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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TonyHem on April 09, 2009, 09:58:51 pm

Title: Anti-Dif inverse circ functions
Post by: TonyHem on April 09, 2009, 09:58:51 pm: Yeah, just started this today so I suck.
How do I do:

"A curve has a gradient given by $\frac{dy}{dx}=2+\frac{1}{\sqrt{1-x^2}}$ and its graph passes through the origin. Find the equation of the curve"

What I thought was x=0 since it passes through the origin, so dy/dx = 3. but it's wrong so yeah :{
Title: Re: Anti-Dif inverse circ functions
Post by: pHysiX on April 09, 2009, 10:06:20 pm: They are asking for the original function, not the tangent to the curve or the gradient, which was what u did ;)
So do:

if dy/dx = that thing then
y=antidiff of 2 + antidiff of (1/SQRT(1-x^2)) +C

*I'm assuming that u can antidiff the first part. The second part: we know that the derivative of Sin^-1(x) is exactly "that", which implies the antidiff of "that" must be Sin^-1(x) =]

Since it passes the origin:

2x + Sin^-1(x) + C = y

y(0) ==> 0=C

Therefore:

y= 2x + Sin^-1(x)
Title: Re: Anti-Dif inverse circ functions
Post by: TonyHem on April 09, 2009, 10:09:26 pm: zzzz -_-"
I get it now, thanks.