consider the curve given by the equation ( y^{3}-xy = 2 ). it can be shown that ( \frac{dy}{dx}=\frac{y}{3y^{…

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

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

Answer

(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})

Answers:

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