Question

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

Answer 1

\[y=\frac{3}{\sqrt{16+x^2}}\] what are you solving for?

Answer 2

The area of the region between 0 and 2.

Answer 3

you are looking for
\[\int^2_0 \frac{3}{\sqrt{16-x^2}}dx \]?

Answer 4

yes.

Answer 5

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

Answer 6

x = 4sin(t)

Answer 7

dx = 4cos(t) dt

Answer 8

antiderivative of \[\frac{1}{\sqrt{a^2-x^2}}\] is \[sin^{-1}(\frac{x}{a})\]

Answer 9

3*4 cos(t)
[S] --------- dt right?
4 cos(t)

Answer 10

\[1/a *\arcsin(x/a) ?\]

Answer 11

[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 :)

Answer 12

3 * sin^-1(x/4)

Answer 13

I pulled out the three and had (1/4)arcsin(x/4) evaluated from 0 to 2

Answer 14

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 \[\frac{\pi}{6}\]

Answer 15

Did you multiply by the 3?

Answer 16

oops forgot about the ''3"
answer is \[\frac{\pi}{2}\] sorry

Answer 17

So when I get a decimal, how can I revert it make to degrees/radians so that the computer will except it?

Answer 18

\[frac{test}{ing}\]

Answer 19

lol... that didnt work out like i thought

Answer 20

what's that?

Answer 21

pi 180
-- * ----- = 90 degrees...
2 pi

Answer 22

pi/2 is right

Answer 23

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.

Answer 24

okay

Answer 25

Thanks everyone for your help!

Answer 26

youre welcome :)

Answer 27

I hope this is clear. the area is \[\frac{\pi}{2}\] square units.

Answer 28

Yes, and your answer was right.