Mathematical Document: Plane Equation, Function Behavior, and Line Intersection - Prof. E., Exams of Mathematics

Download Mathematical Document: Plane Equation, Function Behavior, and Line Intersection - Prof. E. and more Exams Mathematics in PDF only on Docsity!

Solutions to Second Exam, Math 114, Fall 2002

Question 1 Which of the following points lies on the same plane as (1, 2 , 0), (2, 2 , 1),

and (0, 1 , 1)?

(A) (1, 1 , 1) (B) (4, − 1 , 1) (C) (4, 1 , 1)

(D) (1, 1 , −1) (E) (2, − 1 , 3) (F) (2, 1 , 3)

Answer 1 We use P 0 = (1, 2 , 0) as the base point. Then vectors parallel to the plane

are given by

~u = (2 − 1)̂ı + (2 − 2)̂ + (1 − 0)̂k

= ̂ı + ̂k

~v = (0 − 1)̂ı + (1 − 2)̂ + (1 − 0)̂k

= −̂ı − ̂ + ̂k.

The normal vector to the plane is given by the cross-product, ~n = ~u × ~v.

~n =

∣

∣

∣

∣

∣

∣̂

ı ̂ ̂k 1 0 1 − 1 − 1 1

∣ ∣ ∣ ∣ ∣ ∣

= ̂ı − 2 ̂ − ̂k

So the plane is given by the equation ~n ·

−−→

P 0 P = 0, which is

(x − 1) − 2(y − 2) − (z − 0) = 0

or more simply

x + 3 = 2y + z.

By inspection we see that the only solution is the point (2, 1 , 3) corresponding to choice (F).

Question 2 Consider the function

f (x, y) = x 2 − 2 x cos (xy)

Which of the following statements correctly describes the behavior of f (x, y)?

(I) f has infinitely many critical points.

(II) f has a saddle point at (1, 2 π).

(III) f has a local maximum at (1, 0).

(IV) f has a local minimum at (− 1 , π).

(A) only (II) (B) (I) and (II) (C) (III) and (IV)

(D) only (III) (E) (I) and (IV) (F) (I), (II) and (IV)

Answer 2 First we compute the critical points. The equation fx = 0 implies

2 x − 2 cos (xy) + 2xy sin (xy) = 0

while fy = 0 implies

2 x 2 sin (xy) = 0

From this second equation, we have either x = 0 or sin (xy) = 0. If x = 0, then the first

equation implies cos 0 = 0, which is impossible. So the only solutions are when sin (xy) = 0.

This happens when xy = nπ for some integer n. Then we have cos (xy) = cos (nπ) = (−1)n.

So fx = 0 becomes

2 x − 2(−1) n = 0

which implies x = (−1)n, so y = (−1)nnπ. Thus we have infinitely many critical points, so

part (I) is true.

Now we compute the second derivatives to determine information the local behavior of

the critical points.

fxx = 2 + 4y sin (xy) + 2xy 2 cos (xy), fxy = 4x sin (xy) + 2x 2 y cos (xy), fyy = 2x 3 cos (xy)

At the point x = (−1) n , y = (−1) n nπ, we have

fxx = 2 + 2n^2 π^2 , fxy = 2nπ, fyy = 2

Since fxx > 0 for any n, and since

fxxfyy − f 2 xy = 4 + 4n

2 π 2 − 4 n 2 π 2 = 4 > 0

for any n, every critical point is a local minimum.

So part (IV) is true and parts (II) and (III) are false. Thus the correct choice is (E).

Question 3 Find the point at which the plane passing through P = (1, 0 , −1) and

perpendicular to the line

−→

r (t) = (2 − t)̂ı + (2 + t)̂ + t̂k intersects the line.

