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C3 exponential differentiation

I would appreciate some help with this question thanks!

Given that T is a measure of temperature in degrees C, t is the time in minutes and that T = 20 + 60e^-0.1t, t > or equal to 0

a) Find dT/dt
I worked it out and I got -6e^-0.1t

b) Find the value of T at which the temperature is decreasing at a rate of 1.8 degrees C per minute

I thought I would do -6e^-0.1t = 1.8
but then I am stuck, the answer I should get is 38 :confused:
Reply 1
i thaught it would be -60 not -6
Reply 2
Hrov
i thaught it would be -60 not -6

I did it wrong with latex, I needed to differentiate 60e to the power of -0.1t
which is -6e to the power of -0.1t (same as the answer in the book)
kinglrb
I would appreciate some help with this question thanks!

Given that T is a measure of temperature in degrees C, t is the time in minutes and that T = 20 + 60e^-0.1t, t > or equal to 0

a) Find dT/dt
I worked it out and I got -6e^-0.1t

b) Find the value of T at which the temperature is decreasing at a rate of 1.8 degrees C per minute

I thought I would do -6e^-0.1t = 1.8
but then I am stuck, the answer I should get is 38 :confused:

ln ?
I answered this EXACT question a few hours ago.
I'll copy out my solution for you :smile:


your first answer is correct. :smile:

Then, find out the value for "t" and set the expression you found in a) equal to -1.8. (It's a decrease remember, so negative)

so basically -6e^0.1t = -1.8, take logs of both sides and solve for t.

Using your value of t (I get ~12.04 ish) substitute this into the original expression for T (not dT/dt).

So T = 60e^(-0.1 x 12.04) + 20

should give 38 *C

:smile:
me, myself and I
I answered this EXACT question a few hours ago.
I'll copy out my solution for you :smile:


your first answer is correct. :smile:

Then, find out the value for "t" and set the expression you found in a) equal to -1.8. (It's a decrease remember, so negative)

so basically -6e^0.1t = -1.8, take logs of both sides and solve for t.

Using your value of t (I get ~12.04 ish) substitute this into the original expression for T (not dT/dt).

So T = 60e^(-0.1 x 12.04) + 20

should give 38 *C

:smile:

thats it ^^
Reply 6
me, myself and I
I answered this EXACT question a few hours ago.
I'll copy out my solution for you :smile:


your first answer is correct. :smile:

Then, find out the value for "t" and set the expression you found in a) equal to -1.8. (It's a decrease remember, so negative)

so basically -6e^0.1t = -1.8, take logs of both sides and solve for t.

Using your value of t (I get ~12.04 ish) substitute this into the original expression for T (not dT/dt).

So T = 60e^(-0.1 x 12.04) + 20

should give 38 *C

:smile:

Ahhh brillaint, thanks! I was getting 12.04 but forgot to add it back in so I was doing again and again. Thank you!
kinglrb
Ahhh brillaint, thanks! I was getting 12.04 but forgot to add it back in so I was doing again and again. Thank you!


:biggrin: glad I could help! Just thought it was such a coincidence when I saw the question lol, I have an interview tomorrow so I've been preparing for it and getting stressed out, haha

:smile:
Reply 8
me, myself and I
:biggrin: glad I could help! Just thought it was such a coincidence when I saw the question lol, I have an interview tomorrow so I've been preparing for it and getting stressed out, haha

:smile:

oooh, interview for what if you don't mind me asking?
kinglrb
oooh, interview for what if you don't mind me asking?

it was for civil engineering :smile: at imperial
fingers crossed it went well, the interview wasn't as daunting as I thought it would be and the interviewer was lovely! but I don't want to think about it too much in case I don't get an offer on ucas! Loved the place though!
Reply 10
me, myself and I
it was for civil engineering :smile: at imperial
fingers crossed it went well, the interview wasn't as daunting as I thought it would be and the interviewer was lovely! but I don't want to think about it too much in case I don't get an offer on ucas! Loved the place though!

Oh brillaint! I am sure you did great, Good luck with that :wink:
kinglrb
Oh brillaint! I am sure you did great, Good luck with that :wink:

Thanks! :smile:
are you applying this year?
Reply 12
me, myself and I
Thanks! :smile:
are you applying this year?

Yupp :yep: , I've applied for medicine at Newcastle, Leicester, Aberdeen, Glasgow and Chemical Engineering at UCL. I still haven't got any interviews so am still waiting :woo:
kinglrb
Yupp :yep: , I've applied for medicine at Newcastle, Leicester, Aberdeen, Glasgow and Chemical Engineering at UCL. I still haven't got any interviews so am still waiting :woo:


Ooooh of course, just realised it says so in your sig! lol, good luck with them all though :smile:
Reply 14
me, myself and I
Ooooh of course, just realised it says so in your sig! lol, good luck with them all though :smile:

Thank you! :smile:

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