in a sine wave: v=100sin(200π*t+π/4) 10.7) Find the value of the greatest voltage rate of changee – you have already found when this is and this value of t can be used in the expression for the rate of change that you have also found (t=-0.00125 or 0.00125) 10.8) Using integral calculus, calculate the RMS value of the voltage

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in a sine wave: v=100sin(200π*t+π/4) 10.7) Find the value of the greatest voltage rate of changee – you have already found when this is and this value of t can be used in the expression for the rate of change that you have also found (t=-0.00125 or 0.00125) 10.8) Using integral calculus, calculate the RMS value of the voltage

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equation edited...big hands and little keyboards don't mix well :-(
hint 200(.00125)=.25=1/4 \[-.00125\leq t \leq 0.00125\\-.25\leq 200t \leq .25\\-\frac{\pi}{4}\leq 200t\pi \leq \frac{\pi}{4}\\0\leq 200t\pi +\frac{\pi}{4}\leq \frac{\pi}{2}\]
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can you go on ?
I thought that the greatest voltage rate of change is simply where the line crosses the x-axis and is found by 2π*frequency*amplitude = The bit that's throwing me is having to use differential calculus to find it. Derivative of 100sin(200πt+4/π) = 20,000πcos(200πt+π/4). 2π*frequency*amplitude = 62831.8 but 20,000πcos(200πt+π/4) = 62808.2 surely these should equal the same? Also just realised that my original question didn't mention using differential calculus. Sorry about that.
10. 7) If you already know at what 't' the greatest rate of change occurs , then what I would do is take the derivative of the function and plug in +t. 10.8) Equation for RMS using an integral X_rms = \[\sqrt{1/T \int\limits_{0}^{T}(x)^2dt}\]

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