题目

Problem

The curve $C$ has equation

y=(3x-2)^6

(a) Find

\frac{dy}{dx}

(2)

Given that the point $P\left(\dfrac13,1\right)$ lies on $C$ ,

(b) find the equation of the normal to $C$ at $P$ . Write your answer in the form $ax+by+c=0$ where $a$ , $b$ and $c$ are integers to be found.

(4)

题目中文翻译

曲线 $C$ 的方程为

y=(3x-2)^6

(a) 求

\frac{dy}{dx}

已知点 $P\left(\dfrac13,1\right)$ 在曲线 $C$ 上，

(b) 求曲线 $C$ 在点 $P$ 处法线的方程。把答案写成 $ax+by+c=0$ 的形式，其中 $a$ 、 $b$ 和 $c$ 是需要求出的整数。

解答

We have

y=(3x-2)^6

求导得

\frac{dy}{dx}=6(3x-2)^5\cdot 3

\boxed{\frac{dy}{dx}=18(3x-2)^5}

P\left(\frac13,1\right)

we have

3x-2=3\left(\frac13\right)-2=-1

So the gradient of the tangent is

\frac{dy}{dx}=18(-1)^5=-18

Therefore the gradient of the normal is

\frac{1}{18}

Using the point-gradient form,

y-1=\frac1{18}\left(x-\frac13\right)

Multiply by $18$ :

18y-18=x-\frac13

Multiply by $3$ :

54y-54=3x-1

Bring all terms to one side:

3x-54y+53=0

Therefore the equation of the normal is

\boxed{3x-54y+53=0}