Active Calculus

1. Since the population \(P(t)\) of the town is growing at an annual rate of \(3\%\) per year, this means that \(\frac{dP}{dt} = 0.03 P\)

2. We use separation of variables and write \(\frac{1}{P} \frac{dP}{dt} = 1.03\text{.}\) We then have

   \begin{equation*} \int \frac{1}{P} \frac{dP}{dt} dt = \int 0.03 dt \end{equation*}

   and thus \(\int \frac{1}{P} dP = \int 0.03 dt\text{.}\) Evaluating the integrals on each side, we have \(\ln|P| = 0.03 t + c\) By definition of the logarithm function, it follows that

   \begin{equation*} |P| = e^{0.03t + c} = e^c e^{0.03t} = C e^{0.03t}\text{.} \end{equation*}

   Incorporating the \(\pm\) that follows from the abslute value sign into the constant \(C\text{,}\) we conclude that \(P = Ce^{0.03t}\text{.}\)

3. Given that \(P(0) = 10000\text{,}\) we see that \(10000 = C e^0 = C\text{.}\) So

   \begin{equation*} P = 10000 e^{0.03t}\text{.} \end{equation*}

4. The population will have doubled when \(P = 200000\text{.}\) To determine the time it takes the population to double, we solve the equation

   \begin{equation*} 10000 e^{0.03t} = 20000\text{.} \end{equation*}

   Thus \(e^{0.03t} = 2\text{,}\) so \(t = \frac{\ln(2)}{0.03} \approx 23.105\text{.}\) It will take about 23.1 years for the population to double in size.

5. If the annual growth rate is \(k\text{,}\) then the differential equation for the population is

   \begin{equation*} \frac{dP}{dt} = kP\text{.} \end{equation*}

   We can use separation of variables to show that \(P = P_0 e^{kt}\text{,}\) where \(P_0\) is the population at time \(t = 0\text{.}\) We then find the doubling time by solving the equation \(P_0 e^{kt} = 2P_0\text{.}\) We first solve for \(e^{kt}\) and get \(e^{kt} = 2\text{.}\) Taking natural logs, \(kt = \ln(2)\text{,}\) and dividing by \(k\text{,}\) \(t = \frac{1}{k} \ln(2)\text{.}\) The doubling time for the population is thus \(\frac{1}{k} \ln(2)\text{.}\)

We have

where \(k\) is a constant. We also have \(T(0) = 105\text{.}\)

1. Since the coffee is cooling at a rate of one degree per minute at time \(0\text{,}\) it follows that \(\frac{dT}{dt} = -1\) at this instant. From the given differential equation, we know \(-k(T(0) - 75) = -1\text{.}\) Next, \(T(0) = 105\text{,}\) so \(-k(105 - 75) = -1\text{,}\) and thus \(k = \frac{1}{30}\text{.}\)

2. We use separation of variables to solve the differential equation \(\frac{dT}{dt} = -\frac{1}{30} (T - 75)\text{.}\) Separating variables and integrating both sides with respect to \(t\text{,}\)

   \begin{equation*} \int \frac{1}{T - 75} dT = -\int \frac{1}{30} dt \end{equation*}

   Integrating both sides,

   \begin{equation*} \ln \left| T - 75 \right| = -\frac{1}{30} t + c \end{equation*}

   Using the definition of the logarithm function and writing \(e^c = C\text{,}\) we see that

   \begin{equation*} |T - 75| = e^{-t/30 + c} = C e^{-t/30} \end{equation*}

   Finally, including the absolute value in \(C\) and solving for \(T\text{,}\) \(T - 75 = C e^{-t/30}\text{,}\) so \(T = 75 + Ce^{-t/30}\text{.}\)

3. As \(t \to \infty\text{,}\) \(e^{-t/30} \to 0\text{.}\) So

   \begin{equation*} \lim_{t \to \infty} \left(75 + Ce^{-t/30} \right) = 75\text{.} \end{equation*}

   This means that as \(t \to \infty\text{,}\) \(T \to 75\text{,}\) or the temperature of the coffee tends to 75 degrees.

4. In order to determine the temperature of the coffee after 20 minutes, we need to first determine the value of the constant \(C\) in the solution \(T = 75 + Ce^{-t/30}\text{.}\) To do this, we use the fact that the temperature of the coffee at \(t = 0\) is \(105^\circ\)F. Thus, \(105 = 75 + Ce^0\text{,}\) so \(C = 30\text{.}\) Now we know that \(T = 75 + 30e^{-t/30}\text{,}\) and when \(t = 20\text{,}\) \(T(20) = 75 + 30e^{-2/3}\text{.}\) The temperature of the coffee after 20 minutes is approximately \(90.4^\circ\)F.

5. To determine how long it will take the coffee to cool to \(80^\circ\)F, we solve the equation \(80 = 75 + 30 e^{-t/30}\) for \(t\text{.}\) Solving first for \(e^{-t/30}\text{,}\) we subtract \(75\) from both sides and divide by \(30\) to get \(e^{-t/30} = \frac{5}{30}\text{.}\) Taking logarithms and solving for \(t\text{,}\) \(-\frac{t}{30} = \ln \left( \frac{1}{6} \right)\text{,}\) and thus \(t = -30 \ln \left( \frac{1}{6} \right)\text{.}\) We conclude that it will take about 53.75 minutes for the coffee to cool to \(80^\circ\)F.

