Question

the temperature, T, in degrees celsius, of a yam put into a 200 degrees C oven is given as a function of time, t, in minutes, by:
T = a(1-e^kt)+ b.
a)if the yam starts at 20 degrees C, find a and b.
b)if the temperature of the yam is initially increasing at 2 degrees C per minute, find k.

Answer 1

k 589002

Answer 2

duh

Answer 3

?

Answer 4

idk

Answer 5

hope i helped

Answer 6

First off, your equation should be

\[T(t)=a(1-e^{-kt})+b\]
You need the exponent to be negative, or else the equation blows up at infinity.

A) you want to look at the equation at t=0 and t= infinity

It's temperature when it starts out, at t=0, is 20ºC, so we can solve for b
\[T(0)=20ºC=a(1-\cancel{e^{k*0}}^1)+b
\\ \hspace{2.1cm} =a(0)+b\]

At t=infinity, the temperature of the cooked yam will be 200ºC
\[T(\infty)=200ºC = a(1-\cancel{e^{k*\infty }}^0)+b \]
Since we already know b, we can solve for a now.


B) So if we know that the yam is increasing it's temperature 2ºC per minute, we know what the temperature of the yam is at t=1 min

\[T(1)=22ºC=a(1-e^{-k*1})+b\]

And since we know both a and b from the other part, you can solve for k.


Remember, to solve exponentials in base e, you take the natural log
\[c=e^{-x}\]
\[\ln c = \ln e^{-x}\]
\[ \ln c = -x\]