Re: what is the tangent line to xy^2 + x^2y = 2?

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Re: what is the tangent line to xy^2 + x^2y = 2?

Catherine,
Can you please show us the algebra you did to arrive at y = 1/sqrt x from xy^2 + x^2y = 2?
This seems to me to be the hardest part of the problem. Finding the slope of the tangent line is
cake if you can actually get y by itself on one side of the equation.

Thanks!

Posted 227 day ago

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Answers (2)

atta

The tangent line to xy^2 + x^2y = 2 is derived by differentiating the function and then using algebraic formulas. The type of problem implies a type of differentiation called or referred to as Implicit Differentiation, which is needed when both xy terms are involved in the equation. When performing implicit differentiation, both sides of the equation are differentiated and then solved for dy/dx. By differentiating the above function of xy^2 + x^2y = 2 yields

d/dx[xy^2] + d/dx[x^2y] = d/dx[2] => 2xydy/dx + y^2(1) + 2xy + x^2dy/dx = 0. By solving for dy/dx yields the implicit form of differentiation, which is dy/dx[2xy + x^2] = -2xy - y^2 => dy/dx = -2xy - y^2 / 2xy + x^2 = f'(x), where f'(x) is the slope of the line tangent to xy^2 + x^2y = 2 at a specified point. Therefore, given f'(x) = slope of the line, all that is needed to find the line tangent to the graph of f is merely the algebraic formula called the point-slope form defined formally as y-y1 = m(x - x1). By letting (m = f'(x)), yields the substitution of y-y1 = f'(x)(x - x1). I leave the rest for you.

J.C

Posted 224 day ago

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pecks_tutoring

We found the slope of the tangent line at (1,1) in an earlier post. The slope was -1.
This gives:
y = -x + b

Use the given point(1,1) to get b:
1 = -1 + b ==> b = 2

The line tangent to the curve xy² + x²y = 2 at the point (1,1) is:
y = -x +2

Posted 227 day ago

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