Calculus II, Spring 2014

Brief Solutions to Midterm 2

May 08, 2014 Show all details.

1. (10%) Evaluate

\frac{d}{dy} ∫_{1}^{2+y^{2}} \frac{cos(xy)}{x}dx

Ans:

= \frac{cos(xy)}{x}|_{x=2+y^{2}} + ∫_{1}^{2+y^{2}} \frac{d}{dy} \frac{cos(xy)}{x}dx

= \frac{cos((2 + y^{2})y)}{2 + y^{2}} + ∫_{1}^{2+y^{2}} − sin(xy)dx (8%)

= \frac{cos((2 + y^{2})y)}{2 + y^{2}} + \frac{1}{y}( cos((2 + y^{2})y) − cos y) (2%)

2. (10%) True or false? Prove it or give a counter example.

Assume f(x, y), f_{x}(x, y) and f_{y}(x, y) are all continuous in R^{2}. Let C = {(x, y), f(x, y) = f(0, 0)} and T be a tangent vector of C at (0, 0). Then ∇f(0, 0) · T = 0. Ans: True (3%). Proof: C is a curve. We may assume it is parametrized by (x(t), y(t)) with (x(0), y(0)) = (0, 0). Therefore T = (x′(0), y′(0)). Since f is constant on C, we have

0 = \frac{d}{dt}f(x(t), y(t))|_{t=0} = ∇f(0, 0) · T (7%).

3. (10%) Use Lagrangian multipliers (and only Lagrangian multipliers) to find extreme values of

f(x, y, z) = xy + 2z^{2} on

\{ \begin{matrix}x^{2} + y^{2} + z^{2} = 9 \\ x − y = 0 \end{matrix}

Ans: Let g_{1}(x, y, z) = x^{2} + y^{2} + z^{2} − 9, g_{2}(x, y, z) = x − y. Solve from

\{ \begin{matrix}g_{1}(x, y, z) = 0, \\ g_{2}(x, y, z) = 0, \\ ∇f(x, y, z) = λ_{1}∇g_{1}(x, y, z) + λ_{2}∇g_{2}(x, y, z) \end{matrix} (6%).

Max: (x, y, z) = (0, 0, ±3). f_{max} = 18, (2%). Min: (x, y, z) = ±(\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0). f_{min} = \frac{9}{2}, (2%).

4. (10%) Find the absolute maximum and minimum of f(x, y) = 2 + 2x + 2y − x^{2} − y^{2} in the region bounded by x = 0, y = 0 and x + y = 6. Ans:

(a) Interior points.

\{ \begin{matrix}f_{x}(x, y) = 2 − 2x = 0 \\ f_{y}(x, y) = 2 − 2y = 0 \end{matrix} ⇒ \{ \begin{matrix}x = 1 \\ y = 1 \end{matrix}

The critical point of f is (1, 1). f(1, 1) = 4. (2%)

(b) Boundary points. (i) Line segment (0, 0)–(0, 6) (x = 0).

f(0, y) = 2 + 2y − y^{2}, 0 ≤ y ≤ 6.

\frac{d}{dy}f(0, y) = 2 − 2y = 0 implies y = 1. f(0, 1) = 3. (1%) (ii) Line segment (0, 0)–(6, 0) (y = 0).

f(x, 0) = 2 + 2x − x^{2}, 0 ≤ x ≤ 6.

\frac{d}{dx}f(x, 0) = 2 − 2x = 0 implies x = 1. f(1, 0) = 3. (1%) (iii) Line segment (0, 6)–(6, 0) (y = 6 − x).

f(x, 6 − x) = 2 + 2x + 2(6 − x) − x^{2} − (6 − x)^{2} = −2x^{2} + 12x − 22, 0 ≤ x ≤ 6.

\frac{d}{dx}f(x, 6 − x) = −4x + 12 = 0 implies x = 3. f(3, 3) = −4. (1%) (iv) Endpoints of above three line segents.

f(0, 0) = 2, f(0, 6) = −22, f(6, 0) = −22. (3%) Therefore,

absolute maximum = f(1, 1) = 4,
absolute mimimum = f(0, 6) = f(6, 0) = −22.  (2%)

5. (10%) Let f(x, y) = x^{3} + y^{3} and g(r, θ) = f(r cos θ, r sin θ). Evaluate ∂_{r}^{2}g + (∂_{r}g)/r + (∂_{θ}^{2}g)/r^{2} Ans:

