Question

consider the curve given by the equation ( y^{3}-xy = 2 ). it can be shown that ( \frac{dy}{dx}=\frac{y}{3y^{2}-x} ).
(a) write an equation for the line tangent to the curve at the point ( (-1,1) ).
(b) find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
(c) evaluate ( \frac{d^{2}y}{dx^{2}} ) at the point on the curve where ( x = -1 ) and ( y = 1 ).

Explanation:

(a)

Step1: Find the slope of the tangent line

We are given \(\frac{dy}{dx}=\frac{y}{3y^{2}-x}\). Substitute \(x = - 1\) and \(y = 1\) into the derivative:
\(\frac{dy}{dx}\big|_{x=-1,y = 1}=\frac{1}{3(1)^{2}-(-1)}=\frac{1}{3 + 1}=\frac{1}{4}\)

Step2: Use the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(-1,1)\) and \(m=\frac{1}{4}\)
\(y-1=\frac{1}{4}(x + 1)\)
\(y=\frac{1}{4}x+\frac{1}{4}+1\)
\(y=\frac{1}{4}x+\frac{5}{4}\)

(b)

Step1: Set the denominator of \(\frac{dy}{dx}\) equal to zero

A vertical tangent line occurs when \(\frac{dy}{dx}\) is undefined, i.e., when \(3y^{2}-x = 0\), so \(x = 3y^{2}\)

Step2: Substitute \(x = 3y^{2}\) into the original equation

The original equation is \(y^{3}-xy=2\). Substitute \(x = 3y^{2}\) into it:
\(y^{3}-(3y^{2})y=2\)
\(y^{3}-3y^{3}=2\)
\(-2y^{3}=2\)
\(y^{3}=-1\)
\(y=-1\)

Step3: Find the corresponding \(x\) value

If \(y=-1\), then \(x = 3y^{2}=3(-1)^{2}=3\)

(c)

Step1: Use the quotient rule to find \(\frac{d^{2}y}{dx^{2}}\)

The quotient rule states that if \(u = y\) and \(v=3y^{2}-x\), then \(\frac{d^{2}y}{dx^{2}}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}\)
We know that \(\frac{du}{dx}=\frac{dy}{dx}=\frac{y}{3y^{2}-x}\) and \(\frac{dv}{dx}=6y\frac{dy}{dx}-1\)

Substitute \(x=-1,y = 1,\frac{dy}{dx}=\frac{1}{4}\) into \(\frac{d^{2}y}{dx^{2}}\)
\(v=3(1)^{2}-(-1)=4\), \(u = 1\), \(\frac{dv}{dx}=6(1)\times\frac{1}{4}-1=\frac{3}{2}-1=\frac{1}{2}\)

\(\frac{d^{2}y}{dx^{2}}=\frac{4\times\frac{1}{4}-1\times\frac{1}{2}}{4^{2}}=\frac{1-\frac{1}{2}}{16}=\frac{\frac{1}{2}}{16}=\frac{1}{32}\)

Answer:

s:

• (a) \(y=\frac{1}{4}x+\frac{5}{4}\)
• (b) \((3,-1)\)
• (c) \(\frac{1}{32}\)