Recent content by clairaut

I've tried X=(52)^.5 [cos(@)] Y=(52)^.5 [sin(@)] I can't get the substitution to match

Thanks. I'll try it. It looks nasty

And why can't I assume the trig substitutions here? Is it because the substitution results in that zero?

Where am I supposed to start?

Double integral of (52-x^2-y^2)^.5 2<_ x <_ 4 2<_ y <_ 6 I get up to this simplicity that results in a zero! 1-cos^2(@) - sin^2(@) = 0 This identity seems to be useless. HELP PLEASE.

Oh I dunno... The wonderful symmetry that Hess found to make his matrix...? The universal application of isomorphism found in his matrix... Stuff like that

Never mind. I already know about the determinant from hessian matrix. Some so called helpers are very useless. Does it hurt you to have a discussion?

That link brings up 2 variable fxns.

Lol. How have you helped? You've merely agreed with me... No more

How would you find the critical points of a function with three variables? Let's assume that there are no constraints.

Yes. No problems with gradient. It's the CRITICAL POINTS that are hard to find.

[A(x,b)] = (x^2-b^2)^.5 (w + b - 2x) Where b is the leg of each of the right triangles in the corners. [A(x,@)] = (w)(x)[sin(@)] + (x^2)[sin(@)][cos(@)] - 2x^2[sin(@)] Above are the cross sectional surface area equations.

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Thanks! When i do the first and second derivative test, I can find the local minimums. However, I can only deduce the local maximum without a formal derivative test. Is there a way to mathematically prove the local maxima?

Also, volume of trapezoid is not important here. I only need the cross sectional area of this trapezoid.