How Do You Determine the Angle Between Gradient Vectors in Parametric Formulas?

oahsen
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Homework Statement


Find the angle between (grad)u and (grad)v at all points with x!=0 and y!= 0 if
x =( e^u)*(cos v) and y = (e^u) (sinv) .


The Attempt at a Solution



is not here x and y a function of u and v? How are we going to find grad of u and v? Should we pull out u and y from the equations (I mean if x =( e^u)*(cos v)) then u=ln(x/cosv) etc. )
 
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No, you write u and v in terms of x and y. Here's a hint to start with: sin2x + cos2x = 1
 
neutrino said:
No, you write u and v in terms of x and y. Here's a hint to start with: sin2x + cos2x = 1

thank you very much, ı understand. (ı found the answer as pi/2. I hope it is true)
 
oahsen said:
ı found the answer as pi/2. I hope it is true
It's certainly true.
 
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...
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