Question

Find the equation of the tangent and normal to the curve y=cosx at x=π/3?

Answer 1

Tangent's equation is `y=-sqrt3/2x+(pi)/(2sqrt3)+1/2`

Normal's equation is `y=2/sqrt3x-(2pi)/(3sqrt3)+1/2`

Explanation:

First, you find the first derivative of the function then substitute with the given point `(x_1,y_1)` to get the slope`(m)` and then substitute in this function

`color(green)((y-y_1)=m(x-x_1)`

apply for the given function

`y=cosx`

At `x=pi/3` `rarr` `y=cos(pi/3)=1/2`

`(x_1,y_1)=` `(pi/3,1/2)`

`dy/dx=-sinx`

Substitute with the point `(pi/3,1/2)`

`dy/dx=m_1=-sin(pi/3)=-sqrt3/2`

Now Substitute in the formula in green with `(x_1,y_1),m_1`
to get the equation of the tangent to the curve at the given point

`y-1/2=-sqrt3/2(x-pi/3)`

`y=-sqrt3/2x+(pi)/(2sqrt3)+1/2`

the normal to the curve would be at the same point but its slope is different and in order to get its slope `m_2`

We use this

`m_1*m_2=-1`

`m_2=2/sqrt3`

Now Substitute with `(x_1,y_1),m_2` in the equation in green to find the equation of the normal

`y-1/2=2/sqrt3(x-pi/3)`

`y=2/sqrt3x-(2pi)/(3sqrt3)+1/2`