Max Distance: Find Largest Difference Between Function & XY Plot

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Homework Statement



Hey.
I need to find the biggest distance between this function and the XY plot, any ideas?

Homework Equations





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asi123 said:

Homework Statement



Hey.
I need to find the biggest distance between this function and the XY plot, any ideas?
You mean the xy-plane.
The surface is given by 2x^3+ 3y^2+ 2z^2+ 2xz= 6 and the distance from any point (x, y, z) to (x, y, 0) (the xy-plane) is just z. To minimize of maximize that, look at the derivatives of z with respect to x and y. Using the chain rule to differentiate with respect to x, 6x^2+ 4zz_x+ 2z+ 2xz_x= 0 so z_x= -(6x^2+ 2z)/(4z+ 2x)= -(3x^2+ z)/(2z+x)=0 and differentiating with respect to y, 6y+ 2zz_y+ 2xz_y= 0 so z_y= -6y/(2z+ 2x)= -3y/(z+ x)= 0. Find values of of x, y, z that satisfy those as well as 2x^3+ 3y^2+ 2z^2+ 2xz= 6.
 
HallsofIvy said:
You mean the xy-plane.
The surface is given by 2x^3+ 3y^2+ 2z^2+ 2xz= 6 and the distance from any point (x, y, z) to (x, y, 0) (the xy-plane) is just z. To minimize of maximize that, look at the derivatives of z with respect to x and y. Using the chain rule to differentiate with respect to x, 6x^2+ 4zz_x+ 2z+ 2xz_x= 0 so z_x= -(6x^2+ 2z)/(4z+ 2x)= -(3x^2+ z)/(2z+x)=0 and differentiating with respect to y, 6y+ 2zz_y+ 2xz_y= 0 so z_y= -6y/(2z+ 2x)= -3y/(z+ x)= 0. Find values of of x, y, z that satisfy those as well as 2x^3+ 3y^2+ 2z^2+ 2xz= 6.

10x a lot
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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