Recent content by Lyn

ok i think I'm getting there. thenk you so much for all your help, i'll try finish it off now

Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2

I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest. I have done the same for dz/dy d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy...

ok i tried to expand your expression, not really sure but i get d^2z/dx^2=d/dx(dz/du.du/dx) =d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx) then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles

i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?

Thank you for replying. Yes I have set u=sqrt(x^2+y^2) Would d^2z/dx^2=d/dx.(dz/dx)? So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation

I am stuck on the question, 'If f is a twice differentiable function of a single variable, find f = z(sqrt(x^2+y^2)) that satisfies d^2z/dx^2 +d^2z/dy^2 = x^2 +y^2 (ALL d's ARE MEANT TO BE PARTIAL DERIVATIVES) i know dz/dx=(dz/du).(du/dx) i can find du/dx but i don't know how to find dz/du

Hi i am struggling too with this same question, http://img132.imageshack.us/img132/5736/wathh1.jpg I have tried to follow what you have said and i get my (d's are partial derivatives) dz/dx = 2. I am unsure if this is correct because just like the problem the other person had with this...