The differential equation below models the temperature of a 87°C cup of coffee in a 17°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 67°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with
t = 0
corresponding to the time when the temperature was 87°C.)
dy
dt = -(1/50)(y-17)
I also have a little variation of this question but why do you take the ln(y-17)?
You take the ln because you integrate it and have log base e
To solve the given differential equation, we can separate the variables and integrate both sides.
First, let's rewrite the given equation:
dy/dt = -(1/50)(y - 17)
To separate the variables, we need to move the terms involving y to one side and the terms involving t to the other side:
dy / (y - 17) = -(1/50) dt
Now, we can integrate both sides.
∫ (dy / (y - 17)) = ∫ (-(1/50) dt)
To integrate the left side, we use the substitution u = y - 17, du = dy:
∫ (1/u) du = - (1/50) ∫ dt
ln|u| = - (1/50) t + C1
Now, substituting back u = y - 17, we have:
ln|y - 17| = - (1/50) t + C1
Next, let's solve for y:
|y - 17| = e^(- (1/50) t + C1)
We can remove the absolute value by considering two cases:
Case 1: y - 17 > 0
y - 17 = e^(- (1/50) t + C1)
y = e^(C1) * e^(- (1/50) t) + 17
Case 2: y - 17 < 0
- (y - 17) = e^(- (1/50) t + C1)
y = 17 - e^(C1) * e^(- (1/50) t)
Now, let's simplify by combining constants:
Let A = e^(C1)
Case 1: y = A * e^(- (1/50) t) + 17
Case 2: y = 17 - A * e^(- (1/50) t)
Therefore, the general solution to the differential equation is:
y = A * e^(- (1/50) t) + C
Where A and C are constants that depend on the initial conditions. To find the particular solution, we need a specific initial condition, such as the initial temperature of 87°C at time t = 0.
Using the given initial condition, we substitute y = 87 and t = 0 into the general solution:
87 = A * e^(- (1/50) * 0) + C
87 = A + C
Now, we need another piece of information to determine the values of A and C. Specifically, we need to know the rate of cooling at y = 67.
Using the given rate of cooling, substitute y = 67 into the general solution:
1 = (-1/50) * (67 - 17)
Simplifying this equation, we find:
1 = -50/50
This equation is not true, so there must be a mistake in the problem setup. Please double-check the given rate of cooling at y = 67 and the initial temperature.
just separate the variables as usual
dy/dt = -1/50 (y-17)
dy/(y-17) = -1/50 dt
ln(y-17) = -1/50 t + C
y = c e^(-t/50) + 17
y(0) = 87, so c=70
y(t) = 17+70e^(-t/50)