# Question b15b8

Jul 31, 2017

The value is about $233333 (or$230000 if you round to two significant digits).

#### Explanation:

We are assuming that $\frac{\mathrm{dV}}{\mathrm{dt}} = - \frac{k}{t + 3} ^ 2$ for some constant $k > 0$. Upon separation of variables, this can be written as $\mathrm{dV} = - \frac{k}{t + 3} ^ 2 \mathrm{dt}$.

Integration of both sides leads to $V = \frac{k}{t + 3} + C$.

We are given that $V \left(0\right) = 500000$ and $V \left(1\right) = 400000$ (the decrease in value is $100000 from time $t = 0$to time $t = 1$). Therefore, we get the following system of equations to solve for $k$and $C$: $500000 = \frac{k}{3} + C$and $400000 = \frac{k}{4} + C$. Solving the first equation for $C$and substituting into the second equation gives $400000 = \frac{k}{4} + \left(500000 - \frac{k}{3}\right)$, or $100000 = \frac{k}{3} - \frac{k}{4} = \frac{k}{12}$. Hence, $k = 1200000$and $C = 500000 - \frac{1200000}{3} = 500000 - 400000 = 100000$. Therefore, $V = V \left(t\right) = \frac{1200000}{t + 3} + 100000$and V(6)=1200000/9+100000 approx$233333.

Jul 31, 2017

$V = \frac{1200000}{t + 3} + 100000$

 V(6) = $233,333 # #### Explanation: We must decode the given text to form an appropriate Differential Equation which we must then solve. Depreciation reduces the value of an object. $\frac{\mathrm{dV}}{\mathrm{dt}}$represents the rate of change of the value of the machine. Thus the depreciation is $- \frac{\mathrm{dV}}{\mathrm{dt}}$. We are told that the depreciation is inversely proportional to the square of $t + 3$. Thus: $- \frac{\mathrm{dV}}{\mathrm{dt}} \propto \frac{1}{t + 3} ^ 2 \implies - \frac{\mathrm{dV}}{\mathrm{dt}} = \frac{K}{t + 3} ^ 2$Where $K$is the constant of proportionality. The above is a First order linear separable Ordinary Differential Equation. And we can "seperate the variables" to get: $\int \setminus \mathrm{dV} = - K \int \setminus \frac{1}{t + 3} ^ 2 \setminus \mathrm{dt}$Which we can integrate to get: $V = \frac{K}{t + 3} + C$Note that we have have two unknown constants and two conditions; thus: "The initial value of the machine was$500,000":

$\implies V = 500000$ when $t = 0$

$\therefore 500000 = \frac{K}{3} + C$ ..... [A]

"its value decreased $100,000 in the first year": $\implies V = 400000$when $t = 1$$\therefore 400000 = \frac{K}{4} + C$..... [B] Eq[A] - Eq[B]: $100000 = \frac{K}{3} - \frac{K}{4}$$\therefore \frac{K}{12} = 100000 \implies K = 1200000$Subs $K = 1200000$into Eq [A]: $500000 = \frac{1200000}{3} + C$$\therefore 500000 = 400000 + C$$\therefore C = 100000$Thus the solution to the DE is: $V = \frac{1200000}{t + 3} + 100000$We seek the value when $t = 6$: $V = \frac{1200000}{9} + 100000$$\setminus \setminus = \frac{400000}{3} + 100000$$\setminus \setminus = \frac{400000}{3} + \frac{300000}{3}$$\setminus \setminus = \frac{700000}{3}$$\setminus \setminus = 233333.33333 \ldots$$\setminus \setminus = 233333 \setminus \setminus\$ rounded to the nearest integer