Calc Help plz explain it as well?

Question

An electric current, I, in amps, is given by
I=cos(6 t)+sqrt(6) sin(6 t)
What is the maximum value of I?

Answer 1

The maximum/minimum value of any function/equation can be found by making the derivative of that function equal to zero, and then solving. This is because the derivative is more or less the slope of the function/equation, and when the slope is equal to zero, you have a minimum of maximum value.

With Trig/Algebra you solve:

I' = -6sin(6t) + 6sqrt(6)cos(6t)
-6sin(6t) + 6sqrt(6)cos(6t) = 0
sqrt(6)cos(6t)=sin(6t)
tan(6t)=sqrt(6)
tan-1(sqrt(6))=6t
t= tan-1(sqrt(6))/6
t= 0.197 Radians

Answer 2

You want to maximize the I function. To find max/min, take the derivative of I.

Set I'=0 and solve for t. All values of t that work are inflection points.
Then, find I". Plug in the value(s) of t you got. Whenever I" is negative you have a max, and whenever it's positive, you have a min value for I.


This is how you approach ALL max/min problems. I hope that made sense.