Solve Word Problem: Find a and b in T(t)=250-ae^(-bt)

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SUMMARY

The discussion focuses on solving the equation T(t) = 250 - ae^(-bt) to find the constants a and b, given the temperature of a yam in an oven at 250°C. After 30 minutes, the yam's temperature is 150°C and increasing at a rate of 3°C/min. Using Newton's law of cooling, the values of b can be derived as b = ln(10/7)/10, while a can be calculated using the equation a = 70e^(40b). The discussion emphasizes the application of differential equations in thermal dynamics.

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Homework Statement



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

Homework Equations





The Attempt at a Solution



I have no idea how to even start. Please help
 
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Just by looking at the question as stated, what are the values of:

T(30)
T'(30) ?

How can you use these to find a and b?
 
betsinda said:
A yam is pust in an oven maintained at a constant temp of 250degrees C. Suppose that after 30 min the temp of the yam is 150degrees C and is increasing at a rate of 3degrees C/min. If the temp of the yam t minutes after it is put in the oven is modeled by
T(t)=250-ae^(-bt), find a and b.

Hi betsinda! :smile:

Hint: what is Newton's law of cooling?
 
I think I have found b

180 = 250 - ae^-40b
150 = 250 - ae^-30b


-70 = -ae^-40b
70/e^-40b = a

-100 = -70/e^-40b * e^-30b
100/70 = e^40b * e ^-30b
10/7 = e^10b
ln 10/7 = 10 b
b = ln(10/7)/10

but how do I find a?
 
betsinda said:
70/e^-40b = a
:
10/7 = e^10b

but how do I find a?

a = 70e40b = 70(e10b)4 :wink:
 
Thank you !
 

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