Mathematics 1

Root mean squared (RMS) average value

• For the function f(t) the RMS average over the interval from t = a to t = b is defined by
• RMS=dt……..(1)
• Usually used for oscillating quantities like voltage and current. For such functions, the RMS value is usually defined over a complete cycle.
• Example: Find the R.M.S. value of y = sin (3πt).
• Solution:
• We haven’t been given any limits for the integration so we need to calculate the period.
• Period=
• where ω is the angular velocity which is the number multiplying t in our function, so 3π in this case. Therefore,
• Period==
• Substituting in (1) what we know (y = sin (3πt), a = 0 and b = ) we have;
• R.M.S. = dt)=dt)=dt

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Root mean squared (RMS) average value –continuation of example

R.M.S. = dt=dt)=dt

To integrate ) use the formula for 2:

sin2(3πt) = (1 − cos(2 × 3πt)) = (1 − cos (6πt)).

Then R.M.S.: :dt=dt=== =)=

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Curve Length

• We want to determine the length of the continuous function on the interval
• Assume that the derivative is also

continuous on [a, b].

• Divide the interval up into n equal

subintervals each of width and

denote the point on the curve at

each point by Pi. We can then

approximate the curve by a series

of straight lines connecting the points.

• L ≈ where the length of each of the line segments. Assuming that
• We have:

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Curve Length -continuation

• Using linear approximation:
• f(xi) -f(xi-1) ≈ f′(xi-1 )(xi −xi-1). we obtain:
• = =
• The exact length of the curve is then,
• L==
• Using the definition of the definite integral we get:
•  L=
• In other notation we get the following:

•  

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Curve Length - Example 1.58.

• Find the length of the curve
• f(x) = x between x = 0 and
• x = 1.
• For y=x we have the curve
• length = 1.414
• It’s clear from simple