Question #262851

  1. Calculate the turning points of the function y=sin3t using differential calculus
  2. Show which are maxima, minima or points of inflexion using the second derivative

Expert's answer

1. y=sin⁡(3t)y=\sin(3t)

Domain: (−∞,∞)(-\infin, \infin)

Find the first derivative


y′=(sin⁡(3t))′=3cos⁡(3t)y'=(\sin(3t))'=3\cos(3t)

Find the critical number(s)


y′=0=>3cos⁡(3t)=0y'=0=>3\cos(3t)=0

3t=π2+πn,n∈Z3t=\dfrac{\pi}{2}+\pi n, n\in \Z

t=π6+πn3,n∈Zt=\dfrac{\pi}{6}+\dfrac{\pi n}{3}, n\in \Z

y(π6+2πm3,m∈Z)=1y(\dfrac{\pi}{6}+\dfrac{2\pi m}{3}, m\in \Z)=1

y(π2+2πk3,k∈Z)=−1y(\dfrac{\pi}{2}+\dfrac{2\pi k}{3}, k\in \Z)=-1

The turning points are

(π6+2πm3,1),(π2+2πk3,−1),m,k∈Z\big(\dfrac{\pi}{6}+\dfrac{2\pi m}{3}, 1\big), \big(\dfrac{\pi}{2}+\dfrac{2\pi k}{3},-1\big), m,k\in \Z

2.

Find the second derivative


y′′=(3cos⁡(3t))′=−9sin⁡(3t)y''=(3\cos(3t))'=-9\sin(3t)

y′′(π6+2πm3)=−9<0y''(\dfrac{\pi}{6}+\dfrac{2\pi m}{3})=-9<0

y′′(π2+2πm3)=9>0y''(\dfrac{\pi}{2}+\dfrac{2\pi m}{3})=9>0

The points (π6+2πm3,1),m∈Z\big(\dfrac{\pi}{6}+\dfrac{2\pi m}{3}, 1\big), m\in \Z are maxima.


The points (π2+2πk3,1),k∈Z\big(\dfrac{\pi}{2}+\dfrac{2\pi k}{3}, 1\big), k\in \Z are minima.



y′′=0=>−9sin⁡(3t)=0y''=0=>-9\sin(3t)=0

3t=πl,l∈Z3t=\pi l, l\in \Z

t=πl3,l∈Zt=\dfrac{\pi l}{3}, l\in \Z

y(πl3)=0,l∈Zy(\dfrac{\pi l}{3})=0, l\in \Z

The points (πl3,0),l∈Z\big(\dfrac{\pi l}{3}, 0\big), l\in \Z are points of inflection.



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