The dropping at the rate of g (t)=10e^(-0.1t) for 0 is less than or = to t less than or = 10, where g is measured in degrees in Fahrenheit and t in minutes.
If the metal is initially 100 degrees Fahrenheit, what is the temperature to the nearest degree Fahrenheit after 6 minutes? I'm stuck with this problem.
The equation is:
g (t)=10e^(-0.1t), 0 <= t <= 10
g(t) is the dropping rate, which I assume means the rate at which the temperature drops.
Let T(t) be the temperature of the metal over time.
Then dT/dt = g(t)
Now, a differential quotient, such as dT/dt, is by definition, the rate of increase of T wrt t.
Since we know that the Temperature is actually dropping over time, the dropping rate, g(t), must be negative.
i.e. g (t) = -10e^(-0.1t), 0 <= t <= 10
So,
dT/dt = -10e^(-0.1t)
Integrating wrt t,
T(t) = 100e^(-0.1t) + const
We are given that initially (at t = 0), the temperature of the metal is 100 degrees. i.e.
100 = 100e^(0) + const
100 = 100 + const
Const = 0
Our expression for the temp of the metal then is
T(t) = 100e^(-0.1t)
At t = 6,
T(6) = 100e^(-0.6)
T(6) = 100*0.5488
T(6) = 54.88
The temperature of the metal after 6 minutes is 55 degrees Fahrenheit