Solution - University of Toledo

[Pages:4]Review Test 1, Math 286

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Solutions

Name

1. Find an equation of the tangent plane to the surface 0 = x sin(x + y) - sin z at

(2,-2,0).

(13)

Solution: The formula for the tangent surface at (2,-2,0) to the graph 0 =

f (x, y, z) is

0 = fx(2, -2, 0)(x - 2) + fy(2, -2, 0)(y + 2) + fz(2, -2, 0)(z - 0) Compute the first partials of f (x, y, s) = x sin(x + y) - sin z.

fx(x, y) = sin(x + y) + x cos(x + y) and fy(x, y) = x cos(x + y) and fz = cos z

At (2,-2,0), fx(2, -2, 0) = sin 0 + 2 cos 0 = 2 and fy(2, -2) = 2 cos 0 = 2 and fz(2, 2, 0) = 1. Therefore the tangent plane is

0 = 2(x - 2) + 2(y + 2) - z = 2x + 2y - z

2. Test the function f (x, y) = x4 + y3 + 32x - 27y for local maxima, minima and

saddle points.

(18)

Solution: Find the critical points.

f (x, y) = (4x3 + 32)i + (3y2 - 27)j

Set f = 0: 4x3 + 32 = 0 and 3y2 - 27 = 0. Simplify: x3 + 8) = 0 and (y - 3)(y + 3) = 0. The first equation has only one solution x = -2, and the second has two y = ?3. There are 2 points: (-2, -3), (-2, 3). Since f is differentiable everywhere, these are the only critical points. Check whether they are local max, min or saddles. Compute the second partials.

fxx = 12x2, fxy = 0 fyy = 6y

At (-2,3), fxx(-2, 3) = 48, fxy(0, 3) = 0 and fyy(0, 3) = 18 so that D = fxxfyy - fx2y = (48)(18) > 0. Since fxx > 0 this means (0,3) is a local minimum. Check next (-2,-3). Here fxx(-2, -3) = 48, fxy(-2, -3) = 0 and fyy(-2, -3) = -18. Therefore D = 32(-18) < 0 so that this is a saddle.

3. Use Lagrange multipliers (or otherwise) to find the maximum and minimum

values of f (x, y) = xy2 on the curve x2 + y2 = 27.

(15)

Solution: We solve for x, y and in the equations

f (x, y) = g

where g(x, y) = x2 + y2. Differentiating we have

y2i + 2xyj = (2xi + 2yj)

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which gives two equations

y2 = 2x and 2xy = 2y

in addition to the constraint equation x2 + y2= 27. One solution is y = 0 so that x = ? 27. This gives the two points ( 27, 0) and (- 27, 0). If y = 0 then we may divide by y in the equation 2xy = 2y so that x = . Substituting in the first equation, gives y2 = 2x2. In the Constraint equation thisgives 3x2 =27 or x =?3. Therefore y = ? 18. This gives four more points (3, 18), (3, - 18) (-3, 18) (-3, - 18) . Evaluate f(x, y) = xy2 at each of these points: f (? 27, 0) = 0; f (3, ? 18) = 54; f (-3, ? 18) = -54. The maximum value is 54 and it occurs at two points (3, ? 18). The minimum value is -54 and it occurs at two points (-3, ? 18).

4. Find the absolute maximum and minimum values of f (x, y) = 2xy - x - y on

the closed triangular plate bounded by the lines x = 0, y = 0, and x + y = 4 and

in the first quadrant. (Show all your work.)

(22)

Solution: We need the derivative f = (2y - 1)i + (2x - 1)j.

(a) Check f for critical points. Set f = 0 to get 2y - 1 = 0 and 2x - 1 = 0 so that (1/2,1/2) is a critical point. Since f is differentiable everywhere, this is the only critical point.

