Explore real-world applications modeled by separable differential equations, including exponential growth/decay, Newton's Law of Cooling, and the concept of orthogonal trajectories.
A quantity \(y\) changes at a rate proportional to its current size. What is the differential equation that models this situation?
The statement "rate of change of \(y\) is proportional to \(y\)" is written as \(\frac{dy}{dt} = ky\), where \(k\) is the constant of proportionality.
What is the general solution to the exponential growth/decay equation, \(\frac{dy}{dt} = ky\)?
By separating variables (\(\frac{1}{y}dy = k dt\)) and integrating, the general solution is found to be \(y(t) = Ce^{kt}\).
In the solution \(y(t) = Ce^{kt}\), what does the constant \(C\) typically represent?
\(C\) represents the initial value of the quantity \(y\) at time \(t=0\). This is because \(y(0) = Ce^{k \cdot 0} = Ce^0 = C\). It is often denoted as \(y_0\).
In the model \(\frac{dy}{dt} = ky\), what is the significance of the sign of the constant \(k\)?
If \(k > 0\), the quantity \(y\) is increasing, which models exponential growth. If \(k < 0\), the quantity \(y\) is decreasing, which models exponential decay.
A population \(P\) of an endangered species is decreasing at a rate of \(3\)% per year. If the current population is \(1,200\), what is the particular solution that models this population?
The DE is \(\frac{dP}{dt} = -0.03P\). The general solution is \(P(t) = Ce^{-0.03t}\). The initial condition is \(P(0)=1200\), so \(C=1200\). The particular solution is \(P(t) = 1200e^{-0.03t}\).
The half-life of Carbon-14 is approximately \(5730\) years. What is the value of the decay constant \(k\) in the model \(A(t) = A_0 e^{kt}\)?
At \(t=5730\), \(A(t) = 0.5A_0\). So, \(0.5A_0 = A_0e^{k(5730)}\). This gives \(\ln(0.5) = 5730k\), so \(k = \frac{\ln(0.5)}{5730}\), which is approximately \(-0.000121\).
What is the differential equation for Newton's Law of Cooling?
\(\frac{dT}{dt} = k(T - T_s)\), where \(T(t)\) is the temperature of the object at time \(t\), \(T_s\) is the constant temperature of the surroundings, and \(k\) is a cooling constant (\(k<0\)).
Show that Newton's Law of Cooling, \(\frac{dT}{dt} = k(T - T_s)\), is a separable differential equation.
Yes, it is separable. We can divide by \((T - T_s)\) and multiply by \(dt\) to arrange it as \(\frac{1}{T - T_s} dT = k \, dt\).
What is the general solution to Newton's Law of Cooling, \(\frac{dT}{dt} = k(T - T_s)\)?
Integrating \(\frac{1}{T - T_s} dT = k \, dt\) gives \(\ln|T - T_s| = kt + C\). The explicit solution is \(T(t) = T_s + A e^{kt}\), where \(A\) is a constant determined by the initial temperature.
What is an orthogonal trajectory to a family of curves?
An orthogonal trajectory is a curve that intersects every curve in a given family at a right (\(90^{\circ}\)) angle. The family of orthogonal trajectories is itself a family of curves.
If a family of curves is defined by a differential equation \(\frac{dy}{dx} = f(x, y)\), what is the differential equation for its orthogonal trajectories?
Since perpendicular lines have negative reciprocal slopes, the differential equation for the orthogonal trajectories is \(\frac{dy}{dx} = -\frac{1}{f(x, y)}\). Solving this new DE gives the family of orthogonal curves.
Find the differential equation for the orthogonal trajectories to the family of parabolas \(y = cx^2\).
First find the slope of the family: \(\frac{dy}{dx} = 2cx\). Eliminate the constant \(c\) using \(c = \frac{y}{x^2}\), so \(\frac{dy}{dx} = 2(\frac{y}{x^2})x = \frac{2y}{x}\). The orthogonal slope is the negative reciprocal: \(\frac{dy}{dx} = -\frac{x}{2y}\).
