Question 1189998
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The current i (in amperes) at time t in a particular circuit is given by
i =12sin t + 5cost.
Find the maximum current and the first time that it occurs.
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<pre>
Doing it in the Calculus frame, take the derivative  {{{(di)/(dt)}}};  equate it to zero and find the time


The derivative is  {{{(di)/(dt)}}} = 12cos(t) - 5sin(t).


The equation  {{{(di)/(dt)}}} = 0  is  12cos(t) - 5sin(t) = 0,  or  12cos(t) = 5sin(t),  which gives

    {{{sin(t)/cos(t)}}} = {{{12/5}}},  or  tan(t) = {{{12/5}}},


Hence,  t = arctan(12/5) = 1.176 units of time.   It is the first occurrence time of the maximum.     <U>ANSWER</U>



           Since everything happens in the 1st quadrant, 
       where both sin(t) and cos(t) are positive, it is clear
          that the found extremum is indeed the maximum
       (not a minimum), and a special check is not needed.



Next, from tan(t) = {{{12/5}}},  we have  sin(t) = {{{12/sqrt(12^2+5^2)}}} = {{{12/13}}};

                                   cos(t) = {{{5/13}}}.


Substituting these values for sin(t) and cos(t) into the formula for "i", we obtain the value of  {{{i[max]}}}


    {{{i[max]}}} = {{{12*(12/13)}}} + {{{5*(5/13)}}} = {{{(144+25)/13}}} = {{{169/13}}} = 13 amperes.    <U>ANSWER</U>.
</pre>

Solved.