Answer to Question #148809 in Calculus for Eucarius Walker

Question #148809
Consider the curve defined by 2y^2-x^2y=3.

a. Show that dy/dx=2xy/3y-x^2

b. Write an equation for the line tangent to the curve at the point (1, -1).

c. Show that there is a point P with x=coordinate 0 at which the line tangent to the curve P is horizontal. Then find the y coordinate of P.

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Expert's answer
2020-12-08T11:15:54-0500

(a)



"\\left.\\frac{d}{dx}\\right|2y^2-x^2\\cdot y =3\\to 2\\cdot 2y\\cdot y'-2x\\cdot y-x^2\\cdot y'=0\\\\[0.3cm]\n\\left.\\left(4y-x^2\\right)\\cdot y'=2xy\\right|\\div\\left(4y-x^2\\right)\\\\[0.3cm]\n\\boxed{y'=\\frac{2xy}{4y-x^2}}"

Q.E.D.


(b)

Note : according to the condition of the problem, we need to write the equation of the tangent to the point "P(1,-1)" - this means that "x_0=1\\,\\,\\,\\text{and}\\,\\,\\,y(1)=-1" .

From the theory we know that the equation of the tangent line at a point "x=x_0" has the form



"y_{tang}(x)=y(x_0)+y'(x_0)\\cdot(x-x_0)"

In our case,



"x_0=1\\to y(1)=-1\\to y'(1)=\\frac{2\\cdot1\\cdot y(1)}{4\\cdot y(1)-1^2}=\\\\[0.3cm]\n=\\frac{2\\cdot(-1)}{4\\cdot(-1)-1}=\\frac{-2}{-4-1}=\\frac{2}{5}\\longrightarrow\\\\[0.3cm]\ny_{tang}(x)=y(1)+y'(1)\\cdot(x-1)=-1+\\frac{2}{5}\\cdot(x-1)=\\\\[0.3cm]\n=-\\frac{5}{5}+\\frac{2x}{5}-\\frac{2}{5}=\\frac{2x}{5}-\\frac{7}{5}\\\\[0.3cm]\n\\boxed{y_{tang}(x)=\\frac{2x}{5}-\\frac{7}{5}\\,\\,\\,,\\,\\,\\,\\text{at}\\,\\,\\,x=1}"

(c)

Note : We have to show that the tangent at "x=0" is horizontal - this means that the derivative is 0. To do this, first, we find the value of the function at "x=0" . Let's use the equation



"\\left\\{\\begin{array}{l}\nx=0\\\\[0.3cm]\n2y^2-x^2\\cdot y=3\n\\end{array}\\right.\\to2y^2-0^2\\cdot y=3\\to 2y^2=3\\\\[0.3 cm]\n\\boxed{y(0)=\\pm\\sqrt{\\frac{3}{2}}}"

Note : For our purpose, it is enough to understand that "y(0)=y_0\\neq0" . Then,



"\\text{from part (a)}\\,\\,\\, :\\,\\,\\,y'(x)=\\frac{2xy}{4y-x^2}\\to y'(0)=\\frac{2\\cdot0\\cdot\\overbrace{y(0)}^{y_0}}{4\\cdot\\underbrace{y(0)}_{y_0}-0^2}=0\\\\[0.3cm]\n\\boxed{y'(0)=0\\longrightarrow\\text{the tangent is horizontal}}"

Q.E.D.

ANSWER



"y_{tang}(x)=\\frac{2x}{5}-\\frac{7}{5}\\,\\,\\,,\\,\\,\\,\\text{at}\\,\\,\\,x=1"


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