Arc Length of Sine Curve

Answer 1

I have a problem with the equation y = a*sin(b*x), when the particle is going at v m/s. If you think about it, the faster the particle goes, the shorter time it will take to complete one period of the sine wave. However, whoever wrote the problem did not tell you what depended on the velocity, and if it changed the amplitude as well as the period. For now, let's assume that the only reaction for a change in velocity is the period. Well, right now, b is the only thing that is affecting the period of the sine wave, so we'll leave it alone. So the period of the sine wave right now is (2*pi)/b. As velocity increases, the period decreases, as mentioned earlier. So we can say that it is a direct relationship within the function, and our new period, dependent on velocity, is (2*pi)/(b*v). So our new equation is y = a*sin(b*v*x).

Now we've settled that problem, let's focus on what this function is telling us. It is merely describing the behavior of the particle - it's pattern. To determine a displacement versus time function, we need to figure the particles behavior with respect to some origin, which we can say reasonably is at the point (0, 0). So the displacement of the particle from the origin can be calculated with two points, right? (distance formula)

Our two points are (0, 0), and (x, a*sin(b*v*x)). Now with the distance formula:

sqrt((x2-x1)^2 +(y2-y1)^2)

We obtain:

sqrt((x^2) + (a^2)(sin^2(b*v*x))). This is the object's displacement from the origin with respect to x. To turn this into a displacement vs. time fuction, (which is what you want to determine the x and y coordinates as a function of time) you can do a simple substitution: replace x with t, time.

We can do this because the location of the particle, or point, depends on x. Well, in a real life situation the location depends on time. So the x coordinate is the same as time in graphing.

Our final equation yields:

Y = sqrt((t^2) + (a^2)(sin^2(b*v*t))).

So any point you can describe as (t, sqrt((t^2) + (a^2)(sin^2(b*v*t))))

 

There was absolutely now calculus involved with this question, as many would foolishly assume - Gettin the arc length with respect to time would only give you the distance traveled - and that is NOT the same thing as x and y coordinates, which thankfully is what you needed.