\frac{∂g}{∂r} = \frac{∂f}{∂x} \frac{∂x}{∂r} + \frac{∂f}{∂y} \frac{∂y}{∂r} = 3x^{2} · cos θ + 3y^{2} sin θ = 3r^{2} cos^{3} θ + 3r^{2} sin^{3} θ,  (2%)
\frac{∂^{2}g}{∂r^{2}} = \frac{∂}{∂r}(\frac{∂g}{∂r}) = 6r cos^{3} θ + 6r sin^{3} θ,  (2%)
\frac{∂g}{∂θ} = \frac{∂f}{∂x} \frac{∂x}{∂θ} + \frac{∂f}{∂y} \frac{∂y}{∂θ} = −3r^{3} cos^{2} θ sin θ + 3r^{3} cos θ sin^{2} θ,  (2%)
\frac{∂^{2}g}{∂θ^{2}} = \frac{∂}{∂θ}(\frac{∂g}{∂θ}) = 3r^{3}(3 sin^{2} θ cos θ + 3 cos^{2} θ sin θ − cos θ − sin θ).  (2%)

Therefore,

\frac{∂^{2}g}{∂θ^{2}} + \frac{1}{r} \frac{∂g}{∂r} + \frac{1}{r^{2}} \frac{∂^{2}g}{∂θ^{2}} = (6r cos^{3} θ + 6r sin^{3} θ) + (3r cos^{3} θ + 3r sin^{3} θ)
+ 3r(3 sin^{2} θ cos θ + 3 cos^{2} θ sin θ − cos θ − sin θ)
= 6r cos θ + 6r sin θ.  (2%)

6. (10%) Derive the Taylor expansion of f(x, y, z) around (0, 0, 0) up to quadratic terms of x, y and z and an expression of the remainder term, R_{2}. You may assume that f and all its first and second derivatives are all continuous in R^{3}.

Ans: Assume that f and all its first and second derivatives are continuous on R^{3}. Parametrize the line segment joining O(0, 0, 0) to P (h, k, ℓ) by

\{ \begin{matrix}x = th \\ y = tk \\ z = tℓ \end{matrix} , 0 ≤ t ≤ 1.

Let F (t) = f(th, tk, tℓ). Then (2%)

F′(t) = \frac{∂f}{∂x} \frac{∂x}{∂t} + \frac{∂f}{∂y} \frac{∂y}{∂t} + \frac{∂f}{∂z} \frac{∂z}{∂t} = hf_{x} + kf_{y} + ℓf_{z},  (1%)
F′′(t) = h(hf_{xx} + kf_{yy} + ℓf_{xz}) + k(hf_{yx} + kf_{yy} + ℓf_{yz}) + ℓ(hf_{zx} + kf_{zy} + ℓf_{zz})
= h^{2}f_{xx} + k^{2}f_{yy} + ℓ^{2}f_{zz} + 2hkf_{xy} + 2kℓf_{yz} + 2hℓf_{xz},  (2%)
F′′′(t) = h^{2}(hf_{xxx} + kf_{xxy} + ℓf_{xxz}) + k^{2}(hf_{yyx} + kf_{yyy} + ℓf_{yyz})
+ ℓ^{2}(hf_{zzx} + kf_{zzy} + ℓf_{zzz}) + 2hk(hf_{xyz} + kf_{xyy} + ℓf_{xyz})
+ 2kℓ(hf_{yzx} + kf_{yzy} + ℓf_{yzz}) + 2hℓ(hf_{xzx} + kf_{xzy} + ℓf_{xzz})
= h^{3}f_{xxx} + k^{3}f_{yyy} + ℓ^{3}f_{zzz} + 3h^{2}kf_{xxy} + 3h^{2}ℓf_{xxz}
+ 3hk^{2}f_{xyy} + 3k^{2}ℓf_{yyz} + 3hℓ^{2}f_{xzz} + 3kℓ^{2}f_{yzz} + 6hkℓf_{xyz},  (2%)

where all the partial derivatives of f are evaluated at (th, tk, tℓ). By Taylor’s theorem, there exists c ∈ (0, 1) such that

F (1) = F (0) + F′(0)(1 − 0) + \frac{1}{2!}F′′(0)(1 − 0)^{2} + \frac{1}{3!}F′′′(c)(1 − 0)^{3},
f(h, k, ℓ) = (f + (hf_{x} + kf_{y} + ℓf_{z})

+ \frac{1}{2!}(h^{2}f_{xx} + k^{2}f_{yy} + ℓ^{2}f_{zz} + 2hkf_{xy} + 2kℓf_{yz} + 2hℓf_{xz}))|_{(0,0,0)}

