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

Physics is a correct section for this question, and there is a link for you.

http://openstudy.com/study#/groups/Physics

Answer 2

thanks...

Answer 3

\[\Large\bf T(t)\quad=\quad a(1-e^{kt})+b\]

`a) If the yam starts at 20 degrees C, find a and b.`
So the way to interpret this is, at time t=0 the temperature T of the yam is 20.
\(\Large\bf T(0)=20\)

Answer 4

\[\Large\bf 20\quad=\quad a(1-e^{0k})+b\]Understand how I set that up?
I set it equal to a `temperature` of 20, and then plugged in 0 for the `time`.

Answer 5

okay i understand that so far. so now we solve for a?
would it be a = (20-b) / (1-e^0k) ?

Answer 6

No it looks like we'll start by solving for b.
the exponential simplifies down a bit, \(\Large \bf e^{0k}=1\)

Answer 7

So our expression is:\[\Large\bf 20\quad=\quad a(1-1)+b\]

Answer 8

so b = 20. then plug it back into the equation?

Answer 9

yah let's do that and see if we can figure out a.. hmm

Answer 10

I'm confused! ): how do you find a if a(0) = 0?

Answer 11

We'll have to find a `without` plugging in the T(20)=0.
We can safely plug in b=20 though.

Answer 12

so you mean 20 = a(1-e^kt) + (20) ?

Answer 13

\[\Large\bf T(t)\quad=\quad a(1-e^{kt})+b\]
We used T(20)=0 to find b. b=20.
So now we plug that into our function.\[\Large\bf T(t)\quad=\quad a(1-e^{kt})+20\]

Answer 14

but now how do we solve for a if we are missing t?

Answer 15

wait. I'm sorry. i just realized that i typed the wrong equation. it should be T= a(1 - e^-kt) + b
does that change things?

Answer 16

\[\Large\bf T(t)\quad=\quad a(1-e^{-kt})+20\]No, luckily it doesn't change anything we've done so far.

Answer 17

`The yam is put into a 200 degree C oven,`
Hmm I'm trying to figure out how this plays into our problem :(
It probably helps us figure out `a` somehow. hmmm

Answer 18

I'm not sure why Solomon tried to kick you over to the Physics section :P lol
This is as much a math problem as it is physics XD

Answer 19

yea, thats what i thought lol.

Answer 20

So the `initial` difference in temperature between the yam and the oven is 200 - 20 = 180 degrees C.

I think we need to use this somehow.. hmmm thinkingggggggggg

Answer 21

@hartnn @Zarkon @Loser66

Answer 22

D: I'm sorry i have no idea what to do next

Answer 23

ya I'm a little stumped here :d grr

Answer 24

hmm

Answer 25

@satellite73

Answer 26

the @ paging thing isn't working right now :( grr

Answer 27

`a yam put into a 200 degrees C oven`
means the max value the yam can achieve in the oven is = 200 degreesC

T = a(1-e^-kt)+ b.
will be maximum when you do not subtract anything from 1
that is, when e^-kt = 0

so, Tmax = 200 = a(1-0)+b
from this you can find 'a' :)
see if it makes sense.

Answer 28

hint for part b)
` increasing at 2 degrees C per minute` ----> dT/dt = +2

Answer 29

okay so i got 1 = 180.

Answer 30

now for b.
i know you plug in 2 for dT/dt and find the derivative but I'm kind of confused about it :\

Answer 31

so you know you have to find the derivative, did you try finding it ? what did u get ?