93ç-ç‚Bø ‚25

93ç-ç‚Bø ‚25

1. Find the arc length function s(t), unit tangent vector T (t), unit normal vector N(t), and the curvature κ(t) of the curve

R(t) =

e^t, e^−t,√ 2t

, t≥ 0, with initial point t = 0.

Solution:

S(t) = e^t− e^−t

−

→T(t) = 1 e^t+ e^−t

D

e^t,−e^−t,√ 2E

or 1 e^2t+ 1

D

e^2t,−1,√ 2e^tE

−

→N(t) = 1 e^t+ e^−t

D√ 2,√

2, e^−t− e^tE

or 1 e^2t+ 1

D√ 2e^t,√

2e^t,1 − e^2tE κ(t) =

√2

(e^t+ e^−t)² or

√2e^2t (e^2t+ 1)².

2. Let z = f (x, y) be a function with continuous second-order partial derivatives, and x = u²cos v, y = u²sin v. Express the Laplacian ∂²z

∂x² +∂²z

∂y² in terms of u, v,

∂z

∂u, ∂z

∂v, ∂²z

∂u², ∂²z

∂v². Solution:

∂z

∂u = 2u cos vfx+ 2u sin vfy

∂z

∂v = −u²sin vfx+ u²cos vfy

∂²z

∂u² = 4u²cos²vf_xx+ 4u²sin²vf_yy+ 8u²sin v cos vfxy+ 2 cos vfx+ 2 sin vfy

∂²z

∂v² = u⁴sin²vf_xx + u⁴cos²vf_yy− 2u⁴sin v cos vfxy − u²cos vfx− u²sin vfy

∂²z

∂u∂v = −2u³sin v cos vfxx+2u³sin v cos vfyy+2u³ cos²v− sin²v
fxy−2u sin vf^x+2u cos vfy

1 u⁴

∂²z

∂v² + 1 4u²

∂²z

∂u² = fxx+ fyy− 1

u² cos vfx− 1

u² sin vfy +1 2

1

u²cos vfx+1 2

1

u²sin vfy

= fxx+ fyy− 1 4u³

∂z

∂u

⇒ 1 u⁴

∂²z

∂v² + 1 4u²

∂²z

∂u² + 1 4u³

∂z

∂u = fxx+ fyy

3. Suppose f (x, y) is differentiable at (a, b), u =< 1, 0 >, v =

 1

√2, 1

√2

, Duf(a, b) = 3, and Dvf(a, b) =√

2.

1

(a) Find ∇f(a, b).

(b) What is the maximum of Dwf(a, b) for any w.

(c) Find all unit vectors w = hw¹, w2i such that D^wf(a, b) = 0.

Solution:

(a) ∇f(a, b) = (3, −1) = 3i + (−1)j (b) ç −→w =

 3

√10,√−1 10

 v, D−→wf(a, b) =√

10, |×M

(c) ç −→w = ±

 1

√10, 3

√10

 v, D_wf(a, b) = 0

4. Given f (x, y) = 3xy − 2xy²− x²y.

(a) Find all critical points, and determine its nature.

(b) Use the method of Lagrange multipliers to find the maximum and minimum of f (x, y) on the plane region: xy ≥ 1

8, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 (You must use the indicated method in order to receive any partial credits.)

5. Consider the iterated integral Z ¹

0

Z √

1−y²

1−y

1

px²+ y² dx

! dy

(a) Sketch the region of integration.

(b) Evaluate the integral by reversing the order of integration.

Solution:

(a)

2

(b)

Z ¹

0

Z √

1−y²

1−y

1

px²+ y²dx dy = Z ^π₂

0

Z ¹

1 cos θ+sin θ

1

rr dr dθ

= Z ^π₂

0



1 − 1

cos θ + sin θ

 dθ

= π 2 −

Z ^π₂

0

1

cos θ + sin θ dθ

= π 2 −√

2 ln√ 2 + 1

3