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vf321
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Do you know the power rule?

steel11
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yes

steel11
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i know how to get y' alone. im having problems putting it in the formular yy1=m(xx1)

vf321
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How so?

Coolsector
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(2/3) * x^(1/3) + (2/3) * y^(1/3) * (dy/dx) = 0
dy/dx = x^(1/3) / y^(1/3)

steel11
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yes, got that far. now what coolsector?

vf321
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Fill in y

vf321
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and x

Coolsector
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plug x = 3sqrt(3) , y= 1

steel11
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i filled in both x and y. and got a number. i used that as my slope in the equation yy1=m(xx1)

Coolsector
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you get
dy/dx =  (3sqrt(3))^(1/3)

steel11
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yes, which is 1.732 corect?

Coolsector
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mm i ges 0.577 in fact

Coolsector
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and x1 = 3sqrt(3) y1 = 1...

steel11
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dy/dx = x^(1/3) / y^(1/3)
i replaced x and y with given points in this, and got 1.732

Coolsector
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dy/dx = (3sqrt(3))^(1/3) / 1

Coolsector
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dy/dx = (3sqrt(3))^(1/3) i get 0.577

steel11
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ok. go on with that then. what is the final answer?

steel11
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need an equation in y= form

Coolsector
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y =1 0.577(x+3sqrt(3))
y= 1 0.577x 3
y = 0.577x 2

Coolsector
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in fact dy/dx = 1/sqrt(3)
so :
y = x/sqrt(3) 2

Coolsector
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\[y = \frac{ x }{ \sqrt{3} } 2\]


steel11
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says incorrect :/

Coolsector
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if we plug x = 3sqrt(3) we get y = 1 right ?

steel11
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yes

Coolsector
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well weird

steel11
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yeah, i have no idea how to do this problem :/ thanks anyways

Coolsector
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im sure it is correct

steel11
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just tried putting in again, says wrong. :

steel11
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going to close post and make another one.