Question

A super bread dough increases in volume at a rate proportional to the volume \(V\) present. If \(V\) increases by a factor of 10 in 2 hours and \(V(0)=V_{0},\) find \(V\) at any time \(t .\) How long will it take for \(V\) to increase to \(100 \mathrm{~V}_{0}\) ?

Short answer

Question: Find the time it takes for the volume of bread dough to increase to 100 times its initial volume, given that the volume increases by a factor of 10 in 2 hours. Answer: It will take approximately 4 hours for the volume to increase to 100 times its initial volume.

Step-by-step solution

Write the rate of increase formula

According to the problem, the rate of increase is proportional to the volume. We can represent this fact using differential equations that describe the rate of change of the volume with respect to time: \(dV/dt = kV\), Where \(k\) is a constant of proportionality.

Solve the differential equation

To solve the differential equation, we can use separation of variables. Divide both sides by \(V\), and multiply both sides by \(dt\): \(\frac{dV}{V} = k dt\). Now integrate both sides: \(\int \frac{dV}{V} = \int k dt\). The result is: \(ln(V) = kt + C\) , where \(C\) is the constant of integration.

Determine the initial condition and constant of integration

Now, we use the initial condition \(V(0) = V_{0}\) to find the constant of integration \(C\). By plugging in \(V = V_{0}\) and \(t = 0\), we get: \(ln(V_{0}) = k(0) + C \Rightarrow C = ln(V_{0})\).

Write the general solution of the differential equation

Substitute the value of \(C\) back into the equation: \(ln(V) = kt + ln(V_{0})\). By exponentiating both sides, we get the general solution: \(V(t) = V_{0}e^{kt}\).

Determine the value of k using the given data

We are given that the volume increases by a factor of 10 in 2 hours. So, when \(t = 2\), \(V = 10V_{0}\). Plug these values into the general solution: \(10V_{0} = V_{0}e^{2k}\). Divide both sides by \(V_{0}\), we get: \(10 = e^{2k}\). Now, take the natural logarithm of both sides to find \(k\): \(ln(10) = 2k\). Therefore, \(k = \frac{1}{2} ln(10)\).

Determine the time for the volume to reach 100V0

We want to find the time \(t\) when the volume reaches \(100V_{0}\). So, let \(V(t) = 100V_{0}\) and \(k = \frac{1}{2} ln(10)\). Plug these values into the general solution: \(100V_{0} = V_{0}e^{\frac{1}{2} ln(10)t}\). Divide both sides by \(V_{0}\) and take the natural logarithm of both sides: \(ln(100) = \frac{1}{2} ln(10)t\). Solve for \(t\): \(t = \frac{2 ln(100)}{ln(10)}\) Thus, it will take approximately \(t = 4\) hours for the volume to increase to \(100V_{0}\).

Key concepts

Rate of Increase

Understanding the rate of increase is crucial when it comes to the study of differential equations, specifically in contexts involving growth or decay over time. Imagine a baker observing their bread dough rising. The rate at which the volume of the dough changes is not random but instead based on how much dough there already is—the larger the dough, the more it can rise.

This phenomenon is described mathematically by differential equations like the one found in our exercise: dV/dt = kV. Here, V represents the volume of the dough, t is time, and k is a constant expressing the proportionality between the volume and its rate of change over time. To put it simply, the speed of the dough’s increase in volume directly relates to the volume present at any given moment.

Exponential Growth

The concept of exponential growth describes a situation where the increase in size or number of a certain quantity is proportional to its current value—meaning it grows faster as it gets larger. Our dough example perfectly reflects this: as the volume increases, the rate at which it grows also increases.

This is captured by the solution to our differential equation: V(t) = V_0e^{kt}. This formula indicates that the volume at any time t is the initial volume V_0 multiplied by the exponential function e to the power of the product of our growth rate k and time t. The exponential function here is key to representing the runaway effect of growth we observe in nature, economics, and many other fields.

Separation of Variables

Separation of variables is a mathematical method used to solve a wide array of differential equations, including the one in our problem. The core idea is to take an equation in which the variables are mingled together and rearrange it so each variable is with its differential—effectively separating them.

In practice, we take our initial rate equation dV/dt = kV and manipulate it to get dV/V = k dt. By moving all terms involving V to one side and all terms involving t to the other, we set the stage for the integration process, which will allow us to find a general solution that we can then tailor to specific situations using initial conditions.

Integration

The concept of integration might sound intimidating, but it's essentially just the process of finding the whole from the parts—a sort of mathematical 'summation'. After we've separated our variables, we need to integrate both sides to undo the differentiation and get back to an equation in terms of V and t only.

In the context of our exercise, we integrate dV/V with respect to V, and k dt with respect to t, leading us to ln(V) and kt plus a constant, respectively. Integration is what transforms the initially impenetrable rate of change into a manageable equation that describes the volume of dough at any given time in relation to its exponential rate of growth.