imoHarlinlilsonece
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Find the points on the lemniscate 2(x^2 + y^2)^2 = 25(x^2 - y^2) where the tangent is horizontal. The derivative using implicit differentiation is (-4x^3 - 4xy^2 + 25x)/(4x^3 + 4x^2 y + 25y)

Respuesta :

caylus
Hello,

f(x,y)=2(x²+y²)-25(x²-y²)=0(1)

@f/@x=2*2*(x²+y²)*2x-50x

@f/@y=2*2*(x²+y²)*2y-50y

dy/dx=-(@f/@x)/(@f/@y)=0
==> 2x*(4x²+4y²-25)=0
==>x=0 or x²+y²=25/4 (2)

As (1)==>2*25/4-25(x²-y²)=0
==> x²-y²=1/2 (3)

(2) and (3)=>2x²=27/4 ==>x²=27/8 and y²=23/8
==>x=-√(27/8) and (y=-√(23/8) or y=+√(23/8)
or x=√(27/8) and (y=-√(23/8) or y=+√(23/8)

meerkat18

You got this 4(x^2 + y^2)(2x - 2yy') = 25(2x - 2yy') I think the 2nd factor on the left-hand side should be (2x+2y y'). So it should be 4(x^2 + y^2)(2x + 2yy') = 25(2x - 2yy') Now solve for y'

you get

y′=−4x3−4xy2+25x25y+4y(x2+y2)=0

It looks like we can solve the numerator for the x values that make it zero.

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