I have a calculus question:
The voltage signal from a standard North
American wall socket can be described by the
equation V(t) = 170sin(120πt), where t is time,
in seconds, and V(t) is the voltage, in volts, at
time t.
a) Find the maximum and minimum voltage
levels, and the times at which they occur.
Basically I understand that my maximum is 170 Volts, my minimum is -170 Volts but it is asking for the exact times at which they occur. So my answer turns out to be :
t = {t | t = (4k+1)/240, k >= 0, k ∈ Z} <- Max
t = {t | t = (4k+1)/240, k >= 0, k ∈ Z} <- Min
I am VERY confused as to how those above intervals came to be and if for example the question was a cosine function instead of sine, would it be different?
Thank you very much everyone!
Best Answer
You are correct to recognize that the amplitude is 170 so the maximum will be 170 and the minimum will be -170.
$\sin(t)$ has a period of $2\pi$ and a maximum at every $\frac{\pi}{2}+2\pi *k$ and a minimum at every $\frac{3\pi}{2}+2\pi*k$ where $k$ is a constant.
The major difference between the functions is the period. In your function, $V(t)=170\sin(120\pi t)$, the coefficient of $t$ is $120\pi$, which means that the function travels $120\pi$ times as fast, or completes a period $120\pi$ times faster than usual. Since $\sin(t)$ has a period of $2\pi$, then $V(t)$ will make one period in $\frac{2\pi}{120\pi}$ or every $\frac{1}{60}$.
Recall that $\sin(t)$ has a maximum every $\frac{1}{4}$ of the way along a period. This means that the first maximum will be reached at $\frac{1}{4}*\frac{1}{60}=\frac{1}{240}$. Since the function has a period of $\frac{1}{60}$, then you can continuously add $\frac{1}{60}$ to $\frac{1}{240}$ to obtain more maxima, which would then mean maximums are when:
$$t=\frac{4k+1}{240}$$
where $k$ is an integer greater than or equal to $0$.
Since a minimum occurs every $\frac{3}{4}$ of a period, using similar logic we find that the minima are at:
$$t=\frac{4k+3}{240}$$
again where $k$ is an integer greater than or equal to $0$.