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Solutions to Homework 2

Math. 280, Spring 2008

Problem 1.
A cup of coffee in a room at temperature of $7 0^{\circ}$ F is being cooled according to Newton’s law of cooling:

\frac{d T}{d t} = - k (T - 70), T (0) = 200 . (time is measured in minutes)

(1)

The solution to (1) is

T (t) = 70 + (200 - 70) exp (- k t) .

(2)

The coefficient $k$ can be computed from (2) using the fact that $T (3) = 175$ . Thus,

175 = 70 + 130 exp (- 3 k) \Rightarrow exp (- k) = {(\frac{175 - 70}{130})}^{1 ∕ 3} = {(\frac{21}{26})}^{1 ∕ 3}

(3)

The temperature of the coffee will reach $11 2^{\circ}$ F at $t = t_{1}$ that satisfies the equation

112 = T (t_{1}) = 70 + 130 exp (- k t_{1}) = 70 + 130 {[exp (- k)]}^{t_{1}}, with exp (- k) given in (3)

Thus,

\frac{112 - 70}{130} = {[{(\frac{21}{26})}^{1 ∕ 3}]}^{t_{1}}, or t_{1} = \frac{3 log (\frac{21}{65})}{log (2126)} \approx 15.87 minutes .

Problem 2.
Let $N (t)$ denote the amount of C-14 at time $t$ , with time $t = 0$ corresponding to the moment when the original amount $N_{0}$ of C-14 was present. Time evolution of $N (t)$ is governed by the equation

\frac{d N}{d t} = - λ N, N (0) = N_{0} .

(4)

The amount of C-14 decays in time, thus $λ > 0$ . From (4) , $N (t) = N_{0} exp (λ t)$ and, since the half-life of C-14 is approximately 5568 years, $N_{0} ∕ 2 = N_{0} exp (5568 λ)$ . This gives us $λ = - ln 2 ∕ 5568 \approx - 0.00012448$ . Therefore, the expression for $N (t)$ for all time is given by

N (t) = N_{0} exp (- \frac{ln 2}{5568} t) = N_{0} 2^{- (\frac{t}{5568})} .

(5)

At present time $t$ , $14.5 %$ is left, or equivalently from (5) ,

0.145 N_{0} = N_{0} 2^{- (\frac{t}{5568})} \Rightarrow ln 0.145 = - \frac{t}{5568} ln 2 \Rightarrow t = - \frac{5568 ln 0.145}{ln 2} \approx 15511.753 years .

Problem 3.
For the equation

(1 + x^{2}) d y + (x y + x^{3} + x) d x = 0 or \frac{d y}{d x} + (\frac{x}{1 + x^{2}}) y = - x,

(6)

an integrating factor is $exp (\int \frac{x}{1 + x^{2}} d x) = \sqrt{1 + x^{2}}$ so that

\frac{d}{d x} [y \sqrt{1 + x^{2}}] = - x \sqrt{1 + x^{2}} \Rightarrow y = - \frac{1}{3} (1 + x^{2}) + c {(1 + x^{2})}^{- \frac{1}{2}} .

(7)

Figure 1:

Graphs of

y = - \frac{1}{3} (1 + x^{2}) + c {(1 + x^{2})}^{- \frac{1}{2}}

for various choices of

c

Problem 4.
(a) For the equation

(\frac{3 y^{2} - t^{2}}{y^{5}}) \frac{d y}{d t} + \frac{t}{2 y^{4}} = 0, y (1) = 1,

(8)

let $M = t ∕ (2 y^{4})$ and $N = (3 y^{2} - t^{2}) ∕ y^{5}$ so that $M_{y} = - 2 t ∕ y^{5} = N_{t}$ . From $ϕ_{t} (t, y) = M = t ∕ (2 y^{4})$ we obtain $ϕ (t, y) = t^{2} ∕ (4 y^{4}) + h (y)$ . Thus, using $ϕ_{y} = N = (3 y^{2} - t^{2}) ∕ y^{5}$ , we have $h^{'} (y) = 3 ∕ y^{3}$ , and $h (y) = - 3 ∕ (2 y^{2})$ . The general solution is $t^{2} ∕ (4 y^{4}) - 3 ∕ (2 y^{2}) = c$ . If $y (1) = 1$ then $c = - 5 ∕ 4$ and the solution to initial-value problem (8) is

\frac{t^{2}}{4 y^{4}} - \frac{3}{2 y^{2}} = - \frac{5}{4} .

(9)

Figure 2:

Graph of the integral curve (9)

Figure 3:

Other integral curves of

t^{2} ∕ (4 y^{4}) - 3 ∕ (2 y^{2}) = c

The solutions of the initial value problem (8) and some other integral (dashed) curves of the differential equation

(\frac{3 y^{2} - t^{2}}{y^{5}}) \frac{d y}{d t} + \frac{t}{2 y^{4}} = 0 are shown in Figure 3 .

