Differentiating a function of x wrt time

In summary, the conversation discusses a problem with redundancy and the use of the chain rule to differentiate a displacement function with respect to time. The second function, which depends on two displacement terms, is also examined. The conversation also discusses the use of w as a displacement function for a simply supported beam with x representing the horizontal axis along the beam. The goal is to find ∂^2w/∂t^2 for solving an integral related to the total potential energy functional.
  • #1
c0der
54
0
Hi,

I figured out the only redundancy to my problem is this:

I'll start off with a simple case, where w1,w2 are the displacements at intervals of one third along a beam.

w = 3w1/L.x (Note, x is in the numerator for all cases)

To differentiate this with respect to time, I use the chain rule as follows:

∂w/∂t = ∂w/∂x.∂x/∂t = 3w1/L.w1'

Now for the second function, which depends on two displacement terms w1,w2

w = (2 - 3/L.x)w1 + (-1 + 3/L.x)w2

∂w/∂t = ∂w/∂x.∂x/∂t = -3/L.w1.w1' + 3/L.w2.w2'

I'd imagine this wrong, any hints?
 
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  • #2
c0der said:
Hi,

I figured out the only redundancy to my problem is this:

I'll start off with a simple case, where w1,w2 are the displacements at intervals of one third along a beam.
One third of what? If it's 1/3 of a unit, then there's no need to give them names.
c0der said:
w = 3w1/L.x (Note, x is in the numerator for all cases)
This is extremely difficult to read, with '.' apparently being used to indicate multiplication. The asterisk character, *, is typically used for multiplication.

What does x represent? What does w represent?
c0der said:
To differentiate this with respect to time, I use the chain rule as follows:

∂w/∂t = ∂w/∂x.∂x/∂t = 3w1/L.w1'

Now for the second function, which depends on two displacement terms w1,w2

w = (2 - 3/L.x)w1 + (-1 + 3/L.x)w2

∂w/∂t = ∂w/∂x.∂x/∂t = -3/L.w1.w1' + 3/L.w2.w2'

I'd imagine this wrong, any hints?
 
  • #3
Thanks for your reply.

w = (3w1/L)*x for 0<x<L/3
w = [ 2 - (3/L)*x ]*w1 + [ -1 + (3/L)*x ]*w2 for L/3<x<2L/3
w = 3*w2*[ 1 - x/L ] for 2L/3<x<L

w is a displacement function for a simply supported beam satisfying w=0 at x=0 and x=L, w=w1 at x=L/3 and w=w2 at x=2L/3

w is the displacement function of the beam (breaking up a beam of length L into L/3 sized intervals, such that it can displace vertically). x is the horizontal axis along the beam

Now I'm trying to find ∂^2w/∂t^2, hope this is clear, to solve the integral ∫m*(∂^2w/∂t^2)*wdx, to get the acceleration term for the total potential energy functional (not important at this stage)
 

FAQ: Differentiating a function of x wrt time

What does it mean to differentiate a function of x with respect to time?

When we differentiate a function of x with respect to time, we are finding the rate of change of that function with respect to time. This means we are calculating how the function changes as time passes.

Why is differentiating a function of x with respect to time important?

Differentiating a function of x with respect to time allows us to understand how a variable changes over time. This is crucial in many fields of science, such as physics, chemistry, and biology, where we need to determine rates of change in order to make predictions and understand systems.

What is the notation for differentiating a function of x with respect to time?

The notation for differentiating a function of x with respect to time is d/dt, where d represents the derivative and dt represents the change in time.

How is differentiating a function of x with respect to time different from differentiating with respect to another variable?

Differentiating with respect to time is different from differentiating with respect to another variable because time is constantly changing, whereas other variables may remain constant. This means that when differentiating with respect to time, we are calculating the instantaneous rate of change at a specific moment, rather than the average rate of change over an interval.

Can we differentiate a function of x with respect to time if it is a multivariable function?

Yes, we can differentiate a multivariable function with respect to time. In this case, we would use the partial derivative notation, denoted as ∂/∂t, to indicate that we are only differentiating with respect to time and keeping all other variables constant.

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