Problem 28 - Thomas Calculus in SI Units

Question

Suppose that electricity is draining from a capacitor at a rate that is proportional to the voltage \(V\) across its terminals and that, if \(t\) is measured in seconds, $$\frac{d V}{d t}=-\frac{1}{40} V$$ Solve this equation for \(V\), using \(V_{0}\) to denote the value of \(V\) when \(t=0 .\) How long will it take the voltage to drop to \(10 \%\) of its original value?

Step-by-Step Solution

Answer

The voltage drops to 10% of its original value in approximately 92.1 seconds.

Step 1: Identify the Nature of the Equation

The given equation \( \frac{dV}{dt} = -\frac{1}{40} V \) is a first-order linear differential equation. This type of equation is separable, which means we can separate the variables \( V \) and \( t \) to solve it.

Step 2: Separate Variables

Rearrange the equation to separate the variables:\[ \frac{1}{V} \frac{dV}{dt} = -\frac{1}{40}. \]This allows us to integrate both sides with respect to their variables.

Step 3: Integrate Both Sides

Integrate both sides of the separated equation: \[ \int \frac{1}{V} \, dV = \int -\frac{1}{40} \, dt. \]This results in:\[ \ln |V| = -\frac{1}{40}t + C, \]where \( C \) is the integration constant.

Step 4: Solve for \( V \)

Exponentiate both sides to solve for \( V \):\[ V = e^{C}e^{-\frac{1}{40}t}. \]Let \( e^C = V_0 \), where \( V_0 \) is the initial voltage at \( t = 0 \), giving:\[ V = V_0 e^{-\frac{1}{40}t}. \]

Step 5: Determine Time to Reach 10% Voltage

The voltage \( V \) drops to 10% of its initial value when \( V = 0.1V_0 \):\[ 0.1V_0 = V_0 e^{-\frac{1}{40}t}. \]Divide both sides by \( V_0 \):\[ 0.1 = e^{-\frac{1}{40}t}. \]Take the natural log of both sides:\[ \ln 0.1 = -\frac{1}{40}t. \]Solve for \( t \):\[ t = -40 \ln 0.1. \]

Key Concepts

Separable Differential Equations

Separable differential equations are a type of first-order differential equations where variables can easily be separated to each side of the equation for integration. They take the general form \( \frac{dy}{dx} = g(y)h(x) \). Here, you can move all terms involving \( y \) to one side and all terms involving \( x \) to the other.

• Step 1: Rearrange the equation so each variable term is on one side: \( \frac{dy}{g(y)} = h(x) dx \).
• Step 2: Integrate both sides independently. This allows us to find a solution linking \( y \) and \( x \).

This process makes solving differential equations more straightforward as long as the equation is separable. In our problem, the given equation \( \frac{dV}{dt} = -\frac{1}{40}V \) can be rearranged and separated as \( \frac{1}{V} \frac{dV}{dt} = -\frac{1}{40} \). By integrating each side, you can find a functional form for \( V \) in terms of \( t \).

First-Order Linear Differential Equations

First-order linear differential equations are the simplest type of differential equations involving a function and its first derivative. Their standard form is \( \frac{dy}{dx} + P(x)y = Q(x) \). These equations are highly adaptable to various problems.

• Recognizing Form: Identify if your equation matches the linear form. For our exercise, we identified \( \frac{dV}{dt} = -\frac{1}{40}V \) as a first-order linear equation.
• Use an Integrating Factor: Typically, an integrating factor is used, but if the equation is separable (like in our case), this step is unnecessary.

Once we identify the equation's type, solving it involves either separation of variables as discussed or, usually, inhomogeneous techniques for more complex scenarios. Our problem benefited from being a separable case, simplifying the process.

Exponential Decay

Exponential decay describes processes where a quantity decreases at a rate proportional to its current value. It captures natural decreases over time, commonly seen in radioactive decay and charging in capacitors.

• General Form: The solution formula \( V = V_0 e^{-kt} \) shows how exponential decay works. Here, \( V_0 \) is the initial value, \( k \) is the decay constant, and \( t \) is time.
• Decay Association: In our example \( \frac{dV}{dt} = -\frac{1}{40}V \), the negative sign denotes decay.

Understanding exponential decay helps solve many real-world problems. The exercise illustrated how voltage in a capacitor decays over time. By solving when voltage is 10% of its initial value, we directly applied the concept of exponential decay.