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# C3 exponential differentiation

1. 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
2. i thaught it would be -60 not -6
3. (Original post by 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)
4. (Original post by 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
ln ?
5. I answered this EXACT question a few hours ago.
I'll copy out my solution for you

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

6. (Original post by me, myself and I)
I answered this EXACT question a few hours ago.
I'll copy out my solution for you

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

thats it ^^
7. (Original post by me, myself and I)
I answered this EXACT question a few hours ago.
I'll copy out my solution for you

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

Ahhh brillaint, thanks! I was getting 12.04 but forgot to add it back in so I was doing again and again. Thank you!
8. (Original post by 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!
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

9. (Original post by me, myself and I)
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

oooh, interview for what if you don't mind me asking?
10. (Original post by kinglrb)
oooh, interview for what if you don't mind me asking?
it was for civil engineering 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!
11. (Original post by me, myself and I)
it was for civil engineering 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
12. (Original post by kinglrb)
Oh brillaint! I am sure you did great, Good luck with that
Thanks!
are you applying this year?
13. (Original post by me, myself and I)
Thanks!
are you applying this year?
Yupp , 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
14. (Original post by kinglrb)
Yupp , 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
Ooooh of course, just realised it says so in your sig! lol, good luck with them all though
15. (Original post by me, myself and I)
Ooooh of course, just realised it says so in your sig! lol, good luck with them all though
Thank you!

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