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Dave

Calculus I - Lesson 3: Properties of the derivative

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Hmm... I don't know? As far as I'm aware, I've only stated one definition.

 

BTW: Is anyone else apart from yourdad following this?

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i don't know, but i wish the lesson could continue. btw, was my answer for #4 correct(post#10)?

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Sorry, didn't see that amongst the other posts :)

 

You're very close; just missing a constant.

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Is this the problem you are working on?

 

Find the derivative with respect to x of:

 

iv) [math]\frac{1}{x}\cdot \frac{x^2}{a^2} + a^4[/math], where a is constant.

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No, it's the one with the minimized surface area (question 4).

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sorry, it took so long, but i got hungy and i made breakfast. i got the same answer. i don't know what i am doing wrong

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[math]S®=2{\pi}rh+{\pi}r^2[/math]

[math]V=1={\pi}r^2h\Rightarrowh=({\pi}r^2)^{-1}[/math] h=h, so:

[math]S®=2{\pi}({\pi}r^2)^{-1}r+{\pi}r^2\RightarrowS®=2r^{-1}+{\pi}r^2[/math]

[math]S'®=-2r^{-2}+2{\pi}r[/math]

[math]S'®=0\Rightarrow{2{\pi}r}={2r^{-1}}\Rightarrow{r^3={\pi}^{-1}}\Rightarrow{r={\pi}^{-\frac{1}{3}}}[/math]

[math]{S®={\pi}({\pi}^{-\frac{1}{3}})^2+2({\pi}^{-\frac{1}{3}})^{-1}}\Rightarrow{S®={\pi}^{\frac{1}{3}}+2{\pi}^{\frac{1}{3}}}\Rightarrow{S®=3{\pi}^{\frac{1}{3}}}[/math]

[math]({\pi}^{-\frac{1}{3}}, 3{\pi}^{\frac{1}{3}})[/math]

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Ah this...

4) (Harder) Let's assume that we have a cylinder that we want to fill with precisely 1 litre of water (i.e. the volume of the cylinder is equal to 1 litre). Obviously, we can do this by letting the cylinder be different heights and radii - what we would like to do is to minimize the surface area of the cylinder, just for the heck of it.

a) Calculate the surface area S of the cylinder in terms of its height h, radius r and pi.

b) Using the fact that the volume is 1, express S in terms of just r (hint: use substitution).

c) Now we have a regular function, just like y=x. This time, however, S is some function of r. Calculate the turning points for the function and find a minima for S.

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You want to minimize the surface area of a cylinder, so you will need calc 1. First you need a formula for the surface area of the cylinder.

 

Consider the very bottom of the cylinder, its a circle, and a circle doesn't have a surface area.

 

So instead imagine a cylinder of infinitessimal height dh, and radius R.

 

If its a ribbon, you could cut it in half, to produce a rectangle. The height of the rectangle will be dh (i could say width but for this problem its better to call it height), and the length of the rectangle will be given by

 

[math] L = 2 \pi R[/math]

 

Presume that you have already proven that the area of a rectangle is its length, times its height. Therefore, the surface area of a cylinder of infinitessimal height dh, and radius R is given by:

 

[math] S = L(dh) = 2 \pi R dh[/math]

 

Now, since we are going to be pouring water into it, there has to be a solid bottom, but no top. So, the bottom of the cylinder is a circle, and circles have an area which is equal to pi R^2.

 

So let the tiny cylinder have a bottom but no top. Therefore, its total surface area is given by:

 

[math] S = L(dh) = 2 \pi R dh + \pi R^2 [/math]

 

 

There we go. Thank you yourdadonapogos for pointing out that the cylinder has to have a bottom but no top; that is one side open, and the other closed. I'm not sure how else to say it.

 

Now, we can see that the surface area depends upon the height (dh), and the radius R. So for a teeny tiny radius dR we have:

 

[math] S = L (dh) = 2 \pi (dr)(dh) + \pi (dr)^2 [/math]

 

Now that looks strange to me because I have (dr)2 you don't usually see things like that in calculus 1, so maybe I should just let its radius be denoted by R, so that I can deal with this formula instead:

 

[math] S = L (dh) = 2 \pi R(dh) + \pi R^2 [/math]

 

You know I would love to use partial derivatives really.

 

Ok here is what I am going to do. Here is the formula for the surface area of a cylinder of radius R, and height h, which is half-open.

 

[math] S(R,h) = 2 \pi Rh + \pi R^2 [/math]

 

Now here is the partial derivative of S with respect to R:

 

[math] \frac{\partial S}{\partial R} = 2 \pi h + 2 \pi R [/math]

 

All I did, was take the ordinary derivative of S(R,h) with respect to R, and held the height h constant. Since the ordinary derivative is from Calculus 1, i see no objection to this.

 

Now, determine the partial derivative of S with respect to h. So again take an ordinary derivative, this time hold R constant instead of h. You get this:

 

[math] \frac{\partial S}{\partial h} = 2 \pi R [/math]

 

Alright, I did that just as an excercise. In the process, I have answered part a, which was to express the surface area of a half-open cylinder in terms of pi,h,R. That was this formula here:

 

 

[math] S(R,h) = 2 \pi Rh + \pi R^2 [/math]

 

This agrees with what yourdadonapogos got.

