Question

Use Euler's method to solve dB/dt = 0.05B with initial value B = 800 when t = 0. Delta t = 1 and 1 step: B(1) = Delta t = 0.5 and 2 steps: B(1) = Delta t = 0.25 and 4 steps: B(1) = Suppose B is the balance in a bank account earning interest. Be sure that you can explain why the result of your calculation in part (a) is equivalent to compounding the interest once a year instead of continuously. Then interpret the result of your calculations in parts (b) and (c) in terms of compound interest.

Answer 1

(a) Using Euler's method with ∆t = 0.5, B(1) = 820.

(b) With ∆t = 0.25, B(1) = 830.

(c) Employing ∆t = 0.125, B(1) = 840.

Step-by-step explanation:

(a) Euler's method approximates the solution of a differential equation through discrete steps. Starting with ∆t = 0.5, for one step, the formula B(1) = B(0) + ∆t * dB/dt gives B(1) = 800 + 0.5 * 0.05 * 800 = 820. This represents an estimate of the balance after one year.

(b) Decreasing ∆t to 0.25 enhances the accuracy of the approximation. Applying the same formula, B(1) = 800 + 0.25 * 0.05 * 800 = 830. A smaller ∆t means more frequent recalculations, which aligns closer to continuous compounding.

(c) Further reducing ∆t to 0.125 produces an improved estimate: B(1) = 800 + 0.125 * 0.05 * 800 = 840. As ∆t decreases, the calculated balance approaches a value more akin to continuous compounding. This aligns with the notion that smaller time intervals result in more accurate approximations, resembling continuous compounding where interest is calculated infinitely often.

Interpreting these results in terms of compound interest, larger ∆t values mimic annual compounding, whereas smaller ∆t values resemble more frequent compounding intervals within a year, reflecting a closer approximation to continuous compounding. The decreasing interval ∆t leads to more accurate estimates that closely resemble continuous compounding, capturing a more precise representation of the growth in the account balance.

Answer 2

The given differential equation dB/dt = 0.05B is a simple interest formula, where the interest rate is 0.05 (5%). Euler's method is used to approximate the balance at different time points. The results in parts (a), (b), and (c) represent the balance at t = 0.5,0.25 and 0.125 years, respectively.

Step-by-step explanation:

We can use Euler's method to solve the differential equation dB/dt = 0.05B with the initial value B(0) = 800.

Euler's method estimates the value of the function at subsequent points by taking small steps and using the slope at the beginning of the step to estimate the value at the end of the step.

Part A: One-step approximation

For Delta t = 1 and one step, we calculate B(1) as follows:

B(1) ≈ B(0) + (Delta t) * (dB/dt at t = 0)
B(1) ≈ 800 + (1) * (0.05 * 800)
B(1) ≈ 800 + 40
B(1) ≈ 840

This is equivalent to compounding the interest once a year at a 5% annual rate.

Part B: Two-step approximation

For Delta t = 0.5 and two steps, the approximation is done in two intervals:

1. B(0.5) ≈ 800 + (0.5) * (0.05 * 800) = 820
2. B(1) ≈ 820 + (0.5) * (0.05 * 820) ≈ 841

This mimics compounding interest twice a year.

Part C: Four-step approximation

For Delta t = 0.25 and four steps, it involves the following calculations:

1. B(0.25) ≈ 800 + (0.25) * (0.05 * 800)
2. B(0.5) ≈ B(0.25) + (0.25) * (0.05 * B(0.25))
3. B(0.75) ≈ B(0.5) + (0.25) * (0.05 * B(0.5))
4. B(1) ≈ B(0.75) + (0.25) * (0.05 * B(0.75))

The result of this approximation would be higher than the two-step approximation, representing compounding interest four times a year.