# Derivatives Chain Rule question

1. Oct 9, 2009

### Slimsta

1. The problem statement, all variables and given/known data
http://img21.imageshack.us/img21/6784/probi.jpg [Broken]

2. Relevant equations

3. The attempt at a solution
i have no idea where to begin and my textbook doesnt have any examples that look like this question..
can someone give me hints?
whats the equation that l=10m after some time t, if it reduces by 0.031m/h ?

Last edited by a moderator: May 4, 2017
2. Oct 9, 2009

### Dick

Why didn't you fill in the dl/dt space? Given you filled in V=l^3 correctly, what's dV/dl? Then use the rest of the formula for dV/dt. The homework is really guiding you through the whole thing.

Last edited: Oct 9, 2009
3. Oct 9, 2009

### Slimsta

whats dl ? 0.031m ?
and dt = 1h ?

would dV be 3l^2 ?

Last edited: Oct 9, 2009
4. Oct 9, 2009

### Dick

Yes, and yes.

5. Oct 9, 2009

### Slimsta

whats dl ? 0.031m ?
and dt = 1h ?

would dV be 3l^2 ?

if yes..
for the last part i get 9.3....
becasue (3*10^2)/1 + 0.031 = 9.3

6. Oct 10, 2009

### Staff: Mentor

Not quite. If V = l3, then dV = dV/dl * dl = 3l2 * dl.

There is a whole lot of approximation going on in this problem that seems to be completely glossed over. This is not a complaint about what you are doing, but rather, how the problem is being presented.

You might recall reading that the differentials in dy/dx (or in this case dV/dl) are "infinitesimally small numbers" that are just about indistinguishable from zero.

The equation dV = 3l2*dl is exactly correct. In this problem you don't really have dl; instead you have $\Delta l$, which is not anywhere close to zero. Using $\Delta l$, the goal of this problem is to use derivatives to approximate $\Delta V$.

The real equation is
$\Delta V$ $\approx$ dV = 3l2*dl $\approx$ 3l2*$\Delta l$. If $\Delta l$ is reasonably small, the approximation will be fairly good. In practice, if $\Delta l$ is a small fraction of l, the approximation will probably be good enough.
You probably know what you mean, but you aren't writing what you mean. If you want to divide by 1 + 0.031, you need parentheses around it. Otherwise the expression above would be interpreted as 300/1 + 0.031 = 300.031.

On the other hand, even if you mean to divide by 1.031, how in the world do you get 9.3 out of 300/1.031?

7. Oct 10, 2009

### HallsofIvy

I see no approximation at all here- except for references by Slimsta about "dl" when he should be referring to dl/dt. No, dV is NOT "3l2". In terms of "differentials", which I see no need to use here, dV= 3l2dl- and the "dl" is important.

Am I missing something? Mark44 says, "the goal of this problem is to approximate $\Delta V$" and I see no reference to $Delta V$ in the problem.

Slimsta, do you understand that, the way the derivative is normally defined, "dl/dt" (as well as dV/dl and dV/dt) are NOT fractions? You need to be careful about that. You typically can treat derivatives as if they were fractions but taking that too literally can be dangerous.

Since V= l3, dV/dl= 3l2 and you are told that dl/dt= -3.1. Okay, when l= 10, what is dV/dl? What is dV/dt= (dV/dl)(dl/dt)?

I am also a little uncomfortable with talking about the differentials as "infinitesmally small numbers". That certainly can be done, but just defining "infinitesmally small" requires some very deep math and I think you are better of defining the derivative in terms of the limit as is normally done.

8. Oct 10, 2009

### Staff: Mentor

I should have said, "the apparent goal of this problem is to approximate $\Delta V$"

9. Oct 10, 2009

### Slimsta

okay so now i understand the concept of the question.
V= l3
dV/dl= 3l2 --> l=10 --> 3*102 = 300
dl/dt= -3.1 cm/h = -.031 m/h

dV/dt= (dV/dl)(dl/dt) = 300 * (-.031) = -9.3

sick! i get it... woohoo.
thanks guys!

10. Oct 12, 2009

### Slimsta

i got another question that is similar to this one but a bit more complicated.
i got everything up to the dy/dt
in the last question i had the equation of the cube and from there i took the derivative of it.. here there is no equation.
http://img132.imageshack.us/img132/8048/prob2r.jpg [Broken]
http://img132.imageshack.us/img132/8048/prob2r.jpg [Broken]

Last edited by a moderator: May 4, 2017