Question

How do you find the linearization of `y = sin^-1x` at `x=1/4`?

Answer 1

`y = 4/sqrt(15)x+sin^-1 (1/4)-1/sqrt(15)`

Explanation:

The linearization at any particular point is simply given by the equation of the tangent at that point.

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is `−1`).

We have:

`y=sin^-1x`

When `x=1/4=> y=sin^-1 (1/4)`

Differentiating wrt `x` we have (standard derivative):

`dy/dx = 1/sqrt(1-x^2)`

The slope of the tangent line at `P` is given by:

`m_T = 1/sqrt(1-(1/4)^2)`
`\ \ \ \ \ \ = 1/sqrt(1-1/16)`
`\ \ \ \ \ \ = 1/sqrt(15/16)`
`\ \ \ \ \ \ = 4/sqrt(15)`

So, the equation of the normal using the point/slope form `y-y_1=m(x-x_1)` is;

`\ \ \ \ \ y - sin^-1 (1/4) = 4/sqrt(15)(x-1/4)`
`:. y - sin^-1 (1/4) = 4/sqrt(15)x-1/sqrt(15)`

`:. y = 4/sqrt(15)x+sin^-1 (1/4)-1/sqrt(15)`

And the graph of the function and its approximation are as follows: