Question

How do you find equations for the lines that are tangent and normal to the graph of `y=secx` at `x=pi/4`?

Answer 1

The corresponding y coordinate is `y = sec(pi/4) = 1/cos(pi/4) = sqrt(2)`

We now find the derivative.

`y = secx`

`y= 1/cosx`

`y' = (0 xx cosx - (-sinx xx 1))/(cosx)^2`

`y' = (sinx)/(cos^2x)`

`y' = tanxsecx`

We can now find the slope of the tangent.

`m_"tangent" = tan(pi/4)sec(pi/4)`

`m_"tangent" = 1(sqrt(2))`

`m_"tangent" =sqrt(2)`

The equation is therefore:

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

`y- sqrt(2) = sqrt(2)(x- pi/4)`

`y - sqrt(2) = sqrt(2)x - (sqrt(2)pi)/4`

`y= sqrt(2)x - (sqrt(2)pi)/4 + sqrt(2)`

`y= sqrt(2x) - (sqrt(2)pi + 4sqrt(2))/4`

`y = sqrt(2x) - (sqrt(2)(pi + 4))/4`

Hopefully this helps!