+ (\frac{1}{3!}(h^{3}f_{xxx} + k^{3}f_{yyy} + ℓ^{3}f_{zzz} + 3h^{2}kf_{xxy} + 3h^{2}ℓf_{xxz}

+ 3hk^{2}f_{xyy} + 3k^{2}ℓf_{yyz} + 3hℓ^{2}f_{xzz} + 3kℓ^{2}f_{yzz} + 6hkℓf_{xyz}))|_{(ch,ck,cℓ)}. (1%)

Hence, the Taylor expansion of f(x, y, z) at (0, 0) up to second order is

f(x, y, z) = f(0, 0, 0) + (xf_{x} + yf_{y} + zf_{z})

+ \frac{1}{2!}(x^{2}f_{xx} + y^{2}f_{yy} + z^{2}f_{zz} + 2xyf_{xy} + 2yzf_{yz} + 2xzf_{xz})

+ \frac{1}{3!}(x^{3}f_{xxx} + y^{3}f_{yyy} + z^{3}f_{zzz} + 3x^{2}yf_{xxy} + 3x^{2}zf_{xxz}

+ 3xy^{2}f_{xyy} + 3y^{2}zf_{yyz} + 3xz^{2}f_{xzz} + 3yz^{2}f_{yzz} + 6xyzf_{xyz}),

where all the first and second derivatives are evaluated at (0, 0, 0), and all the third derivatives are evaluated at (cx, cy, cz) for some c ∈ (0, 1). (2%)

7. (10%) Evaluate (\frac{∂U}{∂P})_{V} and (\frac{∂U}{∂T})_{V} at (P, V, T) = (1, 2, 2) where U(P, V, T) = T exp(−P/V) with the constraint P V = T.

Ans:

we have

(\frac{∂U}{∂P})_{V} (1, 2, 2) = \frac{∂}{∂P} U(P, V, T (P, V))_{(1,2,2)} = (\frac{∂U}{∂P} + V \frac{∂U}{∂T} )_{(1,2,2)} (4%)

= − exp(\frac{−1}{2}) + 2 exp(\frac{−1}{2}) = exp(\frac{−1}{2}) (1%)

(\frac{∂U}{∂T})_{V} (1, 2, 2) = \frac{∂}{∂T} U(P (V, T), V, T)_{(1,2,2)} = (\frac{1}{V} \frac{∂U}{∂P} + \frac{∂U}{∂T} )_{(1,2,2)} (4%)

= −(\frac{1}{2}) exp(\frac{−1}{2}) + exp(\frac{−1}{2}) = \frac{1}{2} exp(\frac{−1}{2}) (1%)

8. (10%) Evaluate ∫_{0}^{2} ∫_{y}^{2} \frac{sin x}{x}dxdy.

Ans: we have

∫_{0}^{2} ∫_{y}^{2} \frac{sin(x)}{x}dxdy

= ∫_{0}^{2} ∫_{0}^{x} \frac{sin(x)}{x}dydx (6%)

= ∫_{0}^{2}(\frac{sin(x)}{x})(x)dx (2%)

= ∫_{0}^{2} sin(x)dx = 1 − cos(2) (2%)

9. (20%) Let f(x, y) = \frac{x^{3} + y^{3}}{x^{2} + y^{2}} , for (x, y) ≠ (0, 0) and f(0, 0) = 0. P = (0, 0) and u^{θ} =

(cos θ, sin θ), θ ∈ [0, 2π].

(a) Is f continuous at (0, 0)? Explain.

(b) For fixed θ, write down the definition of the directional derivative (\frac{df}{ds})_{u^{θ},P} and evaluate it.

(c) Does f have a linear approximation at (0, 0)? Explain. Ans:

(a) (6%) Let

x = r cos(u), y = r sin(u)

then

lim_{(x,y)→(0,0)} f(x, y) = lim_{r→0} \frac{r^{3}((cos(u))^{3} + (sin(u))^{3})}{r^{2}} = 0

then f is continuous at 0.

(b) (7%)

(\frac{df}{ds})_{u^{θ},P} = lim_{h→0} \frac{(f(h cos(θ), h cos(θ)) − f(0, 0))}{h}

= lim_{h→0} \frac{(h((cos(θ))^{3} + sin(θ))^{3})}{h} = (cos(θ))^{3} + (sin(θ)))^{3}

(c) (7%) since ∇f(0, 0) · u(θ) = (1, 1) · (cos(θ), sin(θ)) ≠ (cos(θ))^{3} + sin(θ))^{3}) then f is not differentiable at (0, 0) Therefore f does not have linear approximation.