A different related rates problem - no time mentioned

[opus]

Homework Statement

Evening all. I have a related rates problem that I haven't come across which doesn't seem to involve time which is usually the independent variable that we take the derivative with respect to. It has thrown me for a loop.

Find the change in the volume of a right cylinder cone with a height of 2 if the radius r changes from 0.5 to 0.55.

Homework Equations

$V = \frac{1}{3}πr^2h$

The Attempt at a Solution

The set up is where I'm not doing too well. Normally, if the related rates problem had time as an independent variable, I would do as such:
1) Find an equation that relates the variables to each other.
2) Find an equation that shows the rate of change with respect to the independent variable (time usually).
3) Take the derivative of both sides of the equation relating the variables to each other with respect to the independent variable. I would now have an equation that relates the rates of change of everything.
4) Plug in the given rate of change, and solve for the other rate of change.

In this case, as far as I've gotten is to say that I want the change in volume when the change in radius is 0.05 which is given in the question. The change in volume seems to need to be written as $\frac{dV}{dr}$, as we want the change in volume with respect to the radius. But as for writing the change in radius, it doesn't make sense to write $\frac{dr}{dr}$.

Another thing I did is to get the equation in terms of just radius so I'm working with less variables:
$V = \frac{1}{3}πr^2\left(\frac{2r}{r}\right)$ but I don't think this is being much help.

As you can see, I am a mess with this one, so any pointers would be greatly appreciated!

 

[Orodruin]

The way it is phrased your problem does not involve any derivatives at all. It is just asking you for the difference between two volumes.

 

[phinds]

You seem to be trying to shoe-horn this problem into being rate related when it is not. Nowhere in the question you have stated is a rate asked for. It seems to be a simple problem of finding two volumes and taking the difference.

EDIT: I see orodruin beat me to it

 

[opus]

Ok thanks for clearing that up guys. Looks like a curve ball where we were doing a lot of related rates and I just expected this one to be one as well. Oldest trick in the book!

So the book says that $ΔV = \frac{π}{30}$ and I can't quite come up with that.

What I did was take the volume when r = 0.5 and when r = 0.55 and subtract the two to get a total change in volume but somewhere I'm going wrong.
Since the radius is at different lengths, then that would mean that the height is different at these different radii. So to account for this, I said that $\frac{r}{h} = \frac{r}{2}$ and it follows that $r = \frac{2r}{2}$. The reason I did this is so that it takes into account the fact that the height will be different with different radii, and since we have a right triangle in regard to the height and radius, then similar triangles apply.
So I have written:

$ΔV = \frac{1}{3}π\left(0.55\right)^2\left(\frac{2(0.55)}{2}\right) - \frac{1}{3}π\left(0.5\right)^2\left(\frac{2(0.5)}{2}\right)$ which doesn't give a clean number like $\frac{π}{30}$

 

[opus]

This is probably a really dumb mistake. I'm on my 47th calculus problem today preparing for my exam and my brain is a little shot so I apologize for any stupidity that comes across my posts.

 

[Orodruin]

Since the radius is at different lengths, then that would mean that the height is different at these different radii.

Why? Nowhere in the problem is it stated that the height changes.

Based on the given answer the book is approximating the change by the derivative wrt r multiplied by the change in r. To me that is a completely unnecessary approximation the way the problem is phrased.

 

[verty]

The idea is that ${\Delta V \over \Delta r} \approx {dV \over dr}$. So if you know $dV \over dr$, you can find $\Delta V$ approximately given $\Delta r$. Definitely the height is remaining constant. The only change is the radius. See if you can work it out.

 

[opus]

Why? Nowhere in the problem is it stated that the height changes.

Based on the given answer the book is approximating the change by the derivative wrt r multiplied by the change in r. To me that is a completely unnecessary approximation the way the problem is phrased.

I suppose I just thought that it made sense that it was the case. For the cone, there can be only one place where the height is equal to 2, and that will have a specific radius. If we move up, the radius gets smaller and if we move down, the radius gets bigger.
Are you talking about using linear approximations where we could use the tangent line of a function at a point to estimate what value the function will have with a corresponding change in x? --> $y=f(a)+f'(a)(x-a)$

 

[opus]

I thought ΔV was the same as $dV$?
I tried something a little different as it was something I remembered from the book where you can approximate what a function is going to be after a small change in the input values:

$f(a+h) ≈ f(a)+f'(a)h$
$≈\left(\frac{2π}{3}(0.5)^2\right)+\left(\frac{4π}{3}(0.5)\right)0.05$ but this was way off which in hindsight makes sense because it's asking for the change in the volume, not what it's new value is. So I took this value, 1.57, and subtracted it from the volume where r=5.
This gave me 1.04 which is off by a decimal place to what would equate to $\frac{π}{30}$. Must be on the right track but there are no mistakes in the calculator so it must be somewhere else.

