Problem 30 - Calculus of a Single Variable

Question

Find the slope of the tangent line to the graph at the given point. Cissoid: \((4-x) y^{2}=x^{3}\) Point: \((2,2)\)

Step-by-Step Solution

Answer

The slope of the tangent line to the graph at the point (2,2) is -1.

Step 1: Understand the given equation

Firstly, understand the given equation which is an implicit function: \((4-x)y^{2} = x^{3}\). Our goal is to find the derivative of this function using implicit differentiation.

Step 2: Apply the implicit differentiation

In order to find the derivative, implicitly differentiate each side of the equation with respect to x, using the product rule and chain rule where necessary. Differentiating both sides with respect to x would result to: \(-y^{2} - 2(4-x)y \frac{dy}{dx} = 3x^{2}\). Because we are looking for \(\frac{dy}{dx}\) (the derivative of y), let's isolate \(\frac{dy}{dx}\) on one side.

Step 3: Solve for dy/dx

Isolate \(\frac{dy}{dx}\) to one side. First add \(y^2\) to both sides to get \(-2(4-x)y \frac{dy}{dx} = 3x^{2}+y^{2}\). Then, divide both sides by \(-2(4-x)y\) to isolate \(\frac{dy}{dx}\) on one side. That is \(\frac{dy}{dx} = \frac{-3x^{2}-y^{2}}{-2(4-x)y}\). This is the derivative that gives us slope of the tangent.

Step 4: Substitute the coordinates of the given point into dy/dx

Substitute the coordinates of the given point (2, 2) into \(\frac{dy}{dx}\). Then solve \(\frac{dy}{dx}\) at the point (2,2) which gives the slope of the tangent at that point. When x=2 and y=2, \(\frac{dy}{dx} = \frac{-3(2)^2-(2)^2}{-2(4-2)(2)} = -1\). This is the slope of the tangent line to the graph at the point (2, 2).