Remark 1.Please observe that the graph in Figure 2 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition $y (1) = 1$ .

Remark 2.Note that if one multiplies both sides of equation (8) by $y^{5}$ , the resulting equation

(3 y^{2} - t^{2}) \frac{d y}{d t} + \frac{1}{2} t y = 0

(10)

is not exact !!! In other words $1 ∕ y^{5}$ is an integrating factor for (10). (See also Problem 5, below)

(b) The initial-value problem

\frac{d y}{d t} = \frac{y}{2 y^{2} - t}, y (1) = 5, or \frac{d t}{d y} - \frac{t}{y} = 2 y, y (1) = 5,

(11)

is linear in $t$ (as a function of $y$ ). Its integrating factor is $1 ∕ y$ , thus

\frac{d}{d y} [\frac{t}{y}] = 2 \Rightarrow t = 2 y^{2} + c y .

If $y (1) = 5$ then $c = - 49 ∕ 5$ and $t = 2 y^{2} - (49 ∕ 5) y$ .

Figure 4:

Graph of the integral curve

t = 2 y^{2} - (49 ∕ 5) y

Figure 5:

Other integral curves of

t = 2 y^{2} + c y

The solutions of the initial value problem (11) and some other integral (dashed) curves of the differential equation

\frac{d y}{d t} = \frac{y}{2 y^{2} - t}, are shown in Figure 5 .

Remark 3.Please observe that the graph in Figure 4 does not represent a function !!! Which branch of the graph represents the solution satisfying the initial condition $y (1) = 5$ .

Problem 5.
Multiply the equation

(t^{2} + t y) \frac{d y}{d t} - y^{2} = 0 by μ (t, y) = \frac{1}{t^{2} y} .   The result is

(12)

(\frac{1}{y} + \frac{1}{t}) \frac{d y}{d t} - \frac{y}{t^{2}} = 0 .

(13)

With $M = - y ∕ t^{2}$ and $N = 1 ∕ y + 1 ∕ t$ we have $M_{y} = - 1 ∕ t^{2} = N_{t}$ ; therefore equation (13)is exact. From $ϕ_{t} = - y ∕ t^{2}$ we obtain $ϕ (t, y) = y ∕ t + h (y)$ But $ϕ_{y} = N$ ; thus $h^{'} (y) = 1 ∕ y$ , and $h (y) = log | y |$ . The solution of (12) is

ϕ (t, y) = \frac{y}{t} + log | y | = c, where c \in ℝ is a constant.

Problem 6.
Under the change of the independent variable, $t \to y$ , the original equation are transformed into

\begin{aligned} m \frac{d v}{d t} & = - m g - k v^{2}, k > 0 for the “up” motion \\ m \frac{d v}{d t} & = m g - k v^{2}, k > 0 for the “down” motion \end{aligned} \Rightarrow \begin{aligned} m v \frac{d v}{d y} & = - m g - k v^{2}, k > 0 for the “up” motion \\ m v \frac{d v}{d y} & = m g - k v^{2}, k > 0 for the “down” motion \end{aligned}

(14)

Separating variables and integrating the differential equation corresponding to the “up” motion we have

\begin{aligned} \int \frac{m v}{m g + k v^{2}} = - \int d y & \Rightarrow \frac{m}{2 k} ln (m g + k v^{2}) = - y + c_{1} \Rightarrow m g + k v^{2} = c_{2} exp (- \frac{2 k y}{m}) \\ \Rightarrow v^{2} = c_{3} exp (- \frac{2 k y}{m}) - \frac{m g}{k} \end{aligned}

(15)

Using $y (0) = 0$ and $v (0) = v_{0}$ we have that $v = v_{0}$ when $y = 0$ (the initial condition) so that $v_{0}^{2} = c_{3} - m g ∕ k$ and $c_{3} = v_{0}^{2} + m g ∕ k$ . Thus,

v^{2} = (\frac{k v_{0}^{2} + m g}{k}) exp (- \frac{2 k y}{m}) - \frac{m g}{k}

(16)

Setting $v = 0$ in the left hand side of (16) and solving for $y$ we see that the maximum height is

h = \frac{m}{2 k} ln (\frac{k v_{0}^{2} + m g}{m g}) .

(17)

Now, separating variables and integrating the differential equation corresponding to the “down” motion (see (14)) we have

\begin{aligned} \int \frac{m v}{m g - k v^{2}} = \int d y & \Rightarrow - \frac{m}{2 k} ln | m g - k v^{2} | = y + d_{1} \Rightarrow m g - k v^{2} = d_{2} exp (- \frac{2 k y}{m}) \\ \Rightarrow v^{2} = \frac{m g}{k} [1 - d_{3} exp (- \frac{2 k y}{m})] \end{aligned}

(18)

In this case $v = 0$ when $y = 0$ (the initial condition) so $d_{3} = 1$ and

v^{2} = \frac{m g}{k} [1 - exp (- \frac{2 k y}{m})] .

