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December 18, 2014

December 18, 2014

Posted by **Anonymous** on Saturday, May 26, 2012 at 3:07pm.

(4x sqrt)(1 - t^2)(dx/dt) - 1 = 0 and x(0)=-2

I believe that the answer is one of the following two options:

a.) 2x^2 = arcsint + 8

b.) arccost + 8 - (1/2)pi

- CALCULUS -
**Damon**, Saturday, May 26, 2012 at 3:41pm4 x dx = dt/sqrt (1-t^2)

2 x^2 = sin^-1 (t) + c

when t = 0, x = 2

8 = 0 + c

c = 8

so 2 x2 = sin^-1 (t) + 8

or

2 x^2 = cos^-1(t) + c

8 = pi/2 + c

yes, agree

- CALCULUS -
**Anonymous**, Saturday, May 26, 2012 at 5:42pmI'm only permitted to choose one option. And seeing as you came up with arcsin first, would this be a better answer over the option with arccos?

- CALCULUS -
**Damon**, Saturday, May 26, 2012 at 6:51pmThey are the same.

cos (pi/2 - x) = sin x

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