Question

Find the equation of the normal of the curve at the given points :D thankiewss

Answer 1

We want to find the equation of the normal line of
`y=(20-x)/(3x)` at the point
`P=(x_0,y_0)`, where
`y_0=3`. First calculate
`x_0`. We have:


`y_0=f(x_0)=(20-x_0)/(3x_0)\\\\\\3=(20-x_0)/(3x_0)\quad|\cdot3x_0\\\\\\3\cdot3x_0=20-x_0\\\\9x_0=20-x_0\\\\9x_0+x_0=20\\\\10x_0=20\quad|:2\\\\\boxed{x_0=2}`

Now, when we know that
`P=(x_0,y_0)=(2,3)` we can write an equation of the normal line as:


`\boxed{y-y_0=-(1)/(f'(x_0))(x-x_0)}`

Calculate
`f'(x_0)`:


`f'(x)=\left((20-x)/(3x)\right)'=\left((20)/(3x)\right)'-\left((x)/(3x)\right)'=(20)/(3)\cdot\left((1)/(x)\right)'-\left((1)/(3)\right)'=\\\\\\=(20)/(3)\cdot\left(-(1)/(x^2)\right)-0=\boxed{-(20)/(3x^2)}\\\\\\\\f'(x_0)=f'(2)=-(20)/(3\cdot2^2)=-(20)/(3\cdot4)=-(20)/(12)=\boxed{-(5)/(3)}`

and the equation of the normal line:


`y-y_0=-(1)/(f'(x_0))(x-x_0)\\\\\\y-3=-(1)/(-(5)/(3))(x-2)\\\\\\y-3=(3)/(5)(x-2)\\\\\\y-3=(3)/(5)x-(6)/(5)\\\\\\y=(3)/(5)x-(6)/(5)+3\\\\\\\boxed{y=(3)/(5)x+(9)/(5)}`