What is the local linearization of #y = sin^-1x # at a=1/4?

Question

William Abdalla · Accepted Answer

#y=241/256x-1009/4096#

We first have to find in #1/4# the limited development of #Arcsin(x)#

So, we have to find #LD_1(1/4)# #sin^(-1)(x)#

Let #X=x-1/4#

We have to find #LD(0)# #sin^(-1)(X+1/4)#

also, #f'(X)=1/sqrt(1-(X+1/4)^2)=1/sqrt(1-X^2-1/2X-1/16)# #=1/sqrt(15/16-(X^2+1/2X))=16/15*1/sqrt(1-15/16(X^2+1/2X))#

Let #Y=-15/16(X^2+1/2X)#

We have #f'(Y)=16/15*1/sqrt(1+Y)=16/15*(1+Y)^(-1/2)#

And we know a limited development of #(1+Y)^n#

We will only keep term of degree 0 and 1.

#LD_1(0)# #(1+Y)^n=1+nY+o(Y)#

Let replace #Y# by #-15/16(X^2+1/2X)# and #n=-1/2#

#LD_1(0)# #f'(X)=1+15/32(color(red)(cancel(color(black)(X^2)))+1/2X)+o(X)#

#=1+15/64X#

And #LD_(n+1)(0)# #f(X)=f(0)+intLD_n(0)f(X)dX + o(X^(n+1))#

#<=>LD_2(0)sin^(-1)(X)=color(red)(cancel(color(black)(sin^(-1)(0))))^(=0)+int(1+15/64X)dX+o(X^2)#

#LD_2(0)sin^(-1)(X)=X+15/128X^2+o(X^2)#

Then, finally, replace #X=x-1/4#, knowing that we only want a #LD_1(1/4)f(x)#

#LD_2(1/4)sin^(-1)(x)=x-1/4+15/128(x-1/4)^2+o((x-1/4)^2)# #LD_1(1/4)sin^(-1)(x)=x-1/4+color(red)(cancel(color(black)(15/128x^2)))-15/256x+15/4096+o(x)#

#=241/256x-1009/4096#

So, the local linearization of sin^(-1)(x) in #a=1/4# is

#y=241/256x-1009/4096#

Christopher Allers · Answer

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

1. Find the derivative of $ y = \sin^{-1}(x) $ using the chain rule: $ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} $.
2. Evaluate the derivative at $ x = \frac{1}{4} $ to find the slope of the tangent line: $ \frac{1}{\sqrt{1 - (\frac{1}{4})^2}} = \frac{4}{\sqrt{15}} $.
3. Use the point-slope form of a line to find the equation of the tangent line: $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = \left(\frac{1}{4}, \sin^{-1}\left(\frac{1}{4}\right)\right) $.
4. Substitute the values into the equation to find the local linearization.

The local linearization of $ y = \sin^{-1}(x) $ at $ a = \frac{1}{4} $ is given by:

$$ y - \sin^{-1}\left(\frac{1}{4}\right) = \frac{4}{\sqrt{15}}\left(x - \frac{1}{4}\right) $$