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Question
Find the equation to the line tangent to the curve
y = 3arccos(x/2) at the point (1, pi).
Hi Chenne,
y = 3arccos(x/2)
dy/dx = 3.(1/2).-1/√[1 - (x/2)²]
= -3 / 2√(1 - x²/4)
at x = 1
dy/dx = -3 / 2√(1 - 1/4)
= -3 / 2√(3/4)
= -3 / √3
= -√3
This is the slope of the line. As the line passes through (1,π), its equation is
(y-π)/(x-1) = -√3
y-π = -√3(x-1)
y-π = -x√3 + √3
y = -x√3 + (√3 + π)
Regards
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Ahmed Salami
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I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.
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An engineering graduate. I have been doing maths and physics all my life.
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