Question

an object hangs from a string in a stable position and is represented by the function y=4sin(3x)+4, with time in x seconds. what is the maximum height of the object, and when does it occur on the interval 0

Answer 1

Here we have the position of an object represented by the equation:

y = 4*sin(3*x) + 4

Where we can assume that y represents the height of the object.

We want to find the maximum height of the object, and for which value of x it happens in the interval 0 ≤ x ≤ 5

First let's find the maximum height.

y = 4*sin(3*x) + 4

Here the only variation is on the sin( ) part, we know that the maximum value of the sin(x) function is:

sin(x) = 1

when x = 90° = pi/2 = 3.14/2

Replacing that in the equation, we get:

y = 4*1 + 4 = 8

The maximum height of the object is 8.

Now, we know that the maximum of the sin(x) function is when x = pi/2

But in our case, we have the function sin(3*x)

Then this function is maximized when:

3*x = pi/2

x = pi/(2*3) = pi/6 = 3.14/6

And:

0 < 3.14/6 < 5

Then the first maximum occurs at x = 3.14/6 = 0.52

But it can happen again on the interval.

The next maximum of the sin(x) function is when:

x = pi/2 + 2*pi

In our case, we again have sin(3*x), then we need to solve:

3*x = pi/2 + 2*pi

x = (pi/2 + 2*pi)/3 = (3.14/2 + 2*3.14)/3 = 2.62

This is also in the interval 0 < x < 5

The next maximum of sin(x) happens at:

x = pi/2 + 2*pi + 2*pi

In our case, for sin(3*x) we need to have:

3*x = (pi/2 + 2*pi + 2*pi)

x = (3.14/2 + 2*3.14 + 2*3.14)/2 = 4.71

This again is in the interval 0 < x < 5

(is closer to the upper value, so we can expect that the next maximum does not belong to the interval)

Then the maximum height occurs at:

x = 0.52

Then occurs again at:

x = 2.62

And finally occurs again at:

x = 4.71