The Gradient of f(x,y)=ln(2x⁴+ax²y²+2y⁴) is, in each point (x,y)≠(0,0), orthogonal to the circle with center at the origin and radius r=(x²+y²)⁽1/2), then "a" equals to:
I am trying to solve this question, but I just don't understand how to keep going. I read about orthogonality of the gradient, I made some attempts but I got nowhere. Can someone help me, please?
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Sorry, I basically had already solved that, I just didn't pay enough attention to see what was right in front of me.
r(t)= (rcos(t),rsin(t) r'(t)=(-rsin(t),rcos(t)
grad f(x,y) =((8x³+2axy²)/(2x⁴+ax²+2y⁴),(8y³+2ax²y)/(2x⁴+ax²+2y⁴))
for the orthogonality, Grad f(r(t))*r'(t)=0
0 =(8r³cos³+2ar³costsin²t)(-rsint)/(2r⁴cos⁴t+ar⁴cos²tsin²t+2r⁴sint)+(2ar³cos²t+8r³sin³t)*(rcost)/(2r⁴cos⁴t+ar⁴cos²tsin²t+2r⁴sint)
-8r⁴cos³tsint-2ar⁴costsin³t+2ar⁴cos³sint+8r⁴sin³cost=0
-8r⁴cos³tsint+2ar⁴cos³sint+8r⁴sin³cost-2ar⁴costsin³t=0
a=4
The formal way to solve this problem is to: first, find the gradient. To this end we write down $$\nabla f(x,y)=\left({8x^3+2axy^2\over 2x^4+2y^4+ax^2y^2},{8y^3+2ayx^2\over 2x^4+2y^4+ax^2y^2}\right)$$as the gradient vector is orthogonal to the circle $x^2+y^2=r^2$, the for each point on the perimeter of the circle we must have $$\vec{\nabla f}\cdot (y,-x)=0\quad,\quad x^2+y^2=r^2$$from which we obtain$$8x^3y+2axy^3=8xy^3+2ayx^3$$since on the circle, $xy\ne 0$ we conclude that$$8x^2+2ay^2=8y^2+2ax^2$$which is true only if $a=4$ and $$f(x,y)=\ln 2(x^2+y^2)^2$$