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Final Solutions
APPM 2350, Calculus 3, Fall 2008
December 17, 2008
- Answers in table form: a.^ b.^ c.^ d.^ e.^ f.^ g.^ h.^ i.^ j. C B C D A C D D A B
(a) Implicit differentation gives a + b (^) dxdy = 0. Solving for the slope gives dy dx = − a b. The normal slope is (^) ab , so answer C. (b) This is B. (c) The ellipse is parametrized with x(t) = a cos t and y(t) = b sin t from 0 ≤ t ≤ 2 π. The arc length of this curve is given in answer C. (d) Consider the set of paths, y = kx^2 − x, for constants k 6 = 0. Then, xy x + y
kx^3 − x^2 kx^2
= x −
k
which goes to -1/k as x → 0. By the two-path test, this limit does not exist. Pick D. (e) As a function of t, f (x(t), y(t)) = sin^2 (t^2 ) + cos^2 (t^2 ) = 1, and its differential is df = 0dt. Therefore there is no uncertainty. Sorry Heisenberg, its A. (f) For f (x, y, z) =
xyz, we have fx =
√yz 2 √x ,^ fy^ =
√xz 2 √y , and^ fz^ =
√xy 2 √z.^ The linearization about (1,1,1) is given by f (1, 1 , 1) + fx(1, 1 , 1)(x − 1) + fy(1, 1 , 1)(y − 1) + fz (1, 1 , 1)(z − 1) = 1 +
(x + y + z − 3) = −
(1 − x − y − z)
This matches C. (g) f (x, y) = (tan x)(cos y), ∇f = sec^2 x cos y i − tan x sin y j at (0, 0) the direction of fastest decrease is −∇f (0, 0) = −i, which has θ = π, pick D. (h) Switching the ordering gives D. (i) This is A. (j) Answer B passes the test ∂M ∂y = ∂N ∂x , and its potential function is f (x, y) = x
(^2) y 2
The curve is given by C : {x(t) = t^2 , y(t) = t^3 − t, − 1 ≤ t ≤ 1 }.
(a) We have x′(t) = 2t and y′(t) = 3t^2 − 1, so x′(1/2) = 1 and y′(1/2) = − 1 /4. Therefore, dy dx = y′(1/2)/x′^ (1/2) = − 1 /4. The location of the point is (x(1/2), y(1/2)) = (1/ 4 , − 3 /8). The equations are: Tangent: (y + 3/8) = − 14 (x − 1 /4) Normal: (y + 3/8) = 4(x − 1 /4) (b) Set g(t) = f (x(t), y(t)) = t^2 +t^3 −t. Then, use the single-variable first-derivative test: g′(t) = 2t + 3t^2 − 1 = 0. The solutions to this quadratic equation are t = {− 1 , 1 / 3 }, but t = −1 is not on the interior of our interval. A second derivative test, g′′(t) = 6t + 2, g′′(1/3) = 4, shows there is a local minimum at t = 1/3. We also need to test the values of the endpoints t = −1 and t = 1.
t x(t) y(t) f (x(t), y(t)) Classification of Point -1 1 0 1 Global Maximum 1/3 1/9 -8/27 -5/27 Global Minimum 1 1 0 1 Global Maximum
(c) Using Green’s Theorem on the field F = yj, we have the following area formula: ∫ ∫
R
dA =
R
∂N
∂y
dA = −
C
N dx = −
C
y dx. (2)
C
y dx = −
− 1
(t^3 − t)(2t) dt = − 2
t^5 5
t^3 3
− 1
- (a) dP dt
∂P
∂g
dz dt
∂P
∂T
∂T
∂x
dx dt
∂T
∂y
dy dt
∂T
∂z
dz dt
∂P
∂z
dz dt
dP dt
t 1
(b) dP ds
dP dt
dt ds
|v|
dP dt
dP ds
t 1
32 + 4^2
(c) u = −∇T = −j − 2 k and set the other vector orthogonal to ∇T , w = i or w = 2j − k.
- f (x, y) = 2xy + 1
(a) The two-variable, first derivative test gives the linear system
(1) fx = 0 : 2 y = 0 (2) fy = 0 : 2 x = 0
Equation (1) implies y = 0 and Equation (2) shows x = 0. Therefore the only critical point is (0, 0). To classify this point, use the second derivative test. We have fxx = fyy = 0, fxy = 2, and the discriminant fxxfyy − f (^) xy^2 = −4. So (0,0) is a Saddle Point. (b) (1) fx = 0 : 2 y = λ 2 x (2) fy = 0 : 2 x = λ 2 y (12) First, solving Equation (1) for y gives y = λx. Plugging this into Equation (2) and simplifying gives x = λ^2 x. This implies x = 0 or λ^2 = 1. If x = 0, then y = 0 (by either (1) or (2)), and the point (0, 0) is not on the constraint. Thus, λ^2 = 1, meaning either λ = 1 or λ = −1. If λ = 1, then Equation (1) implies x = y. Plugging this into the constraint yields 2x^2 = 1, or x = ± 1 /
2, or (± 1 /
2). If λ = −1, then x = −y, which, plugged into the constraint, also yields x = ± 1 /
2, or (± 1 /
(c) The following table shows the values of f (x, y) and the classification for these five points: (x, y) f (x, y) Classification of Point (0, 0) 1 Saddle Point (1/
- 2 Global Maximum (1/
- 0 Global Minimum (− 1 /
- 0 Global Minimum (− 1 /
- 2 Global Maximum
(d) Let g(x, y) = 2xy−z = −1. Then ∇g = 2y i+2x j−k and |∇g| =
4 x^2 + 4y^2 + 1. This surface is most easily integrated by projecting into the xy plane, so let p = k and let R be the shadow, the unit circle, x^2 + y^2 ≤ 1. Employing the surface integral formula gives:
∫ ∫
S
dσ =
R
|∇g| |∇g · k| dA
R
4 x^2 + 4y^2 + 1 1
dA