(19)

Setting $y = h$ from (17) into (19)and solving for $v$ we obtain that the impact velocity is

v_{i} = \frac{v_{0}}{\sqrt{1 + k v_{0}^{2} m g}} < v_{0} .

(20)

Remark 4.

By taking $y \to \infty$ in the right hand side of (19)one obtains the limiting velocity, $v_{f i n a l}$ , for the problem $m (d v ∕ d t) = m g - k v^{2}$

v_{f i n a l}^{2} = lim_{y \to \infty} \frac{m g}{k} [1 - exp (- \frac{2 k y}{m})] = \frac{m g}{k} \Rightarrow v_{f i n a l} = \sqrt{\frac{m g}{k}} > \sqrt{\frac{m g}{k}} \sqrt{\frac{v_{0}^{2}}{v_{0}^{2} + \frac{m g}{k}}} = v_{i} . (see, (20))

See also (29) , where the same result can be obtained by taking $t \to \infty$ in (29) and using $lim_{x \to \infty} tanh (x) = 1$ .

(b) It is interesting to compare the results of Problem 6 with the same problem but without air resistance. When $k = 0$ the change variables from $t$ to $y$ is not necessary, however, for comparison with the case $k \neq 0$ , I proceed in an exactly the same way as before. The corresponding equations are

\begin{aligned} m \frac{d v}{d t} & = - m g, for the “up” motion \\ m \frac{d v}{d t} & = m g, for the “down” motion \end{aligned} \Rightarrow \begin{aligned} m v \frac{d v}{d y} & = - m g, for the “up” motion \\ m v \frac{d v}{d y} & = m g, for the “down” motion \end{aligned} \Rightarrow

(21)

\begin{aligned} \frac{v^{2}}{2} & = - g y + c_{1}, v |_{y = 0} = v_{0} for the “up” motion \\ \frac{v^{2}}{2} & = g y + c_{2} v |_{y = 0} = 0 for the “down” motion \end{aligned} \Rightarrow \begin{aligned} \frac{v^{2}}{2} & = - g y + \frac{v_{0}^{2}}{2}, for the “up” motion \\ \frac{v^{2}}{2} & = g y for the “down” motion \end{aligned}

(22)

Thus, the maximum height is $v_{0}^{2} ∕ 2 g$ . Observe that the maximum height given in (17) is less than the maximum height in the case without air resistance. Indeed,

\frac{m}{2 k} ln (\frac{k v_{0}^{2} + m g}{m g}) = \frac{m}{2 k} ln (1 + \frac{k v_{0}^{2}}{m g}) < \frac{m}{2 k} (\frac{k v_{0}^{2}}{m g}) \{\begin{matrix} since for x > 0 \\ ln (1 + x) < x \end{matrix}\} = \frac{v_{0}^{2}}{2 g}

Also the impact velocity in the case without air resistance is equal to the initial velocity $v_{0}$ . indeed, substituting $v_{0}^{2} ∕ 2 g$ into $v^{2} ∕ g = g y$ (expression for $v$ for the “down” part of the motion, see, (22)) we obtain

v_{i} = v_{0} (No dissipation of the energy)

Additional remarks to Problem 6

Finally, I show how one can integrate the equations

\begin{align} m \frac{d v}{d t} & = - m g - k v^{2}, v (0) = v_{0}, k > 0 for the “up” motion & (23) \\ m \frac{d v}{d t} & = m g - k v^{2}, v (0) = 0, k > 0 for the “down” motion & (24) \end{align}

without changing the variable $t$ to $y$ . The integrations are slightly more involved. For the “up” motion we have, after separation of variables

\frac{m}{k} (\frac{d v}{m g k + v^{2}}) = - d t \Rightarrow \frac{1}{\sqrt{m g k}} arctan (v ∕ \sqrt{\frac{m g}{k}}) = - \frac{k}{m} t + c \Rightarrow v (t) = \sqrt{\frac{m g}{k}} tan (c_{1} - \sqrt{\frac{k g}{m}} t)

(25)

Applying the initial conditions $v (0) = v_{0}$ we have

v_{0} = \sqrt{\frac{m g}{k}} tan c_{1} \Rightarrow c_{1} = arctan (v_{0} ∕ \sqrt{\frac{m g}{k}}) \Rightarrow v (t) = \sqrt{\frac{m g}{k}} tan [arctan (v_{0} ∕ \sqrt{\frac{m g}{k}}) - \sqrt{\frac{k g}{m}} t] .

