Mathematical Methods for Social Scientists Math 195 (Sec 55), Autumn 2006 Solutions for the Revision Sheet for Mid-term 2

1. (a) A function f: R^{2} → R is continuous at (x, y) if

lim_{(u,v)→(x,y)} f(u, v) = f(x, y)

(b) One example is

f(x, y) = \{ \begin{matrix}\frac{xy}{x^{2}+y^{2}} , & if (x, y) ≠ (0, 0); \\ 0, & if (x, y) = (0, 0). \end{matrix}

One can easily check that f(x, 0) = 0 for all x ∈ R and f(0, y) = 0 for all y ∈ R, however

lim_{(x,y)→(0,0)} f(x, y) ≠ f(0, 0) = 0,

since

lim_{x→0} f(x, x) = 1/2.

(c) We have

|\frac{6x^{2}y^{2}}{x^{2} + y^{2}} | = \frac{6x^{2}y^{2}}{x^{2} + y^{2}} ≤ \frac{6(x^{2} + y^{2})y^{2}}{x^{2} + y^{2}} = 6y^{2} ≤ 6(x^{2} + y^{2}).

For each ε > 0, set δ = \sqrt{ε/6}. So we have that if 0 < \sqrt{(x − 0)^{2} + (y − 0)^{2}} < δ then

|\frac{6x^{2}y^{2}}{x^{2} + y^{2}} − 0| ≤ 6(x^{2} + y^{2}) < 6δ^{2} = 6(\sqrt{ε/6})^{2} = ε.

Therefore, by definition, lim_{(x,y)→(0,0)} \frac{6x^{2}y^{2}}{x^{2}+y^{2}} = 0.

2. (a) For f: R^{2} → R the partial derivative of f with respect to x is

\frac{∂f}{∂x}(x, y) = lim_{h→0} \frac{f(x + h, y) − f(x, y)}{h}

and the partial derivative of f with respect to y is

\frac{∂f}{∂y} (x, y) = lim_{h→0} \frac{f(x, y + h) − f(x, y)}{h}

(b) Clairaut’s Theorem. If f: R^{2} → R is such that ∂^{2}f/∂x∂y and ∂^{2}f/∂y∂x are continuous in on a disc containing a point (x, y), then

\frac{∂^{2}f}{∂x∂y} (x, y) = \frac{∂^{2}f}{∂y∂x}(x, y)

(c) You should find that f_{xy} = 3(1 + x)y^{2}e^{x} = f_{yx}.

(d) Setting u(x, t) = f(x − ct) we can apply the chain rule to find

u_{xx}(x, t) = f′′(x − ct)

and

u_{tt}(x, t) = c^{2}f′′(x − ct).

Thus

u_{tt}(x, t) = c^{2}f′′(x − ct) = c^{2}u_{xx}(x, t)

and so u satisfies the wave equation. The answer is similar for f(x + ct).

3. (a)
z − z_{0} = f_{x}(x_{0}, y_{0})(x − x_{0}) + f_{y}(x_{0}, y_{0})(y − y_{0})

(b)

z − 8 = 4(x − 2) + 8(y − 1)

(c) Note: Point should have been (π, e, π).

z − π = (x − π) + \frac{π}{e} (y − e)

4. (a) If (x, y) ↦ f(x, y) is a differentiable function of x and y, and (s, t) ↦ g(s, t) and (s, t) ↦ h(s, t) are differentiable functions of s and t, then (s, t) ↦ f(g(s, t), h(s, t)) is a differentiable function of s and t. We also have

\frac{∂f}{∂s} (s, t) = \frac{∂f}{∂x}(g(s, t), h(s, t))\frac{∂g}{∂s}(s, t) + \frac{∂f}{∂y} (g(s, t), h(s, t))\frac{∂h}{∂s}(s, t)

and

\frac{∂f}{∂t} (s, t) = \frac{∂f}{∂x}(g(s, t), h(s, t))\frac{∂g}{∂t} (s, t) + \frac{∂f}{∂y} (g(s, t), h(s, t))\frac{∂h}{∂t}(s, t).

