Derivatives Chain Rule question

[Slimsta]

Homework Statement

http://img21.imageshack.us/img21/6784/probi.jpg

Homework Equations

The Attempt at a Solution

i have no idea where to begin and my textbook doesn't 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 ?

 

[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.

 

[Slimsta]

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.

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

would dV be 3l^2 ?

 

[Dick]

Yes, and yes.

 

[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

 

[Mark44]

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

would dV be 3l^2 ?

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.

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

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?

 

[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.

 

[Mark44]

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

 

[Slimsta]

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.

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!

 

[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
http://img132.imageshack.us/img132/8048/prob2r.jpg