Asked by khela
find the points to the curve
y=(cos x)/ (2 + sin x)
at which the tangent is horizontal...
can you explain how get the answer thoroughly and step by step...
y=(cos x)/ (2 + sin x)
at which the tangent is horizontal...
can you explain how get the answer thoroughly and step by step...
Answers
Answered by
Reiny
using the quotient rule,
dydx = [(2+sinx)(-sinx) - cosx(cosx)]/(2+sinx)^2
a horizontal tangent implies the slope is zero, that is, the derivative is zero.
(2+sinx)(-sinx) - cosx(cosx) = 0
-2sinx - sin^2x - cos^2x = 0
-2sinx - (sin^2x + cos^2x) = 0
-2sinx - 1 = 0
sinx = -1/2
x = 7π/6 or x = 11π/6 ..... (210° or 330°)
sub 7π/6 and 11π/6 into the original equation to find the corresponding y values.
Those will give you the points in the domain 0 ≤ x ≤ 2π
dydx = [(2+sinx)(-sinx) - cosx(cosx)]/(2+sinx)^2
a horizontal tangent implies the slope is zero, that is, the derivative is zero.
(2+sinx)(-sinx) - cosx(cosx) = 0
-2sinx - sin^2x - cos^2x = 0
-2sinx - (sin^2x + cos^2x) = 0
-2sinx - 1 = 0
sinx = -1/2
x = 7π/6 or x = 11π/6 ..... (210° or 330°)
sub 7π/6 and 11π/6 into the original equation to find the corresponding y values.
Those will give you the points in the domain 0 ≤ x ≤ 2π
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