Hi,

I am currently carrying out my GCSE maths coureswork, in which I have to use differentiation to find out the smallest surface area of a cylinder box, with a volume of 160cm³ .

If you are reading this thread because you are doing the same coursework, you should note that you need to declare any help you have had (attaching a printout of this discussion is a good way). This will not necessarily affect your mark. In this case, the work explained is beyond the GCSE syllabus, and if you show in your write-up that you understand it, that should gain you credit.

I think in the next step you have got a little confused. I guess you have been given a rule for differentiating a formula $r^n$ with respect to $r$. i.e. $d(r^n)/\mathrm{dr}=n{r}^{n-1}$. We don't have to rearrange anything we just apply this. So for example on differentiation of the first term: $2\pi r^2$ we obtain $4\pi r$. Can you do the second term in exactly the same way?

O.k thanks,
I'm still not quite sure about the second term.

And I'm still not sure how I would then apply this formula to find out the smallest surface area of a cylinder to get a volume of 160cm cubed

The second term is 320/r which is of the form r^-1 we have n=-1, so applying the rule for differentiation we obtain 320nr^n-1 =-320r^-2 =-320/r² .

Perhaps I should have writen the rule more generally for a and b independent of r, i.e. a and b are 'constants':

d(arⁿ + br^m )/dr = ad(rⁿ )/dr + bd(r^m )/dr = anr^n-1 +bmr^m-1

In other words multiplicative constants 'don't matter'; we can just ignore them when we apply the rule for differentiation. This is called the 'linear property of the differential operator' - but I shouldn't worry about that.

Do you know what differentiation is all about? Or are you just working on a rule you've been given? If you tell me what you know I'll give you a few tips about how to proceed...

We basically spent a lesson going over some of the basics of differentiation, but we never went into it in great detail. But in order to get the best grades for my coursework, I was given a rule and told to apply it to my investigation.

The process of differentiation is that of finding a local linear approximation to a function.

That is suppose f(x) is a function, for each number x we want to find a new function g of the form g(h)=a x h. This says that g is a 'linear function' (i.e. if we plotted a graph of g we against h we would get a straight line through the origin.)

We also want g to be so that:

f(x+h)-f(x) = g(h)

for very small values of h.

In other words for each number x we want to find a linear function g which represents a line that is parallel to the curve at x. The function gives a measure of the 'slope' of the curve at the point x. We call the constant a the 'differential'* of f at x and write df/dx(x) = a. (Where g(h)=ah)

Lets have an example. Suppose that f(x) = x² . Now we want to find df/dx(0). That is we want a function g which satisfies g(h) = a x h for a constant number a and so that:

f(0+h)-f(0) = g(h)

for very small values of h.

But this is asking that

h² = g(h)

for very small values of h. If h is very small then h² is going to be very very small, in fact so small it may as well be zero. So it seems that we must choose a=0 i.e. g(x)=0. If you look at the graph of f(x) = x² you'll see its almost flat at x = 0.

The differential of f at x (the number a) gives a measure of the gradient or slope of the function at x. A steep function going up hill (from left to right) has a large positive differential. A steep function going down hill has a large negative one. What is special about the differential when a function is flat? Also think about what happens to a function at a maximum or a minimum.

There are some nice rules for finding the differential of a function f at x (we have seen one already).

If this is horrible I'll try to make it a little simpler.

______
* Properly the function g is itself the differential but since g and a define each other we don't bother about a distinction.

erm...yeah... thanks,I think I understand bits of that!!!!
Does that mean that when the function is flat the differential is zero, and therefore the surface area is at its smallest?
If so, how can I use this in my investigation?

I'm still unsure as to how I would "differentiate A, and set dA/dr(r) = 0" to find "an equation which gives the values of r where A(r) is flat."

I thought the function A(r) = 2pr2+2V/r had already been diferentiated.

is the differential of $A$ at $r$? When you do give us a shout and I'll explain the next bit...

Yeah I think I understand as much as I need to know for that!!!!!
But could you please explain the next bit to me.

Ok, so if A(s) is the minimum of A over all possible values of r what can you say about dA/dr(s) ? and why?

O.k, can you please explain the next stage to me, and how I find the max and min of the function
dA/dr(r) = 4pr-2V/r2

We want the minimum surface area, so we want to find a radius r so that A(r) is a minimum. I don't think there is any point in telling you how to do it especially if is GCSE coursework. Read through what I've written again. I think possibly my explanation was terrible, if you really don't understand it post another message and I'm sure somebody else will give a different perspective.

O.k thanks alot for your help, unfortunately it was a bit too complicated for me to understand!!!!
Can anyone else help explain it in less detail, in regards to my coursework!!!!!!

I'll go ahead and just show you how to finish the problem assuming that you need to have this done by monday..

are you ok up to the point of

Now, $\mathrm{dA}/\mathrm{dr}$ represents the slop, so can you see why the function $A$ is at minimum when $\mathrm{dA}/\mathrm{dr}=0$. If not try drawing a few pictures to convince yourself letting $A=y$ and $r=x$ in the $(x,y)$ plane

So, at minimum $A$,

$0=4\pi r-320/r^2$

and then rearranging

$320/r^2=4\pi r$

Solving for $r^3$

$r^3=80/\pi$

so

$r=(80/\pi {)}^{1/3}$ at the minimum surface area.
Sorry if this has been too hurried,

Brad

Thanks a lot for all your help!!! I understand it now, and was able to work that the radius would be 2.9 when the surface area was at its minimum!!!! (I think that's right, is it!?!)
Thanks again!

Rebecca