Question

90–93. {Use of Tech} Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph.
x(1−y²)+y³=0

Accepted Answer

Start by understanding the problem: We need to find points on the curve where the tangent line is horizontal or vertical. The curve is given by the equation $$ x(1-y^2) + y^3 = 0 . To $$find where the tangent line is horizontal, we need to find where the derivative $$ \frac{dy}{dx} = 0 . $$Use implicit differentiation on the given equation. Differentiate both sides with respect to $$ x . $$Apply the product rule to $$ x(1-y^2) $$ and the chain rule to $$ y^3 . $$This gives: $$ (1-y^2) + x(-2y \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} = 0 . $$Rearrange the differentiated equation to solve for $$ \frac{dy}{dx} $$: $$ \frac{dy}{dx} = \frac{y^2 - 1}{3y^2 - 2xy} . $$Set $$ \frac{dy}{dx} = 0  to $$find horizontal tangents, which implies $$ y^2 - 1 = 0 . $$Solve for $$ y . $$For vertical tangents, set the denominator of $$ \frac{dy}{dx}  to $$zero: $$ 3y^2 - 2xy = 0 . $$Solve for $$ x  in $$terms of $$ y  or $$vice versa. Check if these points satisfy the original curve equation.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Implicit Differentiation

Tangent Lines and Slopes

Graphical Confirmation