Question

1. an equation of the normal line to the curve ( y=(x^{2}-1)^{\frac{1}{3}} ) at the point where ( x = 3 ) is

(a) ( y+12x = 38 )
(b) ( y-4x = 10 )
(c) ( y+2x = 4 )
(d) ( y+2x = 8 )
(e) ( y-2x=-4 )

Explanation:

Step1: Find the derivative of \(y=(x^{2}-1)^{\frac{1}{3}}\)

Use the chain - rule. If \(y = u^{\frac{1}{3}}\) and \(u=x^{2}-1\), then \(\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}\) and \(\frac{du}{dx} = 2x\). So \(\frac{dy}{dx}=\frac{2x}{3(x^{2}-1)^{\frac{2}{3}}}\).

Step2: Evaluate the derivative at \(x = 3\)

Substitute \(x = 3\) into \(\frac{dy}{dx}\). First, \(u=(3)^{2}-1=8\). Then \(\frac{dy}{dx}\big|_{x = 3}=\frac{2\times3}{3\times8^{\frac{2}{3}}}\). Since \(8^{\frac{2}{3}}=(8^{\frac{1}{3}})^{2}=4\), \(\frac{dy}{dx}\big|_{x = 3}=\frac{6}{3\times4}=\frac{1}{2}\).

Step3: Find the slope of the normal line

The slope of the normal line \(m_{n}\) is the negative reciprocal of the slope of the tangent line. If the slope of the tangent line \(m_{t}=\frac{1}{2}\), then \(m_{n}=-2\).

Step4: Find the \(y\) - value when \(x = 3\)

Substitute \(x = 3\) into \(y=(x^{2}-1)^{\frac{1}{3}}\). \(y=(9 - 1)^{\frac{1}{3}}=2\).

Step5: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=3\), \(y_{1}=2\) and \(m=-2\). So \(y - 2=-2(x - 3)\), which simplifies to \(y+2x=8\).

Answer:

D. \(y + 2x=8\)