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### Transcript of Ma 233: Calculus III Solutions to Midterm Examination 2 2011-11-27¢  Ma 233: Calculus III...

• Ma 233: Calculus III Solutions to Midterm Examination 2

Profs. Krishtal, Ravindra, and Wickerhauser 18 questions on 18 pages

Monday, March 14th, 2005

1. Find parametric equations for the tangent line at the point (1/2,− √

3/2,−π/3) on the

curve r(t) = (cos t, sin t, t)

(a) (1/2,− √

3/2,−π/3) + t 〈 1/2,

√ 3/2, 1

〉 (b) (1/2,−

√ 3/2,−π/3) + t

〈√ 3/2, 1/2, 1

〉 ⇐

(c) (1/2,− √

3/2,−π/3) + t 〈 −1/2,

√ 3/2, 1

〉 (d) (1/2,−

√ 3/2,−π/3) + t

〈 − √

3/2, 1/2, 1 〉

(e) (1/2, √

3/2, 1) + t 〈 1/2,−

√ 3/2,−π/3

〉 (f) (

√ 3/2, 1/2, 1) + t

〈 1/2,−

√ 3/2,−π/3

〉 (g) (−1/2,

√ 3/2, 1) + t

〈 1/2,−

√ 3/2,−π/3

〉 (h) (−

√ 3/2, 1/2, 1) + t

〈 1/2,−

√ 3/2,−π/3

Solution: Differentiate to get r′(t) = (− sin t, cos t, 1). The tangent line at t = −π/3

has direction vector r′(−π/3) = (− sin(−π/3), cos(−π/3), 1) = ( √

3/2, 1/2, 1) and base

point r(−π/3) = (1/2,− √

3/2,−π/3), so it may be written as

(1/2,− √

3/2,−π/3) + t 〈√

3/2, 1/2, 1 〉

1

• The parameteric equations of the tangent line are thus

x(t) = 1

2 +

√ 3

2 t

y(t) = − √

3

2 +

√ 3

2 t

z(t) = −π 3

+ t

2

2

• 2. If r(t) = cos(7t)i + sin(−5t)j + 6tk, compute ∫ π/2 0

r(t) dt.

(a) 1 7 i− 1

5 j + 3π

2

4 k.

(b) 1 7 i + 1

5 j + 3π

2

4 k.

(c) 1 7 i− 1

5 j− 3π2

4 k.

(d) −1 7 i− 1

5 j + 3π

2

4 k. ⇐

(e) i− j + 3πk.

(f) i + j + 3πk.

(g) i− j− 3πk.

(h) −i− j + 3πk.

Solution: Integrate the components separately to get the antiderivative

1

7 sin(7t)i +

1

5 cos(−5t)j + 6t

2

2 k

Evaluating the difference between 0 and π/2 gives

−1 7 i− 1

5 j +

3π2

4 k

2

3

• 3. Find the length of the curve

{r(t) = cos(5t)i + sin(5t)j + 6tk : −2 ≤ t ≤ 6}

(a) 8

(b) 48

(c) 2 √

61

(d) 4 √

61

(e) 8 √

61 ⇐

(f) 2 √

86

(g) 4 √

86

(h) 8 √

86

Solution: Find r′(t) = −5 sin(5t)i + 5 cos(5t)j + 6k and

‖r′(t)‖ = √

25 sin2(5t) + 25 cos2(5t) + 36 = √

61.

Integrate this to get the length

∫ 6 −2 ‖r′(t)‖ dt = 8

√ 61.

2

4

• 4. Given that a point has acceleration a(t) = 〈1, 2t,−3t + 1〉, its position is 〈1, 1, 1〉 at

t = 0 and its velocity is 〈−2,−2,−2〉 at t = 0, find its position at all times t.

(a) r(t) = 〈

1 2 t2 − t + 2, 1

3 t3 − t + 2, −1

2 t3 + 1

2 t2 − t + 2

〉 (b) r(t) =

〈 1 2 t2 + t + 2, 1

3 t3 + t + 2, −1

2 t3 + 1

2 t2 + t + 2

〉 (c) r(t) =

〈 1 2 t2 + t− 2, 1

3 t3 + t− 2, −1

2 t3 + 1

2 t2 + t− 2

〉 (d) r(t) =

〈 1 2 t2 − t− 2, 1

3 t3 − t− 2, −1

2 t3 + 1

2 t2 − t− 2

〉 (e) r(t) =

〈 1 2 t2 − 2t + 1, 1

3 t3 − 2t + 1, −1

2 t3 + 1

2 t2 − 2t + 1

〉 ⇐

(f) r(t) = 〈

1 2 t2 + 2t + 1, 1

3 t3 + 2t + 1, −1

2 t3 + 1

2 t2 + 2t + 1

〉 (g) r(t) =

〈 1 2 t2 + 2t− 1, 1

3 t3 + 2t− 1, −1

2 t3 + 1

2 t2 + 2t− 1

〉 (h) r(t) =

〈 1 2 t2 − 2t− 1, 1

3 t3 − 2t− 1, −1

2 t3 + 1

2 t2 − 2t− 1

Solution: Velocity, the antiderivative of acceleration, is

v(t) = 〈 t + cx, t

2 + cy, − 3

2 t2 + t + cz

〉 ,

where cx, cy, cz are constants of integration. Determine these from the condition

〈−2,−2,−2〉 = v(0) = 〈cx, cy, cz〉, so cx = −2, cy = −2, cz = −2 and v(t) =〈 t− 2, t2 − 2, −3

2 t2 + t− 2

〉 .