 

[Mark44]

I thought ΔV was the same as $dV$?

No, they're not the same. $\Delta V$ is the exact change in volume. $dV$ is the differential change in volume, and is equal to $\frac 4 3 \pi r~dr$.

I tried something a little different as it was something I remembered from the book where you can approximate what a function is going to be after a small change in the input values:

$f(a+h) ≈ f(a)+f'(a)h$
$≈\left(\frac{2π}{3}(0.5)^2\right)+\left(\frac{4π}{3}(0.5)\right)0.05$ but this was way off which in hindsight makes sense because it's asking for the change in the volume, not what it's new value is. So I took this value, 1.57, and subtracted it from the volume where r=5.
This gave me 1.04 which is off by a decimal place to what would equate to $\frac{π}{30}$. Must be on the right track but there are no mistakes in the calculator so it must be somewhere else.

Since $f(a + h) \approx f(a) + f'(a)h$, it follows that $f(a + h) - f(a) \approx f'(a)h$. Also, $\Delta f = f(a + h) - f(a)$, by definition. From the first equation, you can approximate $\Delta f$ by $f'(a) h$.
When I put the numbers in, I get the approximate change in volume agreeing with the exact change in volume in the first two decimal places. It could be that your value of 1.57 was rounded off but shouldn't have been.

 

[Ray Vickson]

Ok thanks for clearing that up guys. Looks like a curve ball where we were doing a lot of related rates and I just expected this one to be one as well. Oldest trick in the book!

So the book says that $ΔV = \frac{π}{30}$ and I can't quite come up with that.

What I did was take the volume when r = 0.5 and when r = 0.55 and subtract the two to get a total change in volume but somewhere I'm going wrong.
Since the radius is at different lengths, then that would mean that the height is different at these different radii. So to account for this, I said that $\frac{r}{h} = \frac{r}{2}$ and it follows that $r = \frac{2r}{2}$. The reason I did this is so that it takes into account the fact that the height will be different with different radii, and since we have a right triangle in regard to the height and radius, then similar triangles apply.
So I have written:

$ΔV = \frac{1}{3}π\left(0.55\right)^2\left(\frac{2(0.55)}{2}\right) - \frac{1}{3}π\left(0.5\right)^2\left(\frac{2(0.5)}{2}\right)$ which doesn't give a clean number like $\frac{π}{30}$

I think the book wants you to think about fast and useful approximations. If you use the "differential (tangent-line)" approximation, you get
$$\Delta V_{approx} = V'(r)|_{r=0.5} \:\Delta r = (2 \pi/3) \times 5/100 = \pi/30 \doteq 0.10472$$ If you use the exact difference you get
$$\Delta V_{exact} = V(0.55) - V(0.5) \doteq 0.10996.$$ So, the approximation is not very accurate in this case, but depending on what you want to do, an approximation of 0.105 may be OK compared with the exact 0.110. Note, however, that the approximation would be much better if we had a smaller $\Delta r$, say 0.01 instead of 0.05.

 

[opus]

Ok thanks guys. So are we just using the fact that at some point, if we have the tangent line at that point and shift the x just a little bit, then that will give us an approximate value of what the function is? And to get an exact difference we would have to evaluate at each x value and then subtract them?

 

[Ray Vickson]

Ok thanks guys. So are we just using the fact that at some point, if we have the tangent line at that point and shift the x just a little bit, then that will give us an approximate value of what the function is? And to get an exact difference we would have to evaluate at each x value and then subtract them?

I thought that was exactly what post #11 said explicitly, and other posts hinted at it.

 

[opus]

I thought that was exactly what post #11 said explicitly, and other posts hinted at it.

I don't really understand the notation so I put it in English. Most threads I make people tend to not like me restating what they've said but the reason I do it is to see if I can explain it on my own without looking and see if I get it right. Apologies if it seems unnecessary. Thanks for the help everyone.