Recent content by sebb1e

How would I know, merely from the definition in the book that I need that, for example if we were dealing with some totally arbitrary coordinates? (I completely understand why it is there in this case and know how to derive it)

Homework Statement I know that for a general co-ordinate system, the gradient can be expressed as it is at the bottom of this page: http://en.wikipedia.org/wiki/Orthogo...ree_dimensions However, the book I am working from (A First Course in Continuum Mechanics by Gonzalez and Stuart)...

I know that for a general co-ordinate system, the gradient can be expressed as it is at the bottom of this page: http://en.wikipedia.org/wiki/Orthogonal_coordinates#Differential_operators_in_three_dimensions However, the book I am working from (A First Course in Continuum Mechanics by Gonzalez...

I've reduced it to: dz=2(dz/dx)(dx/dy)(dy/dz)dz+((dz/dy)(dy/dz)+(dy/dx)(dx/dy))dz So I need to show that (dz/dy)(dy/dz)+(dy/dx)(dx/dy)=2 where the d's are partial derivatives. If total derivatives this is obvious, am I missing something obvious here? Also, is using this differential...

I think that you can write it as x=x(y,z) and y=y(x,z)?

Homework Statement Prove that if z=z(x,y) is invertible that: (dz/dx)(dy/dz)(dx/dy)=-1 where the d's represent partial differentiation not total differentiation Homework Equations The Attempt at a Solution I guess you start with the 6 total derivatives and substitute them...

Edit: Ignore

OK, so use the Lorentz transformation to show that two points that are simultaneous in O' and 2L apart are 2Lγ apart in O but not simultaneous. Does this mean that the light rays hitting O at t=0 will imply a length of 2lγ then? What I've done before is draw the actual light rays coming in...

Essentially I've tried to work out the different lengths of time that light rays from the rod take to reach the observer. From the left end it leaves r1 before t=0 and from the right end r2 before t=0. r1>r2 as left end of rod is further from origin. I've drawn two right angled triangles...

I can't seem to get it to come out but am I along the right lines? First ray leaves from (-r1*u-L/g) and second ray leaves from (L/g-r2*u) so apparent length if 2L/g+u(r1-r2). To find r1 and r2 I need to use Pythag theorem on the right angled triangles, (r1c, lamda, -r1u-L/g) and (r2c, lamda...

Providing the ice is floating then during the melting process the level of the water will always stay the same. If you put 10kg of ice in 1000kg of water then the water level is the same as if you put 9kg of ice into 1001kg of water.

Homework Statement Q1 http://www.maths.ox.ac.uk/system/files/private/active/0/b07.2_c7.2.pdf Homework Equations The Attempt at a Solution How do I do the final bit that lead to the discrepancy and why does this occur? I have no idea where to begin.

You need to use Fourier Series. You know the normalised solution to Schrodingers equation is root(2/L)sin(nPix/L) Let the initial wave function be f(x)=1/root(L) So f(x)=root(2/L)[sum of (c_n)sin(nPix/L) So c_n=root(2/L)Integral 0 to L of f(x)sin(nPix/L) by Fourier methods The...

It does take up less volume, but remember some of the ice was above the surface of the water. The volume of ice below the water surface is equal to the volume of 1kg of water.

The new 1kg of water takes up the volume of the icecube that was previously below the surface of the water.