Position, the antiderivative of velocity, is

r(t) = 〈

1

2 t2 − 2t + kx,

1

3 t3 − 2t + ky, −

1

2 t3 +

1

2 t2 − 2t + kz

〉 ,

where kx, ky, kz are constants of integration. Determine these from the condition

〈1, 1, 1〉 = r(0) = 〈kx, ky, kz〉, so kx = 1, ky = 1, kz = 1 and

r(t) = 〈

1

2 t2 − 2t + 1, 1

3 t3 − 2t + 1, −1

2 t3 +

1

2 t2 − 2t + 1

2

5

• 5. Find the curvature of y = cos(3x) at x = π/4

(a) 3/2

(b) 2/3

(c) 11/132/3

(d) 11/133/2

(e) 13/112/3

(f) 13/183/2

(g) 18/112/3

(h) 18/113/2 ⇐

(i) 11/183/2

(j) 13/113/2

Solution: This is the special case of a curve being the graph of a function: y =

f(x) = cos(3x). Therefore, use the formula

κ(x) = |f ′′(x)|

(1 + [f ′(x)]2)3/2

But f ′(x) = −3 sin(3x) and f ′′(x) = −9 cos(3x), so the formula expands to

κ(x) = |9 cos(3x)|

(1 + 9 sin2(3x))3/2

Thus

κ(π/4) = | − 9/

√ 2|

(1 + 9/2)3/2 =

18

113/2

2

6

• 6. Let r(t) = (cos t)i + (sin t)j + 2tk. Find the unit binormal vector B(t) for all t.

(a) (− sin t)i + (cos t)j + 2k

(b) (sin t)i + (− cos t)j + 2k

(c) − sin t√ 5

i + cos t√ 5 j + 2√

5 k

(d) sin t√ 5 i + cos t√

5 j + 2√

5 k

(e) − cos t√ 5

i + − sin t√ 5

j + 0k

(f) − cos t√ 5

i + sin t√ 5 j + 0k

(g) − cos(t)i− sin(t)j + 0k

(h) cos(t)i + sin(t)j + 0k

(i) −2 sin t√ 5

i + −2 cos t√ 5

j + 1√ 5 k

(j) 2 sin t√ 5

i + −2 cos t√ 5

j + 1√ 5 k ⇐

Solution: Tangent vector: r′(t) = (− sin t)i+(cos t)j+2k. Its length is the constant

‖r′(t)‖ = √

sin2 t + cos2 t + 4 = √

5 for all t, so the unit tangent vector is

T(t) = r′(t)

‖r′(t)‖ = − sin t√

5 i +

cos t√ 5

j + 2√ 5 k.

Normal vector:

T′(t) = − cos t√

5 i +

− sin t√ 5

j + 0k,

and its length is

‖T′(t)‖ = √

(− cos t)2/5 + (sin t)2/5 = 1/ √

5.

Hence the unit normal vector is

N(t) = T′(t)

‖T′(t)‖ = − cos(t)i− sin(t)j + 0k.

Binormal vector:

B(t) = T(t)×N(t) = 2 sin t√ 5

i + −2 cos t√

5 j +

1√ 5 k

2

7

• 7. Match the following surfaces with the verbal description of their level curves:

1. z = √

9− x2 − y2 2. z = y2 − x2

3. z = 1

y − 3

X. a collection of parallel lines Y. a collection of circles Z. two lines and a collection of hyperbolas

(a) 1 is X; 2 is Y; 3 is Z.

(b) 1 is Y; 2 is Z; 3 is X. ⇐

(c) 1 is Z; 2 is X; 3 is Y.

(d) 1 is X; 2 is Z; 3 is Y.

(e) 1 is Z; 2 is Y; 3 is X.

(f) 1 is Y; 2 is X; 3 is Z.

(g) All of them are X.

(h) All of them are Y.

(i) All of them are Z.

(j) None of the above.

Solution: Equation 1 is a hemisphere, with circles (Y) as level sets. Equation 2

is a hyperbolic paraboloid, with lines and hyperbolas (Z) as level sets. Equation 3 is

hyperbolic cylinder with parallel lines (X) as level sets. 2

8

• 8. Find lim (x,y)→(0,0)

x2 − 3xy + y2

(x− y)2 , if it exists.

(a) −4

(b) −3

(c) −2

(d) −1

(e) 0

(f) 1

(g) 2

(h) 3

(i) 4

(j) The limit does not exist. ⇐

Solution: The limit does not exist. Along the line y = 0 through (0, 0), the ratio

is constantly 1, while along the line y = −x through (0, 0), the ratio is constantly 5 4 .

Since these do not agree at (0, 0), there can be no limit at (0, 0). 2

9

• 9. Find the partial derivative fxxy for the function f(x, y) = e xy2 .

(a) y6exy 2

(b) 4y3exy 2

(c) 4y3exy 2 + 2y5exy

2

(d) 4y3exy 2 + 2xy5exy

2 ⇐

(e) 2yexy 2 + 3xy5exy

2

(f) 2yexy 2 + 2xy3exy

2

(g) 3xy3exy 2

(h) y5xexy 2

(i) The derivative exists, but is none of the above.

(j) The derivative does not exist.

Solution: fx = y 2exy

2 ; fxx = y

4exy 2 ; fxxy = 4y

3exy 2 + 2xy5exy

2 ; 2

10

• 10. Determine whether each of the following functions is a solution to Laplace’s equation

uxx + uyy = 0.

(I) u(x, y) = x2 + y2

(II) u(x, y) = x2 − y2

(III) u(x, y) = log √

x2 + y2

(a) I only.

(b) II only.

(c) III only.

(d) IV only.

(e) I and II only.

(f) I and III only.