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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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