Need to find variables

[Franky]

T(t) = 250 - ae^(-bt)?
A yam is put in an oven maintained at a constant temperature of 250 degrees Celsius. Suppose that after 30 minutes, the temperature of the yam is 150 degrees Celsius and is increasing at a rate of 3 degrees Celsius/minute. If the temperature of the yam t minutes after it is put in the oven is modeled by T(t) = 250 - ae^(-bt) Find a and b.


I know I need the derivative, which is: T’(t)= ab*e^-bt. And t=30 min. So we should have: T’(30)= ab*e^-30b ... as again t=30. Now how do I get “a” “b”?

 

[Deleted member 4993]

T(t) = 250 - ae^(-bt)?
A yam is put in an oven maintained at a constant temperature of 250 degrees Celsius. Suppose that after 30 minutes, the temperature of the yam is 150 degrees Celsius and is increasing at a rate of 3 degrees Celsius/minute. If the temperature of the yam t minutes after it is put in the oven is modeled by T(t) = 250 - ae^(-bt) Find a and b.


I know I need the derivative, which is: T’(t)= ab*e^-bt. And t=30 min. So we should have: T’(30)= ab*e^-30b ... as again t=30. Now how do I get “a” “b”?

so

150 = 250 - a * e^(-30*b)

a * e^(-30*b) = 100.....................................(1)

taking log_e on both sides

ln(a) - 30 * b = ln(100)...................................(2)

similarly for the derivative:

3 = a*b*e^(-30*b)........................................(3)

dividing (3) by (1) - you can solve for 'b'

Then use (2) to solve for 'a'.

 

[Franky]

Thank you