Problem 32 - Calculus of a Single Variable

Question

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

Step-by-Step Solution

Answer

The slope of the tangent line to the graph of the function at the given point \(\left(\frac{4}{3}, \frac{8}{3}\right)\) is -1

Step 1: Differentiation

First, differentiate the given equation implicitly with respect to \(x\). That is, all terms involving \(y\) should be followed by \(\frac{dy}{dx}\). Carry out differentiation to each term: \(3x^2+3y^2\frac{dy}{dx}-6(y+x\frac{dy}{dx})=0\)

Step 2: Solve for dy/dx

We must now rearrange the equation obtained in step 1 to solve for \(\frac{dy}{dx}\), which represents the slope of the tangent line. We get: \(\frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x}\)

Step 3: Substitute the given point

Now, replace \(x\) and \(y\) in the equation obtained at step 2 by their respective values given: \(\frac{4}{3}\) and \(\frac{8}{3}\). Hence, our equation becomes: \(\frac{dy}{dx} = \frac{6*(\frac{8}{3})-3*(\frac{4}{3})^2}{3*(\frac{8}{3})^2-6*\frac{4}{3}}\)

Step 4: Simplify the expression

On simplifying the above equation, we find the slope \(\frac{dy}{dx}\) to be -1