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You are here: Experts > Teens > Homework/Study Tips > Calculus > Calculus
Expert: Ahmed Salami - 10/31/2009
Question
Find the equation to the line tangent to the curve
y = 3arccos(x/2) at the point (1, pi).
Answer
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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