Sinusoidal Functions: Max/Min Volts in 1s w/ t-Values

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Homework Help Overview

The problem involves analyzing a sinusoidal function representing voltage over time, specifically V(t) = 110sin(5t) + 15. The goal is to determine the maximum and minimum voltage values within the first second and identify the corresponding times.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the periodic nature of the function and how to find the maximum and minimum points based on the derivative. Some participants clarify the relationship between the cycles and the derivative, while others question the meaning of the variable "n" in the context of finding zeros of the cosine function.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the periodicity of the function and the method to find critical points. Guidance has been offered regarding the zeros of the cosine function and the role of the integer "n" in identifying specific points of interest.

Contextual Notes

Participants are working under the constraint of finding solutions only within the first second, which limits the values of "n" that can be considered.

Draggu
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Homework Statement



The voltage, V(t), in volts, of a power supply can be modeled by the function V(t) = 110sin5t+15, where t is the time, in seconds. Find the maximum and minimum voltages, within the first second, and the times they occur.


Homework Equations





The Attempt at a Solution



well, i think since the period is 5? that means there are 5 cycles, which means there will be 5 maximums, and 5 minimums, how do i figure out where the exact points are maximum?

v'(t)=550cos5t

but how do i solve t there?
 
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You are correct to think that there will be 5 cycles. However, that is 5 cycles per 2[tex]\pi[/tex] radians. In order to find the zeros for your derivative, think about when does Cos(5t) = 0? It is equal to 0 whenever the argument inside cosine is [tex]\frac{\pi }{2} + n\pi[/tex]. Thus you can solve for [tex]5t = \frac{\pi }{2} + n\pi \\[/tex] with n = 0, 1, 2, 3... but remember, only with t<1 second.
 
What do you mean with "n"?
 
Draggu said:
What do you mean with "n"?

n is an integer of 0,1,2,3... etc etc. Remember, cosines and sines go on forever and they have an infinite number of zeros. The n tells you which zero you are at. For example, the n = 0 zero is at 5t = [tex]\frac{\pi }{2}[/tex]
 

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