(A)

(

1 2 ,^

7 2 ,^

3 2

)

(B)

(

8 3 ,^

4 3 ,^ −

2 3

)

(C)

(

3 2 ,^

5 2 ,^

1 2

)

(D)

(

5 3

, 7

3

, 1

3

)

(E) (3, 1 , −1) (F) (− 1 , 5 , 3)

Answer 3 A vector parallel to the line

x = 2 − t, y = 2 + t, z = t

is

−→

u = −̂ı + ̂ + ̂k (just take the coefficients of t). So the normal vector to the desired plane

is

−→

n = −̂ı + ̂ + k̂. The plane with normal vector

−→

n = −̂ı + ̂ + ̂k and passing through P = (1, 0 , −1) is

given by

(−1)(x − 1) + (1)(y) + (1)(z + 1) = 0

which simplifies to

−x + y + z + 2 = 0

When does the line intersect this plane? Plug in the equations of the line:

−(2 − t) + (2 + t) + t + 2 = 0

3 t + 2 = 0

t = − 2 3

So x = 2 + 23 = 83 , y = 2 − 23 = 43 , and z = −^23. So the correct choice is (B).

Question 4 Which of these functions satisfies the equation

fxx + fyy = −f

(I) f (x, y) = sin (xy)

(II) f (x, y) = sin x + sin y

(III) f (x, y) = sin (x + y)

(IV) f (x, y) = sin x sin y

(A) (I) only (B) (II) only (C) (II) and (III)

(D) (III) and (IV) (E) (I), (II), (IV) (F) (I), (III), (IV)

Answer 4 We compute fxx and fyy for each choice.

• (I) fxx = −y^2 sin (xy) and fyy = −x^2 sin (xy), so fxx + fyy = −(x^2 + y^2 ) sin (xy) 6 = −f. So (I) is false.
• (II) fxx = − sin x and fyy = − sin y, so fxx + fyy = − sin x − sin y = −f. So (II) is true.
• (III) fxx = − sin (x + y) and fyy = − sin (x + y), so fxx + fyy = −2 sin (x + y) 6 = −f. So (III) is false.
• (IV) fxx = − sin x sin y and fyy = − sin x sin y, so fxx + fyy = −2 sin x sin y 6 = −f. So (IV) is false.

Thus the correct answer is (B).

Question 5 The line of intersection of the two tangent planes to the surfaces xyz = 1

and x^2 + 3y^2 + z^2 = 5 at the point P = (1, − 1 , −1) intersects the yz-plane at the point

(A) (0, 1 , −1) (B) (0, − 1 , 1) (C) (0, 2 , −1)

(D) (0, − 1 , 2) (E) (0, − 1 , −2) (F) (0, − 2 , −2)

Answer 5 To find the line of intersection, we need to find a vector ~v to which it is

parallel. We know this vector lies in both tangent planes, so it must be perpendicular to the

normals of both planes. So if n~ 1 and n~ 2 are the normals to the two planes, ~v = n~ 1 × n~ 2

To find n~ 1 and n~ 2 , we use the fact that the gradient is always perpendicular to the level

surfaces. So

n ~ 1 = ∇(xyz)

∣

∣

∣

(1,− 1 ,−1)

=

(

yẑ ı + xẑ  + xŷ k

)∣∣

∣

(1,− 1 ,−1)

= ̂ı − ̂ − ̂k

and

n ~ 2 = ∇(x 2

• 3y 2
• z 2 )

∣

∣

∣

(1,− 1 ,−1)

=

(

2 x̂ ı + 6ŷ  + 2z ̂k

)∣∣

∣

(1,− 1 ,−1)

= 2̂ı − 6 ̂ − 2 ̂k

The cross product of these two vectors is

~v =

∣

∣

∣

∣

∣

∣̂

ı ̂ ̂k 1 − 1 − 1 2 − 6 − 2

∣ ∣ ∣ ∣ ∣ ∣

= − 4 ̂ı − 4 ̂k

So the line parallel to − 4 ̂ı − 4 ̂k and passing through (1, − 1 , −1) is given parametrically by

the equations

x = 1 − 4 t, y = − 1 , z = − 1 − 4 t

This crosses the plane x = 0 when t = 1

At t = 1 4 , we have^ x^ = 0,^ y^ =^ −1, and^ z^ =^ −2. So the answer is^ (E).

Question 6 Find the length of the shortest segment which joins a point on the curve

x = −s 2 , y = s

to a point on the curve

x = 2t, y = 3 − t.

Hint: Minimize the square of the distance as a function of s and t.

