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.
x²(3y²−2y³) = 4

Accepted Answer

Start by differentiating the given equation implicitly with respect to x. The equation is x²(3y²−2y³) = 4. Use the product rule for differentiation, which states that if you have a product of two functions u(x) and v(x), then the derivative is u'(x)v(x) + u(x)v'(x). Differentiate the left side of the equation: For x²(3y²−2y³), let u = x² and v = 3y²−2y³. The derivative of u with respect to x is 2x, and the derivative of v with respect to x involves implicit differentiation: d/dx(3y²−2y³) = 6y(dy/dx) - 6y²(dy/dx). Apply the product rule: The derivative of x²(3y²−2y³) is 2x(3y²−2y³) + x²(6y(dy/dx) - 6y²(dy/dx)). Set this equal to the derivative of the right side, which is 0, since the derivative of a constant (4) is 0. To find points where the tangent line is horizontal, set dy/dx = 0 and solve the resulting equation for x and y. This will give you the conditions under which the slope of the tangent is zero. To find points where the tangent line is vertical, identify where the expression for dy/dx becomes undefined. This typically occurs when the denominator of the expression for dy/dx is zero. Solve for x and y under these conditions, and verify that these points lie on the original curve by substituting back into the original equation.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Implicit Differentiation

Tangent Lines and Slopes

Graphical Confirmation