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Chapter 4: Problem 13

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

Expert verified
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

01

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.
02

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.
03

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: \(ln(V_{0}) = k(0) + C \Rightarrow C = ln(V_{0})\).
04

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: \(V(t) = V_{0}e^{kt}\).
05

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, \(k = \frac{1}{2} ln(10)\).
06

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 \(100V_{0}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攖he 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鈥檚 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鈥攎eaning 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鈥攅ffectively 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鈥攁 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.

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Most popular questions from this chapter

A process creates a radioactive substance at the rate of \(2 \mathrm{~g} / \mathrm{hr}\) and the substance decays at a rate proportional to its mass, with constant of proportionality \(k=.1(\mathrm{hr})^{-1}\). If \(Q(t)\) is the mass of the substance at time \(t,\) find \(\lim _{t \rightarrow \infty} Q(t)\)

An object with mass \(m\) is given an initial velocity \(v_{0} \leq 0\) in a medium that exerts a resistive force of the form \(a|v| /(1+|v|),\) where \(a\) is positive constant. (a) Set up a differential equation for the speed of the object. (b) Use your favorite numerical method to solve the equation you found in (a), to convince yourself that there's a unique number \(a_{0}\) such that \(\lim _{t \rightarrow \infty} s(t)=\infty\) if \(a \leq a_{0}\) and \(\lim _{t \rightarrow \infty} s(t)\) exists (finite) if \(a>a_{0}\). (We say that \(a_{0}\) is the bifurcation value of \(a\).) Try to find \(a_{0}\) and \(\lim _{t \rightarrow \infty} s(t)\) in the case where \(a>a_{0} .\) HINT: See Exercise \(14 .\)

Suppose water is added to a tank at \(10 \mathrm{gal} / \mathrm{min}\), but leaks out at the rate of \(1 / 5 \mathrm{gal} / \mathrm{min}\) for each gallon in the tank. What is the smallest capacity the tank can have if the process is to continue indefinitely?

Find the decay constant \(k\) for a radioactive substance, given that the mass of the substance is \(Q_{1}\) at time \(t_{1}\) and \(Q_{2}\) at time \(t_{2}\).

A tank initially contains 100 liters of a salt solution with a concentration of \(.1 \mathrm{~g} /\) liter. A solution with a salt concentration of \(.3 \mathrm{~g} /\) liter is added to the tank at 5 liters/min, and the resulting mixture is drained out at the same rate. Find the concentration \(K(t)\) of salt in the tank as a function of \(t\).
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