anonymous
  • anonymous
Find an equation of the tangent line to the curve y=arccos(x/2) at the point (1,4pi)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
Any ideas?
anonymous
  • anonymous
well, the point (1,4pi) is not on the line

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More answers

anonymous
  • anonymous
Oh sorry, it was y=3arccos(x/2)
amistre64
  • amistre64
the slope of the line at any given point is the derivative of the function
amistre64
  • amistre64
is the equation spose to be in point slope or slope intercept?
anonymous
  • anonymous
either way, the point is not on the line, the derivative of y=3arccos(x/2) is y'=-3/(sqrt(4-x^2)) subbing x=1 we get the slope we get the slope of -sqrt(3)
anonymous
  • anonymous
so it narrows down to 2 answers
amistre64
  • amistre64
cos(60)=1/2 so (1,pi/3) is on it i believe
anonymous
  • anonymous
-sqrt(3)=(y-1,x-4pi)
anonymous
  • anonymous
oops, i mean -sqrt(3)=(y-1/x-4pi)
anonymous
  • anonymous
oops, i mean -sqrt(3)=(y-1)/(x-4pi)
anonymous
  • anonymous
So the answer would be?
anonymous
  • anonymous
funny thing is none of them match, so i must have made a mistake somewhere
anonymous
  • anonymous
Got any ideas amistre?
anonymous
  • anonymous
i got it
anonymous
  • anonymous
-sqrt(3)=(y-4pi)/(x-1
anonymous
  • anonymous
mixed up the coordinates
anonymous
  • anonymous
answer is the 3rd one
anonymous
  • anonymous
-sqrt(3)=(y-4pi)/(x-1) isolate y and you'll get the answer
anonymous
  • anonymous
Thanks man, I never would have got that one, I guess I need more studying :P
anonymous
  • anonymous
no prob, by math skills are getting rusty

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