Question

Find the slope of the tangent line to the graph at the indicated point. Folium of Descartes: \(x^{3}+y^{3}-6 x y=0\) Point: \(\left(\frac{4}{3}, \frac{8}{3}\right)\)

Short answer

The slope of the tangent line to the Folium of Descartes at the point \(\left(\frac{4}{3}, \frac{8}{3}\right)\) is \(-\frac{1}{2}\).

Step-by-step solution

Differentiate both sides of the equation

By applying the Chain Rule, we have: \[3x^{2} \frac{dx}{dt} + 3 y^{2} \frac{dy}{dt} - 6 (y \frac{dx}{dt} + x \frac{dy}{dt}) = 0\] We assume \(\frac{dx}{dt} = 1\) as x is the independent variable. Simply solve for \(\frac{dy}{dt}\), which will give the slope of the curve at any point (x, y).

Simplify for \(\frac{dy}{dt}\)

The equation from Step 1 can be rearranged to:\[3 y^{2} \frac{dy}{dt} - 6y = - 3 x^{2} + 6x \frac{dy}{dt} \]Simplify further to isolate \(\frac{dy}{dt}\), which should result in \[\frac{dy}{dt} = \frac{6 y - 3 x^{2}}{3 y^{2} - 6 x} \]

Substitute given point into derivative equation

Now, substitute the given point, \(\left(\frac{4}{3}, \frac{8}{3}\right)\), into the derivative equation. This will give you the slope of the tangent at the specified point. Substituting x = \(\frac{4}{3}\) and y = \(\frac{8}{3}\) to the equation, the slope of the tangent line becomes: \[\frac{6 \times \frac{8}{3} - 3 \times (\frac{4}{3})^{2}}{3 \times (\frac{8}{3})^{2} - 6 \times \frac{4}{3}} \]