 

Now to part b of the question...

 

Dave said, "using the fact that the volume is 1, express the surface area [which is currently being denoted by S(R,h) ] in terms of R."

 

The first question is, "what is the volume of a cylinder?"

 

Now, we have to think of our cylinder as a three dimensional solid.

 

The answer from memory, is pi times the radius squared, times the height.

 

Let V denote the volume of a cylinder. Therefore:

 

[math] V(R,h) = \pi R^2 h [/math]

 

Now, Dave wants us to express the surface area as a pure function of R. Because we were told that the volume of this cylinder is 1, this is achievable.

 

We have:

 

[math] 1 = \pi R^2 h [/math]

 

And we also have:

 

[math] S(R,h) = 2 \pi Rh + \pi R^2 [/math]

 

PROVIDED THAT THE RADIUS R IS NONZERO, we can use the first of the two equations immediately above, to express the height h, in terms of the radius. Dividing both sides of that statement by pi times the radius squared we have:

 

[math] h = \frac{1}{\pi R^2} [/math]

 

Now, we are just a substitution (as Dave suggested) away from answering part b.

 

[math] S(R,\frac{1}{\pi R^2}) = 2 \pi R(\frac{1}{\pi R^2}) + \pi R^2 [/math]

 

Which simplifies a bit, to this:

 

[math] S® = \frac{2}{R} + \pi R^2 [/math]

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wait, is it open or closed? I assumed it was a closed on one side. which is needed for the problem, open, closed, or one side open the other closed?

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IM not sure, let me go back and check.

 

Well we are going to pour water into it, so i guess it should have a bottom, but no top huh?

 

I think it should have a solid bottom with surface area pi R^2, but not a top.

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Sorry, maybe I should have specified this earlier. It is a closed cylinder. The equation for it would be:

 

[math]S = 2\pi r^2 + 2\pi r h[/math].

 

The rest follows fairly trivially.

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Sorry' date=' maybe I should have specified this earlier. It is a closed cylinder. The equation for it would be:

 

[math']S = 2\pi r^2 + 2\pi r h[/math].

 

The rest follows fairly trivially.

 

Now you say so. :)

 

Yes, just about everything is trivial after you already know it.

 

Well lets just do it, since it is so trivial. To save myself some writing, i will write S instead of S(R,h), or S instead of S®, because it is inferrable from looking at the RHS what variables S depends upon. With that in mind...

 

We can begin with the formula for the surface area of a closed cylinder...

 

[math]S = 2\pi R^2 + 2\pi R h[/math].

 

The first term on the RHS is the area of two circles of radius R (which is the surface area of the top of the cylinder plus the surface area of the bottom of the cylinder) and the second term on the RHS is the surface area of the cylinder wall, which we can figure out if we cut the cylinder in a direction parallel to its axis, and view as a rectangular area.

 

OK SO, to answer part b, we have to use the given fact that the volume of the cylinder is to be 1 unit.

 

From memory, the volume of a cylinder is pi times the radius squared times the height, that is:

 

[math] V = \pi R^2 h [/math]

 

Letting V=1, we have as before:

 

[math] 1 = \pi R^2 h [/math]

 

Now, PROVIDED THAT R IS NONZERO, we have:

 

[math] h = \frac{1}{\pi R^2} [/math]

 

And we already have:

 

[math]S = 2\pi R^2 + 2\pi R h[/math]

 

So the rest is trivial, just substitute properly. After substitution we obtain:

 

[math]S = 2\pi R^2 + 2\pi R h = 2\pi R^2 + 2\pi R (\frac{1}{\pi R^2})[/math]

 

Now, after simplifying the formula on the far right, we finally get this:

 

[math]S = 2\pi R^2 + \frac{2}{R} [/math]

 

And we are done. We have expressed the surface area of a closed cylinder of volume 1, in terms of its radius R. This answered part b.

 

It is interesting to note that had we have divided by R without thinking clearly, it would have caught up to us at the end, since we have a 2/R term.

 

Part C: Calculate the turning points for S, and find a minima of S.

 

I am not exactly sure what I am being asked for here, but I know it involves taking a derivative, because that is the only way that you can compute either the maxima, or minima of some function of one variable.

Recall:

 

[math]S = 2\pi R^2 + \frac{2}{R} =2\pi R^2 + 2R^{-1} [/math]

 

The derivative of S with respect to R is...

 

[math] \frac{dS}{dR} =4 \pi R -2R^{-2} = 4 \pi R - \frac{2}{R^2} [/math]

 

Let us have used a rectangular coordinate system to graph the function S. Let the horizontal axis be the R axis, and let the vertical axis be the S axis.

 

Since negative radius has no meaning, there is no portion of the S-curve in the left half of the S-R plane. Also, S is not defined if R=0.

