2a is correct
the period of sin^3(x) is the same as for sin(x). SO, sin^3(x/3) has period 2pi/(1/3) = 6pi
sure. sin(x) does just that. sin(pi)=0, but it takes 2pi to complete the period.
in polar coordinates, sin(theta/2) goes through the origin twice in its period of 4pi.
visit wolframalpha.com and play around with this stuff. It should help.
use the "plot" or "polar plot" command with the functions
1) The period of a trig. function
y=sin kx is 2pi/k. Then period of y=sin^2(pi.x/a) should be 2pi/(pi/a)=2a, but somewhere it is given as a. Which is correct?
2) The period of r=sin^3(theta/3) is given as 3pi. How is it worked out? Is it because after theta=0, the function becomes 0 only at theta=3pi (when it is sin pi=0)?
3) Is it possible for a curve to cross origin (i.e. function value y or r to be 0)more than once before completion of its entire length/cycle?
4 answers
For calculating complete length of curve r=sin^3(theta/3) I got right answer on integrating the arclength integrand from 0 to 3pi. How could that be?
No idea. Did you visit wolframalpha.com and look at the graph?
If you enter
arc length sin^3(x/3), x = 0 .. 3pi
it will give you the arc length, but that's only for 1/2 period.
If you enter
arc length sin^3(x/3), x = 0 .. 3pi
it will give you the arc length, but that's only for 1/2 period.
I did plot the curve on Wolfram and got full curve across 3pi. Curve length is also shown full at 3pi but period is shown as 6pi.I can't understand this.
1 answer