1. We are given that \(\frac{dy}{dt} - (2 - t)y = 2 - t\text{.}\) We first separate the variables by writing \(\frac{dy}{dt} = (2 - t)y + (2 - t) = (2-t)(y+1)\) and then dividing both sides by \((y+1)\) to find

   \begin{equation*} \frac{1}{1+y} \frac{dy}{dt} = 2 - t\text{.} \end{equation*}

   Integrating both sides with respect to \(t\text{,}\) we have

   \begin{equation*} \int \frac{1}{1+y} dy = \int (2 - t) dt\text{.} \end{equation*}

   Evaluating the integrals,

   \begin{equation*} \ln|1 + y| = \left( 2t - \frac{1}{2}t^2 \right) + c \end{equation*}

   Rewriting without logarithms,

   \begin{equation*} |1+y| = e^{\left(2t - \frac{t^2}{2} \right)} e^c \end{equation*}

   Writing \(e^c = C\) and including the sign from the absolute value in this constant,

   \begin{equation*} 1 + y = Ce^{\left(2t - \frac{t^2}{2} \right)}\text{,} \end{equation*}

   and therefore \(y = -1 + C e^{\left(2t - \frac{t^2}{2} \right)}\text{.}\)

2. Given that \(\frac{1}{t} \frac{dy}{dt} = e^{t^2 - 2y}\text{,}\) we first observe that \(e^{t^2 - 2y} = e^{t^2}e^{ -2y}\text{.}\) Separating the variables,

   \begin{equation*} e^{2y} \frac{dy}{dt} = t e^{t^2}\text{.} \end{equation*}

   Integrating with respect to \(t\text{,}\)

   \begin{equation*} \int e^{2y} dy = \int t e^{t^2} dt\text{.} \end{equation*}

   Using standard substitution techniques on both sides, we find that

   \begin{equation*} \frac{1}{2} e^{2y} = \frac{1}{2} e^{t^2} + c \end{equation*}

   Writing \(2c = C\text{,}\) we then have \(e^{2y} = e^{t^2} + C\text{,}\) and taking logarithms to solve for \(y\text{,}\) we find \(2y = \ln \left( e^{t^2} + C \right)\text{,}\) and thus

   \begin{equation*} y = \frac{1}{2} \ln \left( e^{t^2} + C \right)\text{.} \end{equation*}

3. For the initial value problem \(y' = 2y + 2\text{,}\) \(y(0) = 2\text{,}\) we first solve the differential equation. Noting that \(2y + 2 = 2(y+1)\text{,}\) we have

   \begin{equation*} \frac{dy}{dt} = 2(y + 1)\text{,} \end{equation*}

   so

   \begin{equation*} \frac{1}{y + 1} \frac{dy}{dt} = 2 \end{equation*}

   Integrating with respect to \(t\text{,}\)

   \begin{equation*} \int \frac{1}{y + 1} dy = \int 2 dt \end{equation*}

   and thus evaluating the integrals, \(\ln|y+1| = 2t + c\text{.}\) It follows that \(|y + 1| = e^{2t + c} = e^c e^{2t}\text{,}\) so in the usual way we can write \(y+1 = Ce^{2t}\text{,}\) and therefore

   \begin{equation*} y = -1 + Ce^{2t}\text{.} \end{equation*}

   Using \(y(0) = 2\) to solve the IVP, we obtain \(2 = -1 + Ce^0\) and so \(C = 3\text{.}\) The solution to the initial value problem is

   \begin{equation*} y = -1 + 3 e^{2t}\text{.} \end{equation*}

4. For the initial value problem \(y' = 2y^2\text{,}\) \(y(-1) = 2\text{,}\) we first separate variables and write

   \begin{equation*} \frac{1}{y^2} \frac{dy}{dt} = 2\text{.} \end{equation*}

   Next, integrating with respect to \(t\text{,}\)

   \begin{equation*} \int \frac{1}{y^2} dy = \int 2 dt\text{,} \end{equation*}

   and thus \(-y^{-1} = 2t + c\text{.}\) Solving for \(y\text{,}\) it follows

   \begin{equation*} y = -\frac{1}{2t + c}\text{.} \end{equation*}

   We now use the initial value \(y(-1) = 2\) and obtain \(2 = -\frac{1}{-2 + c}\text{.}\) Solving for \(c\) gives \(c = \frac{3}{2}\text{.}\) The solution of the initial value problem is thus

   \begin{equation*} y = -\frac{1}{2t + \frac{3}{2}} = -\frac{2}{4t + 3}\text{.} \end{equation*}

5. For the IVP \(\frac{dy}{dt} = \frac{-2ty}{t^2 + 1}\text{,}\) \(y(0) = 4\text{,}\) we first separate variables and write

   \begin{equation*} \frac{1}{y} \frac{dy}{dt} = \frac{-2t}{t^2 + 1}\text{.} \end{equation*}

   Integrating with respect to \(t\) we find

   \begin{equation*} \int \frac{1}{y} dy = -\int \frac{2t}{t^2 + 1} dt\text{.} \end{equation*}

   Using the substitution \(u = t^2 + 1\) on the right, it follows that

   \begin{equation*} \ln|y| = -\ln\left| t^2 + 1 \right| + c\text{.} \end{equation*}

   By definition and properties of the logarithm,

   \begin{equation*} |y| = e^{-\ln\left| t^2 + 1 \right|}e^c = e^{\ln\left| t^2 + 1 \right|^{-1}}e^c = \left| t^2 + 1 \right|^{-1} e^c\text{.} \end{equation*}

   Writing \(C = e^c\) and absorbing the absolute value from the left side, we have

   \begin{equation*} y = \frac{C}{t^2 + 1}\text{.} \end{equation*}

   Using the initial condition \(y(0) = 4\text{,}\) we obtain \(4 = \frac{C}{1}\) or \(C = 4\text{.}\) Thus, the solution to the initial value problem is

   \begin{equation*} y = \frac{4}{t^2 + 1}\text{.} \end{equation*}