Ask
your own question, for FREE!
Mathematics
8 Online
Find the arc length function for the curve with starting point (0,1).
Still Need Help?
Join the QuestionCove community and study together with friends!
What curve?
\[y=\sin^{-1} x + \sqrt{1-x^2}\]
\[s(x)=\int_{0}^{x} \sqrt{1+\left ( \frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right )^2}\ dx \\ =\int_{0}^{x} \sqrt{1+\left ( \frac{1-x}{\sqrt{1-x^2}}\right )^2}\ dx\\ =\int_{0}^{x} \sqrt{1+\frac{(1-x)^2}{1-x^2}}\ dx\\ =\int_0^{x} \sqrt{1+\frac{1-x}{1+x}}\ dx\\ =\int_0^{x} \sqrt{\frac{2}{1+x}}\ dx\] to be continued....
\[s(x)=\sqrt{2}\int_0^x (1+x)^{-1/2}\ dx\\ =2\sqrt{2} \sqrt{1+x}|_0^x\\ =2\sqrt2(\sqrt{1+x}-1)\] Whew!
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
VERNEISH:
Two objects are moving along separate linear paths where each path is described by position, d, and time, t.
Maci:
what are the best stock apps for people who want to start investing?
Twaylor:
genuienly how do i get better self control i keep watcing youtube and theres this
IzzyCookie:
|dw:1792074056138:dw| Uhm, I don't know where to ask this, but I'm in college welding at my community college, and I need help learning vertical welding for
thehearken:
not sure if this is considered spam, im sorry if it is, but does anyone have tips
21 hours ago
0 Replies
0 Medals
4 days ago
3 Replies
2 Medals
1 week ago
3 Replies
0 Medals
2 days ago
9 Replies
1 Medal
1 week ago
2 Replies
1 Medal