Math. 311 (Fulling) 13 February 2002

Test A – Solutions

Name:

Calculators may be used for simple arithmetic operations only!

1. (12 pts.) Find the inverse (if it exists) of the matrix M = ( \begin{matrix}3 & 8 \\ 1 & 3 \end{matrix} ).

Reduce the augmented matrix:

( \begin{matrix}3 & 8 & 1 & 0 \\ 1 & 3 & 0 & 1 \end{matrix} ) ⟶^{[1]↔[2]} ( \begin{matrix}1 & 3 & 0 & 1 \\ 3 & 8 & 1 & 0 \end{matrix} )

⟶^{[2]→[2]−3[1]} ( \begin{matrix}1 & 3 & 0 & 1 \\ 0 & −1 & 1 & −3 \end{matrix} ) ⟶^{{[1]→[1]+3[2] \atop [2]→−[2]}} ( \begin{matrix}1 & 0 & 3 & −8 \\ 0 & 1 & −1 & 3 \end{matrix} ) .

Therefore,

M^{−1} = ( \begin{matrix}3 & −8 \\ −1 & 3 \end{matrix} ) .

It is easy to check that MM^{−1} = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ).

2. (10 pts.) A function f(x, y) satisfies the equations \frac{∂f}{∂x} = − \frac{2x}{y} f, \frac{∂f}{∂y} = \frac{x2}{y2} f . Calculate

\frac{d}{dx}f(x, x2) .

\frac{d}{dx} f(x, x^{2}) = ∇f · \frac{d}{dx} ( \begin{matrix}x \\ x^{2} \end{matrix} ) = \frac{∂f}{∂x} |_{y=x^{2}} × 1 + \frac{∂f}{∂y} |_{y=x^{2}} × (2x)

= − \frac{2x}{x^{2}} f + \frac{x^{2}}{x^{4}} (2x)f = (− \frac{2}{x} + \frac{2}{x}) f = 0.

Remark: f(x, y) = e^{−x^{2}/y} is a function with these properties.

3. (18 pts.) A curve C in three-dimensional space is specified by the parametric equations

x = t, y = t sin t, z = cos t.

(a) Find the tangent vector to C at the point where t = π.

Let r⃗(t) = ( \begin{matrix}x \\ y \\ z \end{matrix} ). Then r⃗′(t) = ( \begin{matrix}1 \\ sin t + t cos t \\ − sin t \end{matrix} ), so r⃗′(π) = ( \begin{matrix}1 \\ −π \\ 0 \end{matrix} ).

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(b) Find the directional derivative of f(x, y, z) = x + zey at that point, in the direction

of the curve.

The unit vector in the direction of the curve is r⃗′(π) divided by its length:

û = \frac{1}{\sqrt{1 + π^{2}}} ( \begin{matrix}1 \\ −π \\ 0 \end{matrix} ) . Also, r⃗(π) = ( \begin{matrix}π \\ 0 \\ −1 \end{matrix} ) .

So

∇f = (1, ze^{y}, e^{y})|_{r⃗(π)} = (1, −1, 1).

Thus

\frac{∂f}{∂û} = ∇f · û = \frac{1 + π}{\sqrt{1 + π^{2}}} .

4. (15 pts.) Producing a refrigerator requires 0.1 ton of steel and 0.2 ton of plastic. Producing an airplane requires 5 tons of steel and 2 tons of plastic. Producing a ton of steel consumes 3 tons of coal and 10 barrels of water. Producing a ton of plastic consumes 2 tons of coal and 50 barrels of water. Organize these facts into matrices, and find the matrix that tells you how much coal ( c) and water ( w) is needed to make r refrigerators and a airplanes.

Let s and p be the quantities of steel and plastic, and let

( \begin{matrix}s \\ p \end{matrix} ) = B ( \begin{matrix}r \\ a \end{matrix} ) = ( \begin{matrix}0.1 & 5 \\ 0.2 & 2 \end{matrix} ) ( \begin{matrix}r \\ a \end{matrix} ) , ( \begin{matrix}c \\ w \end{matrix} ) = A ( \begin{matrix}s \\ p \end{matrix} ) = ( \begin{matrix}3 & 2 \\ 10 & 50 \end{matrix} ) ( \begin{matrix}s \\ p \end{matrix} ) .

Then ( \begin{matrix}c \\ w \end{matrix} ) = AB ( \begin{matrix}r \\ a \end{matrix} ), where

AB = ( \begin{matrix}3 & 2 \\ 10 & 50 \end{matrix} ) ( \begin{matrix}0.1 & 5 \\ 0.2 & 2 \end{matrix} ) = ( \begin{matrix}0.7 & 19 \\ 11 & 150 \end{matrix} ) .

