −1 8π
=
.
3
Since this is an integral of a positive function, it denitely has to have a positive
answer.
(14) Find the ux integral
S (xi + yj + zk) · n dS , where the surface S is that portion
of the paraboloid z = 4 − x2 − y 2 above the xy -plane, and n is the upward-pointing
normal. (15 points)
Solution: No, you cannot use the divergence theorem here on that surface. The
standard method will be to do this directly. Since, for the ux of a vector eld
F = P i + Qj + Rk over a graph z = f (x, y) with the upward normal can be
evaluated as
∂f
∂f
Q + R dA,
F · ndS =
− P−
∂x
∂y
S
A
where A is the region in the xy -plane which is the domain of f (x, y), and here
f (x, y) = 4 − x2 − y 2 (xi + yj + zk) · n dS = (−(−2x)x − (−2y)y + z) dA
x2 +y 2 ≤4 S 2x2 + 2y 2 + 4 − x2 − y 2 = dA x2 +y 2 ≤4
2π
2 4 + r2 rdrdθ =
0 = 2π 0 1
2r2 + r4
4 2
0 = 24π.
(15) Find the ux of the vector eld F = (x2 + 2x) i + x2 + yz j + cos(x)k outward
across the sphere x2 + y 2 + z 2 = 1. (15 points)
Solution: This one uses the theorems. It is a ux outward, so it has to use the
divergence theorem. F · ndS =
S · FdV
V = ((2x + 2) + z + 0) dV
V = 2xV ol + 2V ol + zV ol
8π
=
,
3
using that fact again that
xdV = xV ol (etc., for y and z ) relating the
integrals to the center of gravity, which in this case is the origin, along with the
volume of the sphere being 4π (radius)3 .
3 MATH 23, SPRING, 2012, FINAL EXAM SOLUTIONS 8 (16) Find the line integral C F · dr, if the vector eld is F = (sin(x) + y 2 ) i + ey j + x k,
and the curve C is the intersection of the paraboloid z = x2 + y 2 with the cylinder
(x − 1)2 + y 2 = 1 in space. The curve C is oriented positively (counter-clockwise)
when viewed from above. (14 points)
Solution: Only one left, so it has to use Stokes' theorem. Well, actually, it uses
Stokes' theorem since it's integrating a vector eld over a closed curve in space
which bounds a surface S the part of the paraboloid z = x2 + y 2 which is
inside the cylinder (x − 1)2 + y 2 = 1. F · dr = × F · ndS, C S where the curl of F, × F, is
i ×F = j k ∂
∂x ∂
∂y
ey ∂
∂z (sin(x) + y 2 )
x
= (0 − 0) i + (0 − 1) j + (0 − 2y) k,
so F · dr =
C × F · ndS
S (−j − 2yk) · ndS =
S (−(0)(2x) − (−1)(2y) − 2y) dA =
(x−1)2 +y 2 ≤1 = 0. ...

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