Solution Keys Page 1

Note: Solutions to Sample Test 3 [#1] (a) Let f(y) = −y^{2}(y − 3). Equilibrium solutions are zero points of f(y) = 0. They are y = 0, y = 3. (b) For any initial point between the to equilibrium points y = 0, y = 3, dy/dt > 0. So the graph of the solution increases, approaching y = 3 asymptotically. (c) By Euler’s method, y_{n+1} = y_{n} + Δxf(y_{n}) = y_{n} + 0.5(−y_{n}^{2}(y_{n} − 3))

\begin{matrix}n & t_{n} & y_{n} & Δxf(y_{n}) \\ 0 & 0 & 1 & 0.5(−1^{2}(1 − 3)) = 1 \\ 1 & 0.5 & 1 + 1 = 2 & 0.5(−2^{2}(2 − 3)) = 2 \\ 2 & 1 & 2 + 2 = 4 & – \end{matrix}

Hence y(1) 4.

[#2] (a) \begin{cases}\frac{dy}{dt} = 5g/L × 100L/min − \frac{y(t)}{1000}g/L × 100L/min = 500 − \frac{y(t)}{10} \\ y(0) = 0 \end{cases}

(b) By the method of separation of variables,

\frac{dy}{dt} = \frac{5000 − y(t)}{10}

\frac{dy}{5000 − y} = \frac{1}{10}dt

∫ \frac{dy}{5000 − y} = ∫ \frac{1}{10}dt = \frac{t}{10} + C_{1}

− ln |5000 − y| = \frac{t}{10} + C_{1}

|5000 − y| = e^{− \frac{t}{10} −C_{1}} = C_{2}e^{− \frac{t}{10}}

5000 − y = ±C_{2}e^{− \frac{t}{10}} = C_{3}e^{− \frac{t}{10}}

y(t) = 5000 − C_{3}e^{− \frac{t}{10}} = 5000 + Ce^{− \frac{t}{10}}

Use the initial condition y(0) = 0, we have

0 = y(0) = 5000 + Ce^{− \frac{0}{10}} = 5000 + C ⟹ C = −5000,

and

y(t) = 5000(1 − e^{− \frac{t}{10}} ).

(c) The time t that happens satisfies

3000 = y(t) = 5000(1 − e^{− \frac{t}{10}} ).

Solving for t to have: 3/5 = 1 − e^{− \frac{t}{10}} , e^{− \frac{t}{10}} = 1 − 3/5 = 2/5, and t = −10 ln 2/5 9.2min.

[#3]

\frac{dy}{dx} = \frac{2xy}{x^{2} + 1}

\frac{dy}{y} = \frac{2x}{x^{2} + 1}dx

∫ \frac{dy}{y} = ∫ \frac{2x}{x^{2} + 1}dx

ln |y| = ln(x^{2} + 1) + C_{1}

|y| = e^{ln(x^{2}+1)+C_{1}} = C_{2}(x^{2} + 1) y = ±C_{2}(x^{2} + 1) = C(x^{2} + 1)

With 3 = y(0) = C(0^{2} + 1) = C we have y(x) = 3(x^{2} + 1). [#4] A = (−1, 1, 2), B = (1, −1, 0), C = (2, 1, 1). (a) A⃗B = (2, −2, −2), B⃗C =

(1, 2, 1). (b)A⃗B · B⃗C = 2(1) − 2(2) − 2(1) = −4. A⃗B × B⃗C = | \begin{matrix}i⃗ & j⃗ & k⃗ \\ 2 & −2 & −2 \\ 1 & 2 & 2 \end{matrix} | =

2i⃗ − 4j⃗ + 6k⃗ = (2, −4, 6). (d) No. Vectors parallel to (−1, 1, 0) have the form k(−1, 1, 0) = (−k, k, 0). Since A⃗B · (−k, k, 0) = 2(−k) − 2(k) − 2(0) = −4k ≠ 0 unless k = 0. (e) (1/2)||A⃗B × B⃗C|| = \sqrt{14}. (f) cos θ = \frac{A⃗B·B⃗C}{||A⃗B||·||B⃗C||} = −\frac{\sqrt{2}}{3} and

θ = arccos −\frac{\sqrt{2}}{3} = 2.06 radians. (g) (x − (−1), y − 1, z − 2) × (A⃗B × B⃗C) = 0 which simplifies to 2(x + 1) − 4(y − 1) + 6(z − 2) = 0 ⟹ x − 2y + 3z = 3. (h) P rojB⃗C A⃗B = \frac{A⃗B·B⃗C}{||B⃗C||^{2}} B⃗C = −\frac{2}{3}⟨1, 2, 1⟩.