(b) Check the boundary. We must check the 3 sides of the triangualr boundary separately. A picture of the triangle with vertices (0,0), (0,4) and (4,0) may help.

i. Boundary: y = 0, 0 x 4. Here f (x, 0) = -x. This is a decreasing function obviously and so has no critical values. The boundary values are x = 0 and x = 4 or in two dimensions the corners (0,0) and (4,0).

ii. Boundary: x = 0, 0 y 4. Here f (0, y) = -y. As above, this is a decreasing function and has no critical points and so the max and min must occur at the endpoints y = 0 and y = 4. These two points correspond to the corners (0,0) and (0,4).

iii. Boundary x + y = 4, 0 x 4. Here f (x, 4 - x) = 2x(4 - x) - x - (4 - x) = -2x2 + 8x - 4. This function is complicated enough that we will do a max min check. We will need the derivative of the function (d/dx)[-2x2 + 8x - 4] = -4x + 8.

A. Find the critical points. Set the derivative to 0: -4x + 8 = 0. So x = 2. Since the function is differentiable everywhere, this is the only critical point. It corresponds to the point (2,2), since y = 4-x.

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B. End points x = 0 and x = 4. These correspond to the vertices (0,4) and (4,0) (which we already located.)

(c) It remains only to evaluate the original function f (x, y) = 2xy - x - y at the points found above, that is at (1/2,1/2), (0,0), (4,0), (0,4) and (2,2). Evaluate f (1/2, 1/2) = -1/2.; f (0, 0) = 0; f (4, 0) = -4; f (0, 4) = -4 and f (2, 2) = 4. So the absolute min value is -4 and it occurs at (0,4) and (4,0). The absolute max is 4 and it occurs at (2,2).

5. Find the volume of the solid bounded by the cylinder y2 + z2 = 4 and the planes

x = 2y, x = 0, z = 0 in the first octant.

(16)

The circular cylinder is centered on the x axis. The solid is bounded above by the

cylinder z = (4 - y2)1/2 below by the xy plane and the projection D of the solid

onto the xy-plane is the triangle with edges x = 2y, x = 0 and the intersection

of the cylinder with the plane z = 0 which gives y2 = 4 or y = 2 (first octant).

The volume is

(4 - y2)1/2 dA =

2 2y

(4 - y2)1/2 dx dy

D

00

=

2 0

2y(4

-

y2)1/2

dy

=

-

2 3

(4

-

y2)3/2|02

=

16 3

6. Evaluate the integral by switching to polar coordinates.

9 - x2 - y2 dA

D

where D = {(x, y)|1 x2 + y2 4, x 0}.

(14)

Solution: Sketch the region D. It is half a disk. In polar coordinates D =

{(r, )|0 r 2, -/2 /2}. Therefore

9 - x2 - y2 dA =

/2 2 9 - r2r dr d

D

/2 1

=

-

1 3

/2

(9 - r2)3/2|12 d

/2

=

1 3

(83/2

-

53/2)

/2 /2

d

=

3

(83/2

-

53/2)

7. Find the mass of the solid region bounded by the parabolic surfaces z = 8-x2-y2 and z = x2 + y2 if the density of the solid is (x, y, z) = x2 + y2

Solution: Sketch the region D. The mass is

dV =

D

z=8-x2-y2 R z=x2+y2

x2 + y2 dz dA

=

z|z=8-x2-y2

z=x2+y2

x2 + y2 dA

R

=

8 - x2 - y2 - (x2 + y2) x2 + y2 dA

R

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Sketch the region R which is the projection of D onto the xy?plane. Since the two paraboloids intersect when 8 - x2 - y2 = x2 + y2 so that x2 + y2 = 4 which is a circle of radius 2 and the region inside (a disk) is the region R. It is clear that it is best described in polar coordinates. The mass is

2 2

(8 - 2r2)rr dr d =

2 2

8r2 - 2r4 dr d

00

00

=

2 0

8 3

r3

-

2 5

r5|20

d

=

2 0

8 3

23

-

2 5

25

d

=

27 15

2 0

1 d

=

28 15

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