(26)

From (26) the time $t_{u p}$ needed for the projectile to attain its maximum height ( $v (t_{u p}) = 0$ ) is

t_{u p} = \sqrt{\frac{m}{k g}} arctan (v_{0} ∕ \sqrt{\frac{m g}{k}})

(27)

For the “down” motion

\frac{m}{k} (\frac{d v}{m g k - v^{2}}) = d t \Rightarrow \frac{1}{\sqrt{m g k}} {tanh}^{- 1} (v ∕ \sqrt{\frac{m g}{k}}) = \frac{k}{m} t + d_{1} \Rightarrow v (t) = \sqrt{\frac{m g}{k}} tanh (d_{1} + \sqrt{\frac{k g}{m}} t)

(28)

Applying the initial condition $v (0) = 0$ one obtains $d_{1} = 0$ and

v (t) = \sqrt{\frac{m g}{k}} tanh (\sqrt{\frac{k g}{m}} t)

(29)

The time $t_{d o w n}$ needed for the projectile to fall from the height (17) can also be computed; from (20)

\begin{aligned} \frac{v_{0}}{\sqrt{1 + k v_{0}^{2} m g}} = \sqrt{\frac{m g}{k}} tanh (\sqrt{\frac{k g}{m}} t) & \Rightarrow \frac{v_{0} \sqrt{k}}{\sqrt{m g + k v_{0}^{2}}} = tanh (\sqrt{\frac{k g}{m}} t) \\ \Rightarrow t_{d o w n} = \sqrt{\frac{m k}{g}} {tanh}^{- 1} (\frac{v_{0} \sqrt{k}}{\sqrt{m g + k v_{0}^{2}}}) \end{aligned}

(30)

One can also show that

t_{u p} < t_{d o w n} .

(31)

Furthermore, when air resistance is neglected

t_{u p} = t_{d o w n} = \frac{v_{0}}{g} . (No dissipation of the energy)

Problem 7.
(a) The equation for velocity at any time and position $y$ is

v^{2} = v_{0}^{2} - 2 g R + \frac{2 g R^{2}}{R + y},

(32)

where $R$ is the radius of the earth, $g$ is the acceleration due to gravity, and $v_{0}$ is the initial velocity of the projectile (in our case, $v_{0} < \sqrt{2 g R}$ ). Now the maximum distance, $y = Y_{m a x}$ , is reached by the projectile when $v = 0$ . In other words, we have the following equation for $Y_{m a x}$ :

0 = V_{0}^{2} - 2 g R + \frac{2 g R^{2}}{R + Y_{m a x}} .

(33)

Its solution is

Y_{m a x} = \frac{R v_{0}^{2}}{2 g R - v_{0}^{2}} .

(b) The solution to the equation $y^{''} = - g$ with the initial conditions $y (0) = 0$ and $y^{'} (0) = v (0) = v_{0} > 0$ is $y (t) = - \frac{1}{2} g t^{2} + v_{0} t$ (with $v (t) = - g t + v_{0}$ ). As in part (a), the maximum distance, $y_{m a x}$ , is reached when $v = 0$ ; this corresponds to $t = v_{0} ∕ g$ . Thus, the maximum distance $y_{m a x}$ is

y_{m a x} = y (\frac{v_{0}}{g}) = - \frac{1}{2} g {(\frac{v_{0}}{g})}^{2} + v_{0} (\frac{v_{0}}{g}) = \frac{v_{0}^{2}}{2 g} .

Finally, since $v_{0} < \sqrt{2 g R}$ , we have

Y_{m a x} = \frac{R v_{0}^{2}}{2 g R - v_{0}^{2}} = \frac{v_{0}^{2}}{2 g - (v_{0}^{2} ∕ R)} > \frac{v_{0}^{2}}{2 g} = y_{m a x} .

v = \frac{d y}{d t} = \sqrt{\frac{2 g R^{2}}{R + y}}, y (0) = 0, (Note the plus sign in front of \sqrt{\frac{2 g R^{2}}{R + y}})

(34)

which is a separable equation. The solution of (34) is

\frac{2}{3} {(R + y)}^{3 ∕ 2} = R t \sqrt{2 g} + c .

With $y (0) = 0$ we have $c = (2 ∕ 3) R^{3 ∕ 2}$ . Thus

y (t) = {(\frac{3 \sqrt{2 g}}{2} R t + R^{3 ∕ 2})}^{2 ∕ 3} - R .

(35)

(d) Solving (35) for $t$ we obtain

t = \frac{2}{3 \sqrt{2 g} R} ({[y (t) + R]}^{3 ∕ 2} - R^{3 ∕ 2}) .

(36)

With $R = 3963$ miles, $y (T_{M}) = 238, 855$ miles, and $g = 32 ∕ 5280 = 0.0061$ mile/s $^{2}$ , we get from (36)
$T_{M} = 181853$ seconds $= 50.52$ hours.