Active Calculus

1. A few steps of Euler's method.

2. Using Euler's method for a solution of \(y'=4y\).

3. Using Euler's method with different time steps.

4.

Newton's Law of Cooling says that the rate at which an object, such as a cup of coffee, cools is proportional to the difference in the object's temperature and room temperature. If \(T(t)\) is the object's temperature and \(T_r\) is room temperature, this law is expressed at

where \(k\) is a constant of proportionality. In this problem, temperature is measured in degrees Fahrenheit and time in minutes.

1. Two calculus students, Alice and Bob, enter a 70\(^\circ\) classroom at the same time. Each has a cup of coffee that is 100\(^\circ\text{.}\) The differential equation for Alice has a constant of proportionality \(k=0.5\text{,}\) while the constant of proportionality for Bob is \(k=0.1\text{.}\) What is the initial rate of change for Alice's coffee? What is the initial rate of change for Bob's coffee?

2. What feature of Alice's and Bob's cups of coffee could explain this difference?

3. As the heating unit turns on and off in the room, the temperature in the room is

   \begin{equation*} T_r=70+10\sin t\text{.} \end{equation*}

   Implement Euler's method with a step size of \(\Delta t = 0.1\) to approximate the temperature of Alice's coffee over the time interval \(0\leq t\leq 50\text{.}\) This will most easily be performed using a spreadsheet such as Excel. Graph the temperature of her coffee and room temperature over this interval.

4. In the same way, implement Euler's method to approximate the temperature of Bob's coffee over the same time interval. Graph the temperature of his coffee and room temperature over the interval.

5. Explain the similarities and differences that you see in the behavior of Alice's and Bob's cups of coffee.

5.

We have seen that the error in approximating the solution to an initial value problem is proportional to \(\Delta t\text{.}\) That is, if \(E_{\Delta t}\) is the Euler's method approximation to the solution to an initial value problem at \(\overline{t}\text{,}\) then

for some constant of proportionality \(K\text{.}\)

In this problem, we will see how to use this fact to improve our estimates, using an idea called accelerated convergence.

1. We will create a new approximation by assuming the error is exactly proportional to \(\Delta t\text{,}\) according to the formula

   \begin{equation*} y(\overline{t})-E_{\Delta t} =K\Delta t\text{.} \end{equation*}

   Using our earlier results from the initial value problem \(dy/dt = y\) and \(y(0)=1\) with \(\Delta t = 0.2\) and \(\Delta t = 0.1\text{,}\) we have

   \begin{align*} y(1) - 2.4883 =\mathstrut \amp 0.2K\\ y(1) - 2.5937 =\mathstrut \amp 0.1K\text{.} \end{align*}

   This is a system of two linear equations in the unknowns \(y(1)\) and \(K\text{.}\) Solve this system to find a new approximation for \(y(1)\text{.}\) (You may remember that the exact value is \(y(1) = e = 2.71828\ldots\text{.}\))

2. Use the other data, \(E_{0.05} = 2.6533\) and \(E_{0.025} = 2.6851\) to do similar work as in (a) to obtain another approximation. Which gives the better approximation? Why do you think this is?

3. Let's now study the initial value problem

   \begin{equation*} \frac{dy}{dt} = t-y, \ y(0) = 0\text{.} \end{equation*}

   Approximate \(y(0.3)\) by applying Euler's method to find approximations \(E_{0.1}\) and \(E_{0.05}\text{.}\) Now use the idea of accelerated convergence to obtain a better approximation. (For the sake of comparison, you want to note that the actual value is \(y(0.3) = 0.0408\text{.}\))

6.

In this problem, we'll modify Euler's method to obtain better approximations to solutions of initial value problems. This method is called the Improved Euler's method.

In Euler's method, we walk across an interval of width \(\Delta t\) using the slope obtained from the differential equation at the left endpoint of the interval. Of course, the slope of the solution will most likely change over this interval. We can improve our approximation by trying to incorporate the change in the slope over the interval.

Let's again consider the initial value problem \(dy/dt = y\) and \(y(0) = 1\text{,}\) which we will approximate using steps of width \(\Delta t = 0.2\text{.}\) Our first interval is therefore \(0\leq t \leq 0.2\text{.}\) At \(t=0\text{,}\) the differential equation tells us that the slope is 1, and the approximation we obtain from Euler's method is that \(y(0.2)\approx y_1= 1+ 1(0.2)= 1.2\text{.}\)

This gives us some idea for how the slope has changed over the interval \(0\leq t\leq 0.2\text{.}\) We know the slope at \(t=0\) is 1, while the slope at \(t=0.2\) is 1.2, trusting in the Euler's method approximation. We will therefore refine our estimate of the initial slope to be the average of these two slopes; that is, we will estimate the slope to be \((1+1.2)/2 = 1.1\text{.}\) This gives the new approximation \(y(1) = y_1 = 1 + 1.1(0.2) = 1.22\text{.}\)

The first few steps look like what is found in Table 7.3.15.

1. Continue with this method to obtain an approximation for \(y(1) = e\text{.}\)

2. Repeat this method with \(\Delta t = 0.1\) to obtain a better approximation for \(y(1)\text{.}\)

3. We saw that the error in Euler's method is proportional to \(\Delta t\text{.}\) Using your results from parts (a) and (b), what power of \(\Delta t\) appears to be proportional to the error in the Improved Euler's Method?