Video Transcript
In a circuit of an alternating current, the current 𝐼, measured in amperes, at any moment 𝑡, measured in seconds, is given by the relation 𝐼 is equal to 16 times the sin of 𝑡 plus 19 times the cos of 𝑡. What is the maximum value of the current in this circuit rounded to two decimal places?
We’re asked to find the maximum value of an alternating current given its expression as a function of time, that is, 𝐼 is equal to 16 times the sin of 𝑡 plus 19 times the cos of 𝑡. To do this, we must find the critical points of the current 𝐼 by taking the first derivative, setting it equal to zero, and solving for 𝑡. We can then determine the nature of the critical points using the second derivative test. Using the known results that the derivative with respect to 𝑥 of sin 𝑥 is the cos of 𝑥 and the derivative with respect to 𝑥 of the cos of 𝑥 is equal to the negative sin of 𝑥, we have the first derivative of 𝐼 with respect to 𝑡 is 16 times the cos of 𝑡 minus 19 times the sin of 𝑡.
And setting this equal to zero, we can rearrange to solve for 𝑡 by first adding 19 times the sin of 𝑡 to both sides. And if we then divide through by 19 times the cos of 𝑡, on our left-hand side, dividing through by the cos of 𝑡 gives us 16 over 19. And on our right-hand side, dividing top and bottom by 19 gives us the sin of 𝑡 divided by the cos of 𝑡. And recalling that the tangent is equal to the sine over the cosine, this gives us 16 over 19 is equal to the tan of 𝑡. If we then take the inverse tangent on both sides, then from the general solution for integers 𝑛, we have the inverse tan of 16 over 19 plus 𝑛𝜋 is equal to 𝑡. That is, 𝑡 is equal to 0.6998 and so on plus 𝑛𝜋.
Now, since the trigonometric functions are periodic, we only need to consider the critical points for 𝑛 is equal to zero and 𝑛 is equal to one. That’s because the current 𝐼 will alternate between the values at these critical points, the maximum and the minimum. Our two critical points are therefore 𝑡 is equal to 0.6998 and so on for 𝑛 equal to zero and 𝑡 is 3.8414 and so on for 𝑛 equal to one.
And now, making some space, we can determine the nature of these critical points using the second derivative test. This tells us that for a twice differentiable function 𝑓 with stationary or critical point at 𝑥 is equal to 𝑥 sub zero, if the second derivative of 𝑓 with respect to 𝑥 at 𝑥 is equal to 𝑥 zero is positive, that’s greater than zero, then 𝑥 sub zero is a local minimum. If the second derivative of 𝑓 with respect to 𝑥 at 𝑥 is equal to 𝑥 sub zero is less than zero, then 𝑥 sub zero is a local maximum. And if the second derivative at 𝑥 is equal to 𝑥 sub zero is equal to zero, then the test is inconclusive.
So now using our known results once more for the derivatives of sine and cosine, we find the second derivative of our current function 𝐼 of 𝑡 is negative 16 times the sin of 𝑡 minus 19 times the cos of 𝑡. So now, substituting our first critical point, 𝑡 is equal to 0.6998 and so on, into our second derivative, we have negative 16 times the sin of 0.6998 and so on minus 19 times the cos of 0.6998 and so on. And this evaluates to negative 24.8394 and so on. And now, doing the same for our second critical point, that’s 𝑡 is equal to 3.8414 and so on, we find the second derivative of 𝐼 with respect to 𝑡 evaluated at 𝑡 is equal to 3.8414 and so on is 24.8394 and so on.
And so we see that the second derivative at our first critical point, 0.6998 and so on, is less than zero. And from our second derivative test, we see that this means that the critical point is a local maximum. And similarly, the second derivative evaluated at our second critical point, that’s 𝑡 is 3.8414 and so on, is greater than zero. And this means our second critical point is a local minimum.
So now making some space, we have for 𝑡 equal to 0.6998 and so on, our current 𝐼 has a local maximum. That is for 𝑛 even in our general solution 𝑡 is 0.6998 and so on plus 𝑛𝜋. And for 𝑡 equal to 3.8414 and so on, our current 𝐼 has a local minimum. And this applies to odd values of 𝑛 in our general solution 𝑡 is equal to 0.6998 and so on plus 𝑛𝜋. Since we’re asked to find the maximum value of the current, we can substitute our value 𝑡 is equal to 0.6998 and so on into our current 𝐼. That is, 𝐼, the current at the maximum value of 𝑡, is equal to 0.6998 and so on is equal to 24.8394 and so on.
We therefore have that the maximum value of the current in this circuit rounded to two decimal places is 24.84 amperes.
