# Differentiating a function of x wrt time

## Main Question or Discussion Point

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?

Mark44
Mentor
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.
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?
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?

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)