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