(A)

√

2 (B)

√

3 (C) 2

(D)

√

5 (E)

√

6 (F)

√

Answer 6 The square of the distance is

f (s, t) = [x(t) − x(s)] 2

• [y(t) − y(s)] 2

= (2t + s 2 ) 2

• (3 − t − s) 2

To minimize f , we find the critical points by setting the partial derivatives fs and ft equal

to zero.

∂f

∂s

= 2(2t + s^2 )(2s) + 2(3 − t − s)(−1) = 0

∂f

∂t

= 2(2t + s 2 )(2) + 2(3 − t − s)(−1) = 0

If we subtract the second equation from the first, we obtain

2(2t + s 2 )(2s − 2) = 0

which implies that either 2t + s^2 = 0 or that s = 1. 2t + s^2 = 0 implies that t + s = 3, so

that s 2 − 2 s + 6 = 0. This equation has no solution for s, so 2t + s 2 = 0 is impossible. So

s = 1.

Plugging s = 1 into the equation fs = 0, we get

2(2t + 1) + (t − 2) = 0

which implies t = 0. So the square of the distance is minimized at t = 0, s = 1. The distance

at t = 0, s = 1 is (^) √

f (1, 0) =

√

12 + 2^2 =

√

So the correct answer is (D).

Question 7 Consider the surface given in spherical coordinates by

ρ =

2 cos φ

cos (2φ)

, 0 ≤ φ ≤

π

4

Which of the following is the cartesian equation of the curve at which this surface intersects

the plane z = 3? Hint: cos 2φ = cos^2 φ − sin^2 φ.

(A) x 2

• y 2 = 0 (B) x 2 − y = 2 (C) x 2 − y 2 = 3

(D) x^2 + y^2 = 3 (E) x^2 − y^2 = 0 (F) (x + y)^2 = 1

Answer 7 First we obviously use the hint, then we look for terms that can be eliminated

using the formulas

x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.

We obtain

ρ =

2 cos φ

cos^2 φ − sin 2 φ

ρ(cos 2 φ − sin 2 φ) = 2 cos φ

ρ 2 cos 2 φ − ρ 2 sin 2 φ = 2ρ cos φ

z 2 − (x 2

• y 2 ) = 2z

Now using the equation z = 3, we get

x 2

• y 2 = z 2 − 2 z = 3

So the correct answer is choice (D).

Question 8 The absolute maximum of the function f (x, y) = xy subject to the con-

straint

9 x 2

• 4y 2 ≤ 36

is

(A) 3 (B) 15 (C) 2

√

(D) 9 (E) 12

√

2 (F) does not exist

Answer 8 To determine the absolute maximum, we first determine the critical points

of f inside the ellipse. So we solve the equations fx = y = 0 and fy = x = 0 to get (0, 0) as

one critical point.

Then we maximize f (x, y) = xy on the curve g(x, y) = 9x^2 + 4y^2 = 36. Using the

Lagrange multiplier method, we get

y = 18λx

x = 8λy

The solution is x = 0, y = 0, which does not satisfy the constraint, or λ = ± 1

If λ = 121 , then y = 32 x, and plugging this into the constraint gives 9x^2 + 9x^2 = 36, so

x = ±

√

2. So the critical points here are (

√

2 , 32

√

2. and (−

√

2 , −^32

√

2).

If λ = − 1 12 , then^ y^ =^ −

3 2 x, and the same computation yields the critical points (

√

2 , −

3 2

√

2)

and (−

√

2 ,

3 2

√

2).

Now we plug the five critical points into f. We get

f (0, 0) = 0, f (±

√

2 , ±

3 2

√

2. = 3, f (±

√

2 , ∓

3 2

√

2) = − 3

So the absolute maximum is 3, and thus the correct choice is (A).

Question 9 Find the directional derivative of

f (x, y, z) = cos (xy) + e yz − ln (xz)

at the point (1, 0 , 2) in the direction from (1, 0 , 2) to (2, 3 , 4).

