Question

An object is taken from a freezer at negative 8 degrees Upper C. Let t be the time in hours after the object was taken from the freezer. At time t the average temperature of the object is increasing at the rate of Upper T prime (t )equals 20 e Superscript negative 0.4 t degrees Celsius per hour. Find the temperature of the object at time t.

Answer 1

`T(t)=-50e^(-0.4t)+42`

Explanation:

We are given that

`T(0)=-8^(\circ) C`

`T'(t)=20e^(-0.4t)`

Integrating on both sides

`T(t)=20\int e^(-0.4 t)`

`T(t)=(20)/(-0.4)e^(-0.4t)+C`

Using the formula

`\int e^(ax)dx=(e^(ax))/(a)+C`

Substitute t=0 and T(0)=-8

`-8=-50+C`

`C=-8+50=42`

Substitute the value of C

`T(t)=-50e^(-0.4t)+42`