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Consider the curve C given by $(x^2+y^2)^2=(x^2−y^2)$

  1. Using implicit differentiation, find $dy/dx$ in terms of $x$ and $y$.
  2. Find all points $(x,y)$ on $C$ such that the tangent line is horizontal. You may assume $(x,y)≠(0,0)$
  3. What is the smallest value y can take on $C$? What is the largest value?

Hint: what should the slope of the tangent line be in each of those cases?

user
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Ruby Liu
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  • This looks like a straight up homework problem to me. Please, share with us what you have done so far and what parts of this problem are giving you trouble. We'd love to help you learn and understand. Solving a question for you does the exact opposite. – Hamed Oct 14 '18 at 20:37
  • I solved y'=2x^3+2xy^2-x/-(2x^2+2y^3+y) – Ruby Liu Oct 14 '18 at 20:39
  • the points I found is ((3/8)^(1/2), (1/8)^(1/2)) – Ruby Liu Oct 14 '18 at 20:40
  • @It would be useful if you also show your work and effort under the question you have posed. – user Oct 14 '18 at 20:47

2 Answers2

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Note: I write $x'$ and $y'$ instead of $dx$ and $dy$ to sort of imply that the derivative is taken with respect to some un-named variable.

From $(x^2+y^2)^2 =(x^2−y^2) $, since $(f^2)' = 2ff'$, the ID (implicit derivative) of the left side is

$\begin{array}\\ 2(x^2+y^2)(x^2+y^2)' &=2(x^2+y^2)(2xx'+2yy')\\ &=4(x^2+y^2)(xx'+yy')\\ &=4xx'(x^2+y^2)+4yy'(x^2+y^2)\\ \end{array} $

and the ID of the right side is $2xx'-2yy'$.

Equating these, dividing by 2, and then grouping the terms with $x'$ and $y'$,

$4xx'(x^2+y^2)+4yy'(x^2+y^2) =2xx'-2yy' $ or $2xx'(x^2+y^2)+2yy'(x^2+y^2) =xx'-yy' $ so that $xx'(2x^2+2y^2-1)+yy'(2x^2+2y^2+1) =0\\ $ or $yy'(2x^2+2y^2+1) =-xx'(2x^2+2y^2-1) $.

Therefore $\dfrac{dy}{dx} =\dfrac{y'}{x'} =\dfrac{-x(2x^2+2y^2-1)}{y(2x^2+2y^2+1)} $.

I'll let you handle the rest.

marty cohen
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HINT

By implicit differentiation we have

$$(x^2+y^2)^2=(x^2−y^2)\implies 4x(x^2+y^2)dx+4y(x^2+y^2)dy=2xdx-2ydy$$

$$\implies \frac{dy}{dx}=\frac{x-2x(x^2+y^2)}{y+2y(x^2+y^2)}$$

user
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