x Turn on thread page Beta
 You are Here: Home >< Maths

# differentiating logs and exponentials watch

1. Hi guys,

I need some help with a differentiation question. I've checked the answer but i cannot figure out how they solved it... =(

These are the problems i am having:

part ic) 'after a long time' therefore means t→∞ so e^kt would tend to 0 right? which means p will equal 0 after a long time? but the answer says it will increase to infinity..... =S

part iia) when they make t= 0, how did they get dp/dt = 6?!

iib) see, this time p=0 after a long time becase as t→∞, e^(at-bt^2) tends to 0....so why is this wrong for part ic?

thank you so so so much for helping.
Attached Images
2. exd9.pdf (17.0 KB, 121 views)
3. 'after a long time' therefore means t→∞ so e^kt would tend to 0 right?
Not sure how you came to that answer. t gets bigger so e^kt gets bigger, seems pretty obvious to me.
4. (Original post by strwbry_short_cake)
Hi guys,

I need some help with a differentiation question. I've checked the answer but i cannot figure out how they solved it... =(

These are the problems i am having:

part ic) 'after a long time' therefore means t→∞ so e^kt would tend to 0 right?
no; for t->Inf exp(kt)->Inf

which means p will equal 0 after a long time? but the answer says it will increase to infinity..... =S

part iia) when they make t= 0, how did they get dp/dt = 6?!
P(t)=3*exp(at-bt^2)
dP/dt=3*(exp(at-bt^2))*(a-2bt)

at time t=0, dP/dt=3*a

iib) see, this time p=0 after a long time becase as t→∞, e^(at-bt^2) tends to 0.....
Because e^(-t)->0 for t->Inf
5. thanks for helping

o yes, thinking about it, a positive exponential will always increase to infinity while a negative exponential will decrease.....

but i still don't understand how they got the '6' in the first place:

6. (Original post by strwbry_short_cake)
thanks for helping

o yes, thinking about it, a positive exponential will always increase to infinity while a negative exponential will decrease.....

but i still don't understand how they got the '6' in the first place:
sorry, I didn't see you had to determine a and b.

I would do it like this: maximum at t=10 => dP/dt=0 at t=10 , so

3*(exp(10a-100b))*(a-20b)=0
i.e.
(1) a=20b
and from before you know that
(2) 3a=6,

so just solve (1) for b and plug in a=2.

Oh and the "6" is given in the introduction (i.e. they measured it experimentally). btw is it normal that they don't write down the units?
7. oh i see, i didn't use the dp/dt = 6 because i thought it belonged to the first model...
yh they sometimes don't write the units in the answers.....
thanks for the help btw.

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:
Updated: January 15, 2010
Today on TSR

### Did he block me?

What should I do?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams