For the curve x^5+2xy^2=yx^2+y^3 : (a) Explain why, except for (0,0), the curve
ID: 2977131 • Letter: F
Question
For the curve x^5+2xy^2=yx^2+y^3 : (a) Explain why, except for (0,0), the curve lies entirely in the first and third quadrants. Hint: factor out x on the left side and y on the right. (b) Find dy/dx by implicit differentiation. (c) What happens when you try to use your answer for (b) to find the slope of the curve at (0,0)? (d) According to (a), x and y must have the same sign, so it makes sense to write y=xt^2. Substitute this in the equation of the curve and solve for x to get x as a function of t. (if it comes up, ignore theExplanation / Answer
x^5+2xy^2=yx^2+y^3
D( x^5)= D(-2xy^2+yx^2+y^3 )
5x^4 = -2D(x)y^2-2xD(y^2)+D(y)x^2+yD(x^2)+D(y^3)
5x^4 = -2y^2-2x(2yy'))+y'x^2+y(2x)+3y^2 y'
5x^4 +2y^2-2xy=-2x(2yy'))+y'x^2+3y^2 y'
5x^4 +2y^2-2xy=y'{-4xy+x^2+3y^2}
y' = {5x^4 +2y^2-2xy}/{-4xy+x^2+3y^2}
y' at (0,0) does not exists
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y=xt^2
x^5+2x^3t^4=t^2x^3+x^2t^6
x^5+2x^3t^4=t^2x^3+x^2t^6
x^3 =t^6-{xt^2(1-2t^2)}