[ilovemath]

The number of people, n, in a queue at a Post Office t minutes after it opens is
modelled by the differential equation

dn/dt = e^0.5t -5
t ≥ 0


Find, to the nearest second, the time when the model predicts that there will be
the least number of people in the queue.

I would have thought the answer is t=0 but the answer is t=3min 13 sec

why?

[Allofthem]

Why would you have thought that?

If I told you that n = f(t) and asked that you find the smallest n, tell me what the process would be.

[ilovemath]

(Original post by Allofthem)
Why would you have thought that?

If I told you that n = f(t) and asked that you find the smallest n, tell me what the process would be.

because t >=0 so that is the smallest POSSIBLE value (i.e: t = 0)

[SecretDuck]

Dn/dt = 0

Remember stationary points are dy/dx equal to zero and can be maximums or minimums.

[thefifthfoo]

This may help:

We have n = f(t) and we want to minimise n (or at least find out the time at which n is at a minimum)

Imagine sketching the function on a graph (relabelling n= f(t) as y, and t as x if this helps).

The minimum value of n (or y if you've relabelled) will be at the vertex of the graph (it only has one). So you need to know what value of t (aka x) this occurs at.

Bear in mind that you have been given dn/dt (which you may have relabelled as dy/dx).

I am sure that now this problem will seem much more familiar to you

[TenOfThem]

(Original post by ilovemath)
The number of people, n, in a queue at a Post Office t minutes after it opens is
modelled by the differential equation

dn/dt = e^0.5t -5
t ≥ 0


Find, to the nearest second, the time when the model predicts that there will be
the least number of people in the queue.

I would have thought the answer is t=0 but the answer is t=3min 13 sec

why?

Are you assuming that there are no people queuing at opening time ... if so ... why

[ilovemath]

(Original post by TenOfThem)
Are you assuming that there are no people queuing at opening time ... if so ... why

well..yes I guess...the MS says use dn/dt = 0
why?

[ilovemath]

(Original post by thefifthfoo)
This may help:

We have n = f(t) and we want to minimise n (or at least find out the time at which n is at a minimum)

Imagine sketching the function on a graph (relabelling n= f(t) as y, and t as x if this helps).

The minimum value of n (or y if you've relabelled) will be at the vertex of the graph (it only has one). So you need to know what value of t (aka x) this occurs at.

Bear in mind that you have been given dn/dt (which you may have relabelled as dy/dx).

I am sure that now this problem will seem much more familiar to you

but for dn/dt = 0 to work surely there has to be a minimum
I was not aware of any point on the e graph where gradient = 0

[SecretDuck]

(Original post by ilovemath)
well..yes I guess...the MS says use dn/dt = 0
why?

It's a rate of change. It is a derivative. Sketch the line by plugging values in. You should see that it intersects the y axis at around 3.

Actually, I've done it for you.

http://m.wolframalpha.com/input/?i=p...5x+-+5&x=0&y=0

[Coulson72]

(Original post by ilovemath)
but for dn/dt = 0 to work surely there has to be a minimum
I was not aware of any point on the e graph where gradient = 0

You have been given an implicit function, if you integrated you'd get n=2e^.5t-5t+c which does have a stationary point, which is a minimum at t= 3 mins 13 seconds (draw the graph if you dont agree)

[ilovemath]

(Original post by Coulson72)
You have been given an implicit function, if you integrated you'd get n=2e^.5t-5t+c which does have a stationary point, which is a minimum at t= 3 mins 13 seconds (draw the graph if you dont agree)

on plotting the lowest point appears at a negative t value?

[TenOfThem]

(Original post by ilovemath)
on plotting the lowest point appears at a negative t value?

You need a local minimum within the range given

[ilovemath]

(Original post by TenOfThem)
You need a local minimum within the range given

oh, so when I plot the function of e^0.5t - 5 I it cross at about t = 3min 13 sec

so just to check why this works:

the original function will have a minimum value (we can see that this is the case by quickly integrating the equation).
The lowest possible n value occurs at this point
at this point dn/dt = 0
hence we can work out the time at which n is lowest

right?

[TenOfThem]

(Original post by ilovemath)
oh, so when I plot the function of e^0.5t - 5 I it cross at about t = 3min 13 sec

so just to check why this works:

the original function will have a minimum value (we can see that this is the case by quickly integrating the equation).
The lowest possible n value occurs at this point
at this point dn/dt = 0
hence we can work out the time at which n is lowest

right?

The turning points are when the derivative=0, yes

[thefifthfoo]

(Original post by ilovemath)
but for dn/dt = 0 to work surely there has to be a minimum
I was not aware of any point on the e graph where gradient = 0

If it was just n= e^f(t) then you're right, the gradient could never equal zero (assuming f(t) is not just a constant) as the derivative (gradient function) would be f'(t)e^f(t) which is always strictly positive.

However, our function has derivative e^f(t) - b and this can equal zero, when e^f(t) = b (with b positive)

[james22]

(Original post by thefifthfoo)
If it was just n= e^f(t) then you're right, the gradient could never equal zero (assuming f(t) is not just a constant) as the derivative (gradient function) would be f'(t)e^f(t) which is always strictly positive.

However, our function has derivative e^f(t) - b and this can equal zero, when e^f(t) = b (with b positive)

what about f(x)=sin(x)? or f(x)=x^2-x?

[thefifthfoo]

(Original post by james22)
what about f(x)=sin(x)? or f(x)=x^2-x?

If f(x) = sinx then y = e^sinx and dy/dx = cosx e^sinx, which could be 0. Ok, so I wasn't completely right.

However, if f(x) = x^2-x then y = e^(x^2 - x) and dy/dx = (2x-1)e^(x^2-x). Which can also be 0.

Ok, so I was completely wrong and what I said only applies (at least in general) if f(x) has the form ax + c. Sorry about that, I wasn't really thinking through my answer. I think I made my point in terms of this question though - its derivative (e^0.5t - 5 or something similar)