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

Question

Christopher Agresta · Accepted Answer

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

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:

Charles Anctil · Answer

undefined To find the linearization of $ y = \sin^{-1}(x) $ at $ x = \frac{1}{4} $, follow these steps:

1. Find the derivative of $ y = \sin^{-1}(x) $, which is $ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} $.

2. Evaluate the derivative at $ x = \frac{1}{4} $ to get $ \frac{dy}{dx} $ at that point.

3. Substitute $ x = \frac{1}{4} $ into the original function to find $ y $ at that point, which is $ y = \sin^{-1}\left(\frac{1}{4}\right) $.

4. Use the formula for linearization: $ L(x) = f(a) + f'(a)(x-a) $, where $ f(a) $ is the function value at the point of tangency, $ f'(a) $ is the derivative at that point, and $ a $ is the point of tangency.

5. Substitute the values you found into the linearization formula to get the linearization of $ y = \sin^{-1}(x) $ at $ x = \frac{1}{4} $.