Calculating Arc Length for a Curved Function with a Starting Point of (0,1)

[ineedhelpnow]

find the arc length for the curve $y=sin^{-1}+\sqrt{1-x^2}$ with starting point (0,1).

$y'=\frac{1-x}{\sqrt{1-x^2}}$

$\int_{0}^{x} \ \sqrt{1+(\frac{1-x}{\sqrt{1-x^2}})^2},dx$

my answer is $\frac{-2\sqrt{2}*(\sqrt{\frac{1}{x+1}}-1)}{\sqrt{\frac{1}{x+1}}}$

i think my answer is wrong though

 

[MarkFL]

I have moved your new question to a new thread. Please don't tag new questions onto existing threads. This makes threads harder to follow. :D

 

[MarkFL]

find the arc length for the curve $y=sin^{-1}+\sqrt{1-x^2}$ with starting point (0,1)...

What is the argument for the inverse sine function?

You go from an integral to a final answer...in order to more easily determine if you are correct and ensure you are doing things correctly, we need to see the intermediary steps. :D

 

[ineedhelpnow]

i only put it under the other thread because i didnt want to start too many new threads and they were the exact same topic. we're supposed to use our calculator for the question so i just put it in there. the answer in the back of the book is $2\sqrt{2}(\sqrt{1+x}-1)$

 

[MarkFL]

i only put it under the other thread because i didnt want to start too many new threads and they were the exact same topic. we're supposed to use our calculator for the question so i just put it in there. the answer in the back of the book is $2\sqrt{2}(\sqrt{1+x}-1)$

It is preferable to have more threads that deal with fewer problems per thread than fewer threads with many problems. The threads are easier to follow and allow for more efficient searching.

The answer you gave is equivalent to the more simplified answer given by your book. Take your answer and multiply by:

\(\displaystyle 1=\frac{\sqrt{1+x}}{\sqrt{1+x}}\)

and then distribute the negative sign out front into the factor in parentheses. :D

 

[ineedhelpnow]

oooh that makes sense.