Dick said: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.
Not quite. If V = l3, then dV = dV/dl * dl = 3l2 * dl.Slimsta said:whats dl ? 0.031m ?
and dt = 1h ?
would dV be 3l^2 ?
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.Slimsta said:if yes..
for the last part i get 9.3...
becasue (3*10^2)/1 + 0.031 = 9.3
HallsofIvy said: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 [itex]\Delta V[/itex]" and I see no reference to [itex]Delta V[/itex] 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.
The chain rule for derivatives is a method for finding the derivative of a composite function. It states that the derivative of a composite function is equal to the derivative of the outer function, multiplied by the derivative of the inner function.
To apply the chain rule, you need to first identify the inner and outer functions. Then, take the derivative of the outer function and multiply it by the derivative of the inner function. You may need to use the product rule or quotient rule as well, depending on the complexity of the function.
Sure, let's say we have the function f(x) = (x^2 + 3)^3. The inner function is x^2 + 3 and the outer function is ( )^3. Using the chain rule, we would first find the derivative of the outer function, which is 3(x^2 + 3)^2. Then, we multiply it by the derivative of the inner function, which is 2x. So the final derivative is 6x(x^2 + 3)^2.
The chain rule is used when finding the derivative of a composite function, where the input of one function is the output of another. It is also used when finding the derivatives of trigonometric, exponential, and logarithmic functions.
Yes, some common mistakes when using the chain rule include forgetting to take the derivative of the outer function, not using the product or quotient rule when necessary, and not properly identifying the inner and outer functions. It is important to carefully follow the steps and double check your work when applying the chain rule.