calculus: Given x^2 - 2xy + 4y^2 = 64, find slope, tangent,

[venialove]

Consider the equation x^2 - 2xy + 4y^2 = 64.

(A) Write an expression for the slope of the curve at any point (x, y).

(B) Find the equation of the tangent lines to the curve at the point at which x = 2.

(C) Find d[sup:2mydg6ju]y[/sup:2mydg6ju]y/dx[sup:2mydg6ju]2[/sup:2mydg6ju] at the point (0, 4).
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Edited by stapel -- Reason for edit: Removing SHOUTING, matching cases

 

[galactus]

Re: calculus help equation

This is another case of implicit differentiation.

Differentiate implicitly to find y'.

After you have y', do it again to find y''. But in that case, when you're finished sub in y' from your first case.

Give it a go and let us see how you're progressing.

 

[venialove]

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Consider the equation: .\(\displaystyle x^2-2xy+4y^2\:=\:64\)

(a) Write an expression for the slope of the curve at any point \(\displaystyle (x,y)\)

\(\displaystyle \text{Differentiate implicitly: }\;2x - 2xy' - 2y + 8yy' \:=\:0\)

. . \(\displaystyle 8yy' - 2xy' \:=\:2y - 2x \quad\Rightarrow\quad 2(4y - x)y' \:=\:2(y-x)\)

\(\displaystyle \text{So: }\;y' \;=\;\frac{y-x}{4y-x}\)

 

[galactus]

Wow, look at you. Very good LaTex. Looks like Sorobans.

Anyway, differentiate again to find y''. Here's what you do regarding a second implicit derivative.

\(\displaystyle \frac{y-x}{4y-x}\)

Using the quotient rule:

\(\displaystyle \frac{(4y-x)(y'-1)-(y-x)(4y'-1)}{(4y-x)^{2}}\)

Simplify:

\(\displaystyle \frac{-3y+3xy'}{(4y-x)^{2}}\)

Now, sub in 'y from before:

\(\displaystyle \frac{-3y+3x\left(\frac{y-x}{4y-x}\right)}{(4y-x)^{2}}\)

\(\displaystyle y''=\frac{3(x^{2}-2xy+4y^{2})}{(x-4y)^{3}}\)