Finding dV/dt in V=4*L^3 at t=0.1s

In summary: The correct answer is \frac{dV}{dt}=120L^2 cm^3/s. In summary, the question is asking to find the value of dV/dt at t=0.1 seconds, given that V is a function of L and dl/dt=10t cm/s. Using the chain rule, it can be determined that dV/dt = 12L^2 cm^3/s. However, without knowing the initial length or any length at a particular time, a numerical solution cannot be obtained.
  • #1
Tymick
7
0
I'm having some trouble with the following question, it was on a test previously and I haven't been able to figure it out :/

Let V=4*L^3 cm^3, where dl/dt=10*t cm/s. Find dV/dt at t=0.1 second
 
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  • #2
Since V is a function of L.

What you can do is use the chain rule and say that

[tex]\frac{dV}{dt}=\frac{dV}{dL} \times \frac{dL}{dt}[/tex]
 
  • #3
Right, following the chain rule I do get dv/dt=120*L^2, I tried before to get L in terms of t, via integration but of course that gives 5t^2+C whereas C would pose as a problem since I've no way of obtaining it, so I can't get a numerical solution I just get a solution in terms of L...
 
  • #4
Tymick said:
Right, following the chain rule I do get dv/dt=120*L^2, I tried before to get L in terms of t, via integration but of course that gives 5t^2+C whereas C would pose as a problem since I've no way of obtaining it, so I can't get a numerical solution I just get a solution in terms of L...

[tex]\frac{dV}{dt}=12L^2[/tex]



You don't need L in terms of t when they tell you that [itex]\frac{dL}{dt}=10t[/itex]
 
  • #5
rock.freak667 said:
[tex]\frac{dV}{dt}=12L^2[/tex]
You don't need L in terms of t when they tell you that [itex]\frac{dL}{dt}=10t[/itex]

so then this question doesn't have a numerical solution, only one in terms of L?
 
  • #6
Is that all the data presented in the question? you don't have the initial length or any length at a particular time?
 
  • #7
rock.freak667 said:
Is that all the data presented in the question? you don't have the initial length or any length at a particular time?

that's about it, my first post states out the entire question, no initial values, at all...and thanks by the way.
 
  • #8
Well without a length at any particular time, you can't find a numerical solution.
 
  • #9
Tymick said:
I'm having some trouble with the following question, it was on a test previously and I haven't been able to figure it out :/

Let V=4*L^3 cm^3, where dl/dt=10*t cm/s. Find dV/dt at t=0.1 second

I would think that as you are given dl/dt and you can work out dv/dl and you are looking for dv/dt

dv/dl [12L^3] dL = 12L^2 dl/dt = 10t

dv/dl x dl/dt... the dl's cancel to leave dv/dt

Wouldn't that be the chain rule?

Perhaps u'v + v'u
or in this case v't + t'v

NO WAIT! So much simpler.

Look at it as...

the change in v over the change in time. The change in V is simply 12L^3 , so dv/dt = 12L^3 / dt ( it would seem something is missing as one is not sure the intial time, I would have to assume time began at zero and so dt would be 0.1 s,
thus dv/dt = 12L^3/0.1 = 120L^3 (I think).

eta: oh I missed he t=0.1 s
 
Last edited:
  • #10
rock.freak667 said:
[tex]\frac{dV}{dt}=12L^2[/tex]



You don't need L in terms of t when they tell you that [itex]\frac{dL}{dt}=10t[/itex]

Isn't dv/dt the differential of v with respect to t? Then surely the answer shouldn't be given in 'L' Also you have worked out dv/dl not dv/dt


I think the chain rule is needed.
 
  • #11
Doctoress SD said:
Isn't dv/dt the differential of v with respect to t? Then surely the answer shouldn't be given in 'L' Also you have worked out dv/dl not dv/dt


I think the chain rule is needed.

Yes that was a typo on my part.
 

1. What is differentiation?

Differentiation is a mathematical process used to find the rate of change of a function. It involves finding the derivative of a function at a specific point.

2. Why is differentiation important?

Differentiation is important because it allows us to analyze and understand the behavior of functions. It is used in various fields such as physics, economics, and engineering to model and solve real-world problems.

3. What is the difference between differentiation and integration?

Differentiation and integration are inverse operations of each other. Differentiation finds the rate of change of a function, while integration finds the original function from its derivative.

4. How is differentiation used in science?

Differentiation is used in science to describe and analyze the behavior of physical systems. It is particularly useful in studying motion, rates of change, and growth of populations.

5. What are some common techniques for differentiating functions?

Some common techniques for differentiating functions include the power rule, product rule, quotient rule, and chain rule. These rules allow us to differentiate functions with multiple variables, products, and quotients.

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