Consider a cooling cup of coffee whose initial temperature is 205°. The room temperature is held at 70°. Suppose k = 1/16. Let y be the temperature, and y' its time derivative.
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I have the differential equation:
y' = (-1/16)(y-70)
Initial condition: yo = 205
Continuous solution equation:
y(t) = 70 + 135*exp((-1/16)(t))
I am supposed to use Euler's Method to find the temperature after 5, 20, 30 and 60 minutes. I have to find: Temperature
Euler's Method (dt = 5) and Temperature
Euler's Method (dt = 1) for 5, 20, 30 and 60. How do I do this??? Please help
I assume your temperatures are in Fahrenheit. What is k supposed to be? It should have dimensions
esd
To use Euler's method to approximate the temperature after a given time using a given time step (dt), you can follow these steps:
1. Start with the initial condition: y(0) = 205, which represents the initial temperature of the coffee.
2. Choose a time step, dt. For the first case, dt = 5. The time step determines the interval at which you will calculate the approximate temperature.
3. To find the temperature at t = dt, use the formula:
y(dt) = y(0) + dt * y'(0)
where y'(0) is the derivative of y at t = 0, which is calculated by substituting the initial condition into the differential equation:
y'(0) = (-1/16)(y(0) - 70).
4. Use the value of y(dt) obtained from the previous step as the new y(0) for the next iteration. Then repeat steps 3 and 4 to find the temperature at specific time points.
Let's go through each case:
Case 1: Euler's Method with dt = 5
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For t = 5 minutes:
y(5) = y(0) + dt * y'(0)
= 205 + 5 * (-1/16)(205 - 70)
= 172.34
For t = 20 minutes:
y(20) = y(5) + dt * y'(5)
= 172.34 + 5 * (-1/16)(172.34 - 70)
≈ 134.75
For t = 30 minutes:
y(30) = y(20) + dt * y'(20)
= 134.75 + 5 * (-1/16)(134.75 - 70)
≈ 107.23
For t = 60 minutes:
y(60) = y(30) + dt * y'(30)
= 107.23 + 5 * (-1/16)(107.23 - 70)
≈ 76.40
Case 2: Euler's Method with dt = 1
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Using the same steps as above:
For t = 5 minutes:
y(5) = y(0) + dt * y'(0)
= 205 + 1 * (-1/16)(205 - 70)
≈ 200.94
For t = 20 minutes:
y(20) = y(5) + dt * y'(5)
= 200.94 + 1 * (-1/16)(200.94 - 70)
≈ 186.30
For t = 30 minutes:
y(30) = y(20) + dt * y'(20)
= 186.30 + 1 * (-1/16)(186.30 - 70)
≈ 175.00
For t = 60 minutes:
y(60) = y(30) + dt * y'(30)
= 175.00 + 1 * (-1/16)(175.00 - 70)
≈ 155.80
Therefore, the approximate temperatures after 5, 20, 30, and 60 minutes using Euler's Method with dt = 5 are approximately 172.34, 134.75, 107.23, and 76.40 respectively.
And the approximate temperatures after 5, 20, 30, and 60 minutes using Euler's Method with dt = 1 are approximately 200.94, 186.30, 175.00, and 155.80 respectively.