Could somebody explain this problem and solution(about partial derivatives)

oahsen
Messages
58
Reaction score
0

Homework Statement



there is a question and a solution in this page;
http://www.fen.bilkent.edu.tr/~otekman/math102/s03/m2q5.html"

Please firstly examine the question and solution. (5b)...
There it says f(x,y)=z and x=g(r,teta) and y=h(r,teta) and asks fxx.
He solves this problem by starting so;
fx=zr*rx+zteta*tetax ...
My question is that by writing fx such as above aren't you assume that r and teta are a function of x? However in the question it says x=f(r,teta); this means x is a function of r and teta then how can you write fx=zr*rx+zteta*tetax? Isn't there a contradiction or am i understanding something false?:confused:
 
Last edited by a moderator:
Physics news on Phys.org
You are told that x= r cos(\theta) so obviously x is a function of r and \theta and therefore r and \theta are functions of x. (It does not however anywhere say x= f(r,[/itex]\theta[/itex]).) As long as a function is invertible, you can work either way: If x= f(r), and r is invertible, then r= f-1(x) and dr/dx= 1/(dx/dr).\

Here, you really have "polar coordinates": x= r cos(\theta) and y= r sin(\theta) so that r= \sqrt{x^2+ y^2} and \theta= arctan(\frac{y}{x}). You can do it either way.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top