anonymous
  • anonymous
how do you solve y = 3/ sqrt(16-x^2) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[y=\frac{3}{\sqrt{16+x^2}}\] what are you solving for?
anonymous
  • anonymous
The area of the region between 0 and 2.
anonymous
  • anonymous
you are looking for \[\int^2_0 \frac{3}{\sqrt{16-x^2}}dx \]?

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anonymous
  • anonymous
yes.
anonymous
  • anonymous
I know that it has to do with the sine identity, but I keep getting the wrong answer using that. My answer is never an angle that is correct
amistre64
  • amistre64
x = 4sin(t)
amistre64
  • amistre64
dx = 4cos(t) dt
anonymous
  • anonymous
antiderivative of \[\frac{1}{\sqrt{a^2-x^2}}\] is \[sin^{-1}(\frac{x}{a})\]
amistre64
  • amistre64
3*4 cos(t) [S] --------- dt right? 4 cos(t)
anonymous
  • anonymous
\[1/a *\arcsin(x/a) ?\]
amistre64
  • amistre64
[S] 3 dt -> 3t ... what does t revert back to tho? x = 4sin(t) x/4 = sin(t) sin^-1(x/4) = t if i did it right :)
amistre64
  • amistre64
3 * sin^-1(x/4)
anonymous
  • anonymous
I pulled out the three and had (1/4)arcsin(x/4) evaluated from 0 to 2
anonymous
  • anonymous
evaluate at upper limit, get \[sin^{-1}(\frac{2}{4})=sin^{-1}(\frac{1}{2})=\frac{\pi}{6}\] plug in 0 get 0 so answer is \[\frac{\pi}{6}\]
anonymous
  • anonymous
Did you multiply by the 3?
anonymous
  • anonymous
oops forgot about the ''3" answer is \[\frac{\pi}{2}\] sorry
anonymous
  • anonymous
So when I get a decimal, how can I revert it make to degrees/radians so that the computer will except it?
amistre64
  • amistre64
\[frac{test}{ing}\]
amistre64
  • amistre64
lol... that didnt work out like i thought
anonymous
  • anonymous
what's that?
amistre64
  • amistre64
pi 180 -- * ----- = 90 degrees... 2 pi
anonymous
  • anonymous
pi/2 is right
anonymous
  • anonymous
the area is the area. As a function this is a function of numbers, not degrees or radians. It corresponds to the radian measure of the angle, not the degree measure. Always work in 'radians' if you are thinking of angles. But they are just numbers.
anonymous
  • anonymous
okay
anonymous
  • anonymous
Thanks everyone for your help!
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
I hope this is clear. the area is \[\frac{\pi}{2}\] square units.
anonymous
  • anonymous
Yes, and your answer was right.

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