Derivatives of Trigonometric Function

Question

Find the coordinates of the two turning points on the curve y=sin³ x cos x for 0<x<pi, Determine which points gives a maximum value of y.

Answer 1

Using the product rule, y' = 3sin²xcos²x + sin³x(-sinx) or 3sin²xcos²x - sin⁴x This is the first derivative.

 

Factor, and set the derivative  = 0 for extrema. So sin²x(3cos²x - sin²x) = 0

 

So sin x = 0, and x = 0 or 3cos ²x = sin²x, so tan²x = 3, so tan x =± √3, and so x = Π/3 and 2Π/3 (tan x = sinx/cosx)

 

Now take the second derivative. I get (after cleaning it up): y'' = 6sinxcos³x - 10sin³xcosx

 

y''(0) = 0, because sin0 =0  So we may have an inflection point here, rather than a max, min. Since it said there were only 2 turning points, I will ignore this one, and assume it's not a max or min.

 

y''(Π/3) = 6(√3/2)(1/2)³ - 10(√3/2)³(1/2)  = 0.65 - 3.25, which is <0, so we assume a maximum here

 

y'' (2Π/3) = 6(√3/2)(-1/2)³  - 10(√3/2)³(-1/2) = -0.65 + 3.25, which is >0, so we assume a minimum here.

 

Now we plug in the x values to find y: y = sin³(Π/3)cos(Π/3) = 0.325, and y = sin³(2Π/3)cos(2Π/3) = -0.325

 

Final answer: MAX AT (Π/3, 0.325), MIN AT (2Π/3, -0.325)