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anonymous

  • 5 years ago

how do you solve y = 3/ sqrt(16-x^2) ?

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  1. anonymous
    • 5 years ago
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    \[y=\frac{3}{\sqrt{16+x^2}}\] what are you solving for?

  2. anonymous
    • 5 years ago
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    The area of the region between 0 and 2.

  3. anonymous
    • 5 years ago
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    you are looking for \[\int^2_0 \frac{3}{\sqrt{16-x^2}}dx \]?

  4. anonymous
    • 5 years ago
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    yes.

  5. anonymous
    • 5 years ago
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    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

  6. amistre64
    • 5 years ago
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    x = 4sin(t)

  7. amistre64
    • 5 years ago
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    dx = 4cos(t) dt

  8. anonymous
    • 5 years ago
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    antiderivative of \[\frac{1}{\sqrt{a^2-x^2}}\] is \[sin^{-1}(\frac{x}{a})\]

  9. amistre64
    • 5 years ago
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    3*4 cos(t) [S] --------- dt right? 4 cos(t)

  10. anonymous
    • 5 years ago
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    \[1/a *\arcsin(x/a) ?\]

  11. amistre64
    • 5 years ago
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    [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 :)

  12. amistre64
    • 5 years ago
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    3 * sin^-1(x/4)

  13. anonymous
    • 5 years ago
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    I pulled out the three and had (1/4)arcsin(x/4) evaluated from 0 to 2

  14. anonymous
    • 5 years ago
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    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 \[\frac{\pi}{6}\]

  15. anonymous
    • 5 years ago
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    Did you multiply by the 3?

  16. anonymous
    • 5 years ago
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    oops forgot about the ''3" answer is \[\frac{\pi}{2}\] sorry

  17. anonymous
    • 5 years ago
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    So when I get a decimal, how can I revert it make to degrees/radians so that the computer will except it?

  18. amistre64
    • 5 years ago
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    \[frac{test}{ing}\]

  19. amistre64
    • 5 years ago
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    lol... that didnt work out like i thought

  20. anonymous
    • 5 years ago
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    what's that?

  21. amistre64
    • 5 years ago
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    pi 180 -- * ----- = 90 degrees... 2 pi

  22. anonymous
    • 5 years ago
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    pi/2 is right

  23. anonymous
    • 5 years ago
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    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.

  24. anonymous
    • 5 years ago
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    okay

  25. anonymous
    • 5 years ago
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    Thanks everyone for your help!

  26. amistre64
    • 5 years ago
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    youre welcome :)

  27. anonymous
    • 5 years ago
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    I hope this is clear. the area is \[\frac{\pi}{2}\] square units.

  28. anonymous
    • 5 years ago
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    Yes, and your answer was right.

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