Find and classify the first four stationary points for t ≥ 0 of the function: f(t) = sin(c1*t)*e^(0.1*t), where c1 = 1

Hi all,

I need to calculate the first four stationary points of the function:

f(t) = sin(c1*t)*e^(0.1*t)

for when t ≥ 0

and c1 = 1 (calculated from previous data)

I know I could differentiate and find the roots but is there another way of finding the four points a quicker/easier/another way on matlab?

IF i do have to differentiate and find roots what's the best way about doing this on matlab?

any help would be much appreciated.

Thanks

Answer 1

[Roger Stafford]

   t = mod(-atan2(c1,0.1),pi)/c1+(0:3)*pi/c1;  % <-- Corrected

[Liam Nolan]

thank you, I may have to explain how/why this was used. I can see how it's differential (let =0) could rearranged to -atan(10)=t and I partially understand the mod function (from research) but have never used it to calculate something like this. could you explain how/why this is used please?

[Roger Stafford]

Well, I confess I did what you stated you didn't want done. I differentiated f(t) and got:

   f'(t) = c1*cos(c1*t)*exp(0.1*t)+0.1*sin(c1*t)*exp(0.1*t)

This will be zero whenever c1*cos(c1*t)+0.1*sin(c1*t) is equal to zero. This is a standard problem in trigonometry. You divide by the square root of the sum of the squares of the two coefficients and you have the equality

   sin(theta)*cos(c1*t)+cos(theta)*sin(c1*t) = sin(theta+c1*t) = 0

where

   theta = atan2(c1,0.1)

Since the sine function is zero for all integer multiples of pi, you get

   theta+c1*t = atan2(c1,0.1)+c1*t = n*pi

or

   t = -atan2(c1,0.1)/c1-n*pi/c1

You stated that you wanted the first four roots of t for t>=0, so I used the 'mod' function to accomplish that:

   t = mod(-atan2(c1,.1),pi)/c1+n*pi/c1;

where n = (0:3).