Question

Find an equation of the normal to the curve at the point where x= 4, giving theanswer in the form ax + by+c = 0, where a, band care integers.

Answer 1

Since the equation of the curve is

`\begin{gathered} y=3x+(48)/(x^2) \\ y=3x+48x^(-2) \end{gathered}`

The slope of the tangent of the curve at the point (x, y) is y'

The normal and the tangent are perpendicular at this point

Then we will find the slope of the tangent, then find from it the slope of the normal

`\begin{gathered} y^(\prime)=3(1)x^(1-1)+48(-2)x^(-2-1) \\ y^(\prime)=3-96x^(-3) \end{gathered}`

Since we need the tangent at x = 4, then substitute x by 4 in y'

`\begin{gathered} m_T=3-96(4)^(-3) \\ m_T=(3)/(2) \end{gathered}`

To find the slope of the normal, reciprocal the m and change its sign

`m_N=-(2)/(3)`

The form of the equation is y = mx + b, where m is the slope and b is the y-intercept

The equation of the normal is

`y=-(2)/(3)x+b`

To find b we need a point on the curve

Since x = 4, substitute it in the equation of the curve to find its corresponding y

`\begin{gathered} y=3(4)+(48)/((4)^2) \\ y=15 \end{gathered}`

Then substitute x and y in the equation of the normal by 4 and 15

`\begin{gathered} 15=-(2)/(3)(4)+b \\ 15=-(8)/(3)+b \end{gathered}`

Add -8/3 to both sides

`\begin{gathered} 15+(8)/(3)=-(8)/(3)+(8)/(3)+b \\ (53)/(3)=b \end{gathered}`

The equation of the normal is

`y=-(2)/(3)x+(53)/(3)`

Multiply all terms by 3 to cancel the denominators

`\begin{gathered} 3y=-(2)/(3)(3)x+(53)/(3)(3) \\ 3y=-2x+53 \end{gathered}`

Add 2x to both sides

`\begin{gathered} 3y+2x=-2x-2x+53 \\ 3y+2x=53 \end{gathered}`

Subtract 53 from both sides

`\begin{gathered} 2x+3y-53=53-53 \\ 2x+3y-53=0 \end{gathered}`

The equation of the normal is 2x + 3y - 53 = 0