Answer 9 First we find a unit vector in the desired direction. The vector is ~v = ̂ı+3̂ +2 ̂k,

and the length is |~v| =

√

12 + 3^2 + 2^2 =

√

14. So the desired unit vector is

~u =

√

(̂ı + 3̂ + 2̂k)

Now we need the gradient of the function:

∇f =

∂f

∂x

̂

ı +

∂f

∂y

̂

 +

∂f

∂z

̂

k

=

[

− y sin (xy) − 1 x

]̂

ı +

[

− x sin (xy) + ze yz

]̂

 +

[

ye yz − 1 z

]̂

k

So at the point (1, 0 , 2), we have

∇f

∣

∣

(1, 0 ,2) = −̂ı + 2̂ − 2 ̂k

Thus the directional derivative is

D~uf = ∇f · ~u =

√

Question 10 Suppose f (x, y) is a smooth function. Suppose f (1, 2) ≈ 3 .1, f (1. 1 , 2) ≈

3 .4, and f (1, 2 .2) ≈ 2 .9. Find the best estimate for f (0. 9 , 2 .1) using linear approximation.

Answer 10 We want to find the linear approximation at (x 0 , y 0 ) = (1, 2) of f , so we

use

L(x, y) = f (1, 2) + fx(1, 2)(x − 1) + fy(1, 2)(y − 2)

We need to estimate fx and fy. To do this, write

fx(1, 2) ≈

f (1. 1 , 2) − f (1, 2)

1. 1

= 3

and

fy(1, 2) ≈

f (1, 2 .2) − f (1, 2)

1. 2

= − 1

Then we have

L(x, y) ≈ 3 .1 + 3(x − 1) − (y − 2)

So our approximation is

f (0. 9 , 2 .1) ≈ 3 .1 + 3(− 0 .1) − (0.1) = 2. 7

Question 11 Match the equations to their graphs. Enter your answers in the table at

the bottom of the page and show your reasoning on the next page.

(1) z = sin (x)y (2) z = y 3 (3) z = x 2

• y 2

(4) 2x + 3y + 4z = 0 (5) z = cos

(

π 1+x^2 +y^2

)

(6) z = x^2

(a)

0 2 4

x

• • 0 2 4

y

0

5

10

15

20

25

30

(b)

0 2 4

x

• • 0 2 4

y

0

2

4

(c)

0 2 4

x

• • 0 2 4

y

–0.

0

1

(d)

0 2 4

x

• • 0 2 4

y

0

2

4

(e)

0 2 4

x

• • 0 2 4

y

0

20

40

60

(f)

0 2 4

x

• • 0 2 4

y

0

2

4

6

8

10

12

14

16

Answer 11

(a) There are only two rotationally symmetric graphs, this one and (c). Since this one clearly resembles a paraboloid of revolution, it must be z = x^2 + y^2 , choice (3).

(b) This is the only planar graph, so it must be 2x + 3y + 4z = 0, choice (4).

(c) This is the other rotationally symmetric graph, so it must be choice (5).

(d) The graph is not obvious, until we look at the intersection with surfaces of constant y and see the sine graph. So it must be z = sin (x)y, choice (1).

(e) The profile is the same for all x, so the function must depend only on y, and must be z = y 3 , choice (2).

(f) The profile is the same for all y, so it depends only on x. So it is z = x 2 , choice (6).

Question 12 True or false. Explain your reasoning.

(a) The line ~r(t) = (1+2t)̂ı+(1+3t)̂+(1+4t)̂k is perpendicular to the plane 2x+3y− 4 z =

(b) The equation r = 3z in cylindrical coordinates, describes a cone.

(c) |~a × ~b| = 0 implies that either ~a = 0 or ~b = 0.

Answer 12

(a) This is false. The line is perpendicular to the plane if and only if the vector parallel to the line is perpendicular to the plane. This happens if and only if the vector along the line is a constant multiple of the normal vector of the plane.

The vector along the line is ~v = 2̂ı + 3̂ + 4̂k. The vector normal to the plane is ~n = 2̂ı + 3̂  − 4 ̂k. Since ~v is not parallel to ~n, this line is not perpendicular to the plane.

(b) This is true. In rectangular coordinates,

r = 3z =⇒ r 2 = 9z 2 =⇒ x 2

• y 2 = 9z 2

which is the equation of a cone through the origin, compressed by a factor of 3 in the z direction.

(c) This is false. |~a × ~b| = 0 may be zero if ~a is parallel to ~b. For example, if ~a = ̂ı and ~b = ̂ı, then ~a × ~b = 0.