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

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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.