Hey! Sign in to get help with your study questionsNew here? Join for free to post
 You are Here: Home >< Maths

# C3 exponential differentiation

Announcements Posted on
Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
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

your first answer is correct.

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

your first answer is correct.

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

your first answer is correct.

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!

Write a reply…

Submit reply

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
that username has been taken, please choose another Forgotten your password?
2. this can't be left blank
this email is already registered. Forgotten your password?
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
your full birthday is required
1. Oops, you need to agree to our Ts&Cs to register
2. Slide to join now Processing…

Updated: November 19, 2008
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.