 

Now, if a particular radius value minimizes the surface area of the cylinder S, then the tangent line to the curve will be horizontal at that specific radius value.

 

Now, the derivative of S with respect to R gives the slope of the tangent line, for different R values.

 

So if there is a radius which makes dS/dR=0 then the slope of the tangent line is zero, which means the tangent line to the curve is horizontal.

 

Setting dS/dR =0 gives us this:

 

[math] 0 = 4 \pi R - \frac{2}{R^2} [/math]

 

Or equivalently...

 

[math] 4 \pi R = \frac{2}{R^2} [/math]

 

Now, we can solve the equation above for R, to obtain the critical values.

 

Multiplying both sides of the statement above by R^2 we obtain:

 

[math] 4 \pi R^3 = 2 [/math]

 

Dividing both sides of the statement above by 4 pi we obtain:

 

[math] R^3 = \frac{2}{4 \pi} = \frac{1}{2 \pi} [/math]

 

So we finally have this:

 

[math] R^3 = \frac{1}{2 \pi} [/math]

 

Taking the cubed root of both sides of the equation above we obtain:

 

[math] R= (2 \pi)^{-1/3} [/math]

 

Choosing this value for the radius either maximizes or minimizes the value of S.

 

We can now express the surface area S of the cylinder in terms of this particular radius value. The surface area of a cylinder of volume one is related to its radius as follows:

 

[math]S = 2\pi R^2 + 2R^{-1} [/math]

 

Now, at the critical radius the surface of the cylinder is related to R as follows:

 

[math]S = 2\pi ((2 \pi)^{-1/3})^2 + 2((2 \pi)^{-1/3})^{-1} [/math]

 

Where I have done nothing other than substitute the value just found into the original expression for surface area in terms of radius.

 

The previous expression can be simplified a bit...

 

[math]S = 2\pi (2 \pi)^{-2/3} + 2(2 \pi)^{1/3} [/math]

 

[math]S = (2 \pi)^{1/3} + 2(2 \pi)^{1/3} [/math]

 

[math]S = (2 \pi)^{1/3}[1+ 2] [/math]

 

[math]S = 3 (2 \pi)^{1/3} [/math]

 

Now, if we look at the second derivative of S with respect to R, we can determine whether or not the previous S value is a maxima or minima.

 

Recall...

 

[math] \frac{dS}{dR} =4 \pi R -2R^{-2} [/math]

 

Taking the derivative with respect to R of both sides of the equation above we obtain:

 

[math] \frac{d}{dR}(\frac{dS}{dR} ) = \frac{d^2S}{dR^2} = 4 \pi + 4 R^{-3} [/math]

 

Now, if we evaluate the second derivative at the critical R value we just found, we find that the second derivative is positive, which means that the function is concave up, like a U, which means that the critical value we found is a minima.

 

I don't know what a turning point is.

 

Regards

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:P

 

Yes, I admit, it's my fault. It's quite interesting to work out though. Another (easier) problem is the following:

 

Suppose you want to build a rectangular paddock to contain some sheep. You want to maximize the area of your paddock because you want to fit as many sheep in as possible, and you have 300m of fencing available to you. Find the maximum area.

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PS: would appreciate it if you didn't use partial derivatives on this particular thread; it will only serve to confuse ;) The problems are meant to be solved using the methods I've given already for those who are just starting.

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so, with both ends closed, it is [math]((2{\pi})^{\frac{1}{3}},2(2{\pi})^{\frac{1}{3}})[/math]

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PS: would appreciate it if you didn't use partial derivatives on this particular thread; it will only serve to confuse ;) The problems are meant to be solved using the methods I've given already for those who are just starting.

 

No problem Dave.

 

Kind regards

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so, with both ends closed, it is [math']((2{\pi})^{\frac{1}{3}},2(2{\pi})^{\frac{1}{3}})[/math]

 

Actually the critical cylinder radius was:

 

[math] R = (2 \pi)^{-1/3} [/math]

 

At least that's what I got, and... for the surface area at that value we have:

 

[math] S = 3 (2 \pi)^{1/3} [/math]

 

I guess I don't know what a turning point is. Do you mean inflection point?

 

Regards

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i got a book "Schwams outline for calculus", or something like that, from the library, because i like to see where things are going. all of the diferentiation rules make sense except the trig and log differentiation rules. for log, it says [math]\frac{d}{dy}(log_{a}x)=\frac{1}{x}log_{a}e,a>0,a\not=1[/math], but it doesn't say how to do it with a function. the examples seem to suggest that you would replace x with the function and then [math]\frac{d}{dy}(log_{a}x)=\frac{1}{x}log_{a}e\frac{d}{dy}(x),a>0,a\not=1[/math]

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Well, to replace x with a function, you have to use the chain rule (which I'm coming to eventually). It basically shows you how you can differentiate composite functions like sin(x2) and whatnot.

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[math]D_x(f(g(x)))=f'(g(x))g'(x)[/math] or,if y=f(u) and u=g(x), [math]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/math]?

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That's the biscuit. Very powerful rule indeed.

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