Question

I need help with problem set 5, 2C-1d.
I think my answer is right but I don't know how to simplify it so it looks like the solution. Then again, my answer could be completely wrong.

Answer 1

find dw for \[w = \sin^{-1}( u/t)\]
bearing in mind that if \[f(x) = \sin^{-1} (x)\] then \[f'(x) = 1/\sqrt{1 - x^2}\]
and the product rule, then my solution is
\[1/\sqrt{1-(u/t)^2}.du/t + 1/\sqrt{1-(u/t)^2}.u.dt\] which simplifies to
\[ (du/t + u.dt)/\sqrt{1-(u/t)^2}\]
The answer provided is
\[(tdu - udt)/t.\sqrt{t^2-u^2}\]
Where am I going wrong?

Answer 2

\[w = \sin ^{-1}(u/t)\]
\[\sin (w)=(u/t)\]
\[\cos (w)dw=\frac{1}{t}du-\frac{u}{t^2}dt\]
since \[\cos (w)=\sqrt{1-\sin ^2(w)}\]
and \[\sin ^2(w)=\frac{u^2}{t^2}\]
then \[\cos (w)=\sqrt{1-\frac{u^2}{t^2}}\]
now, substitute that into the above equation and you get
\[\ \sqrt{1-\frac{u^2}{t^2}} dw=\frac{1}{t}du-\frac{u}{t^2}dt\]
\[dw=\frac{(\frac{1}{t}du-\frac{u}{t^2}dt)}{\sqrt{1-\frac{u^2}{t^2}}}\]
\[dw=\frac{(\frac{t}{t^2}du-\frac{u}{t^2}dt)}{\sqrt{1-\frac{u^2}{t^2}}}\]
\[dw=\frac{(tdu-udt)}{t^2\sqrt{1-\frac{u^2}{t^2}}}\]
\[dw=\frac{(tdu-udt)}{t\sqrt{t^2-u^2}}\]