Consider the curve given by the equation y^(3)-xy=2. It can be shown that (dy)/(dx)=(y)/(3y^(2)-x). Find the point on the curve where the line tangent to the curve is vertical.

Question

Consider the curve given by the equation  y^(3)-xy=2. It can be shown that  (dy)/(dx)=(y)/(3y^(2)-x). Find the point on the curve where the line tangent to the curve is vertical.

Bytelearn · Accepted Answer

Find Derivative Undefined: To find where the tangent line to the curve is vertical, we need to find where the derivative (dy/dx) is undefined, which occurs when the denominator of the derivative is zero.Set Denominator Equal: The derivative is given by (dy/dx) = y / (3y^2 - x). Set the denominator equal to zero to find when the derivative is undefined: 3y^2 - x = 0.Solve for x: Solve for x in terms of y: x = 3y^2.Substitute x into Equation: Substitute x = 3y^2 into the original equation y^3 - xy = 2 to find the corresponding y-value(s): y^3 - (3y^2)y = 2.Simplify Equation: Simplify the equation: y^3 - 3y^3 = 2, which simplifies to -2y^3 = 2.Solve for y: Solve for y: y^3 = -1, which gives y = -1 (since we are looking for real solutions).Substitute y into x: Now that we have y = -1, substitute it back into x = 3y^2 to find the corresponding x-value: x = 3(-1)^2.Calculate x: Calculate the x-value: x = 3(1) = 3.Final Point: The point on the curve where the tangent line is vertical is (3, -1).

Consider the curve given by the equation $y^{3}-x y=2$ . It can be shown that $\frac{d y}{d x}=\frac{y}{3 y^{2}-x}$ . $\newline$ Find the point on the curve where the line tangent to the curve is vertical. $\newline$ $(\square$ , $\square)$

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Consider the curve given by the equation y3−xy=2 y^{3}-x y=2 y3−xy=2. It can be shown that dydx=y3y2−x \frac{d y}{d x}=\frac{y}{3 y^{2}-x} dxdy​=3y2−xy​.\newlineFind the point on the curve where the line tangent to the curve is vertical.\newline(□(\square(□ , □)\square)□)

Consider the curve given by the equation $y^{3}-x y=2$ . It can be shown that $\frac{d y}{d x}=\frac{y}{3 y^{2}-x}$ . $\newline$ Find the point on the curve where the line tangent to the curve is vertical. $\newline$ $(\square$ , $\square)$