5. (10 pts.) Classify each of these integral operators as linear, affine, or fully nonlinear (as a

function of g). ( g is an element of C(0, 1) — that is, a function.)

(a) A(g) = ∫{t \atop 0} e(t−s)g(s) ds. ( A(g) is another element of C(0, 1) — a function of the

variable t. In other words, A: C(0, 1) → C(0, 1) .) Linear. This is clear from the form of the integrand; or, one can easily verify that

A(λg + h) = ∫_{0}^{t} e^{(t−s)}[λg(s) + h(s)] ds = λ ∫_{0}^{t} e^{(t−s)}g(s) ds + ∫_{0}^{t} e^{(t−s)}h(s) ds = λA(g) + A(h).

(b) B(g) = ∫{1 \atop 0} teg(t) dt. ( B(g) is an element of R — a number. B: C(0, 1) → R .)

Nonlinear. Again, this is pretty obvious because the g is up in the exponent. A formal counterexample (to the homogeneity clause of the definition) is

B(λg) = ∫_{0}^{1} te^{λg(t)} dt ≠ ∫_{0}^{1} tλe^{g(t)} dt = λB(g).

( B is not affine, because it’s not of the form of a linear operator plus a fixed vector (which would be a constant number in this case). An example of an affine operator C: C(0, 1) → C(0, 1) is C(g)(t) = A(g)(t) + cos t, A as in part (a).)

6. (10 pts.) Construct the best affine approximation (also known as the first-order approxi-

mation) to T (x, y) = ( \begin{matrix}√ & \overline{x2 + 4y2} \\ x − y \end{matrix} ) in the neighborhood of the point (x0, y0) = (1, 1) .

The matrix of partial derivatives is

JT = ( 1  −1^{\frac{x}{\sqrt{x^{2}+4y^{2}}}  \frac{4y}{\sqrt{x^{2}+4y^{2}}}} ) ,  so Jr⃗_{0}T = ( 1 −1^{\frac{1}{\sqrt{5}} \frac{4}{\sqrt{5}}} ) .

Therefore,

T (x, y) ≈ T (x_{0}, y_{0}) + Jr⃗_{0} ( \begin{matrix}x − x_{0} \\ y − y_{0} \end{matrix} ) = ( \begin{matrix}√ & \overline{5} \\ 0 \end{matrix} ) + ( 1 −1^{\frac{1}{\sqrt{5}} \frac{4}{\sqrt{5}}} ) ( \begin{matrix}x − 1 \\ y − 1 \end{matrix} ) .

7. (25 pts.) Find all solutions (x, y, z) of \{ \begin{matrix}y − z & = 1, \\ x − y − 2z & = 0, \\ 2x + 3y + Az & = B. \end{matrix}\} ( A and B are arbitrary,

but fixed, parameters. Certain special values of A and B will require special attention.)

( \begin{matrix}0 & 1 & −1 & 1 \\ 1 & −1 & −2 & 0 \\ 2 & 3 & A & B \end{matrix} ) ⟶^{[2]↔[1]} ( \begin{matrix}1 & −1 & −2 & 0 \\ 0 & 1 & −1 & 1 \\ 2 & 3 & A & B \end{matrix} )

⟶^{[3]→[3]−2[1]} ( \begin{matrix}1 & −1 & −2 & 0 \\ 0 & 1 & −1 & 1 \\ 0 & 5 & A + 4 & B \end{matrix} ) ⟶^{{[3]→[3]−5[2] \atop 1→[1]+[2]}} ( \begin{matrix}1 & 0 & −3 & 1 \\ 0 & 1 & −1 & 1 \\ 0 & 0 & A + 9 & B − 5 \end{matrix} ) .

Case I: If A = −9 and B ≠ 5 , there are no solutions, because the bottom row gives an inconsistent equation. Case II: If A = −9 and B = 5 , then z is an arbitrary parameter and

y = z + 1, x = 3z + 1.

Case III: If A ≠ −9 , continue reducing:

⟶^{[3]→[3]/(A+9)} ( \begin{matrix}1 & 0 & −3 & 1 \\ 0 & 1 & −1 & 1 \\ 0 & 0 & 1 & \frac{B−5}{A+9} \end{matrix} ) ⟶^{{[1]→[1]+3[3] \atop [2]→[2]+[3]}} ( \begin{matrix}1 & 0 & 0 & 1 + 3C \\ 0 & 1 & 0 & 1 + C \\ 0 & 0 & 1 & C \end{matrix} ) ,

where we define C = \frac{B − 5}{A + 9} to save writing. Therefore, in this case there is the unique solution

x = 1 + 3C,  y = 1 + C,  z = C.