(b) We have

\frac{∂f}{∂s} (s, t) = (e^{y} cos x)(2st) + (e^{y} sin x)(4s^{3}t^{2} + t)

= (e^{s^{4}t^{2}+st} cos(s^{2}t + t^{3}))(2st) + (e^{s^{4}t^{2}+st} sin(s^{2}t + t^{3}))(4s^{3}t^{2} + t)

and

\frac{∂f}{∂s} (s, t) = (e^{y} cos x)(s^{2} + 3t^{2}) + (e^{y} sin x)(2s^{4}t + s)

= (e^{s^{4}t^{2}+st} cos(s^{2}t + t^{3}))(s^{2} + 3t^{2}) + (e^{s^{4}t^{2}+st} sin(s^{2}t + t^{3}))(2s^{4}t + s).

5. (a) For a unit vector u = ⟨a, b⟩, we define

D_{u}(f)(x, y) = lim_{h→0} \frac{f(x + ha, y + hb) − f(x, y)}{h}

and the gradient vector is

∇f = ⟨∂_{x}f, ∂_{y}f⟩

(b) D_{u}(f) = ∇f · u

(c) The directional derivative attains its maximum when u = ∇f/|∇f|.

(d) The vector ∇f and the normal to the tangent plane of a level set of f are parallel.

(e) For f(x, y) = x sin(x−y), we have ∇f(x, y) = ⟨sin(x−y)+x cos(x−y), −x cos(x− y)⟩.

6. (a) The function f: R^{2} → R has a local maximum at (a, b) if there exists a δ > 0 such that f(x, y) ≤ f(a, b) for all (x, y) ∈ B_{δ}(a, b) = {(x, y)|(x − a)^{2} + (y − b)^{2} ≤ δ^{2}}. The function f: R^{2} → R has a local minimum at (a, b) if there exists a δ > 0 such that f(x, y) ≥ f(a, b) for all (x, y) ∈ B_{δ}(a, b).

(b) Suppose all the second-order derivatives of f are continuous on a disc with centre (a,b) and suppose that ∂_{x}f(a, b) = 0 = ∂_{y}f(a, b). Let

D(a, b) = ∂_{xx}f(a, b)∂_{yy}f(a, b) − (∂_{xy}f(a, b))^{2}.

Then (a) if D(a, b) > 0 and ∂_{xx}f(a, b) > 0 then f(a, b) is a local minimum value of f, (b) if D(a, b) > 0 and ∂_{xx}f(a, b) < 0 then f(a, b) is a local maximum value of f, and (c) if D(a, b) < 0, then f(a, b) is neither a local minimum nor a local maximum of f.

(c) The critical points of the function f(x, y) = 4 + x^{3} + y^{3} − 3xy satisfy

\{ \begin{matrix}3x^{2} − 3y & = & 0 \\ 3y^{2} − 3x & = & 0 \end{matrix}

so they are (0, 0) and (1, 1). We find D(x, y) = 36xy − 9. Thus, D(0, 0) = −9 < 0 and so (0, 0) is a saddle point, and D(1, 1) = 36 − 9 > 0 and ∂_{xx}f(1, 1) = 6 > 0 so (1, 1) is a local minimum point.

7. (a) The extreme values of a function f: R^{2} → R subject to the constraint g(x, y) = k occur at the points (x, y) which satisfy

\{ \begin{matrix}∇f(x, y) & = & λ∇g(x, y) \\ g(x, y) & = & k \end{matrix}

for some λ ∈ R. Evaluating f at the points which satisfy this system of equations and taking the largest (smallest) produces the maximum (minimum) of f subject to the constraint g(x, y) = k.

(b) We must solve

\{ \begin{matrix}⟨2xy, x^{2}⟩ & = & ⟨λ2x, 4y⟩ \\ x^{2} + 2y^{2} & = & 6 \end{matrix}

We find the solutions (x, y) are (0, ±\sqrt{3}) and (2, ±1), so

max_{x^{2}+2y^{2}=6} x^{2}y = 4

and

min_{x^{2}+2y^{2}=6} x^{2}y = −4.