Zeros of sinusoidal graph.

[for]

Hi,

I have the following equation:
\(\displaystyle f(t)=50-42sin(\frac{\pi}{2}t)\)

This is a model of a ferris wheel. What I need to figure out is at what times during the first four minutes you will be 65 feet above the ground.

This is what I have so far:
\(\displaystyle t=\frac{arcsin(-\frac{15}{42})}{\frac{\pi}{2}}\)

This is where I get stuck.
I'm going to go ahead and say that someone else showed me this:
\(\displaystyle t=\frac{2[arcsin(-\frac{15}{42})+2\pi]}{\pi}\)
This gives me one of the zeros, and this gives me the other:
\(\displaystyle t=\frac{2[\pi-arcsin(-\frac{15}{42})]}{\pi}\)

I'm having some trouble understanding how he devised this formula.
By looking at a graph, this function is only in quadrant I, and the answer I get is in QII. The thing I'm having trouble with is: How do I move to the next point in QI where the height(y) = 65? There are two I need to find. I understand everything else, and I am by no means asking for the answer, just how to find it. Thanks.

 

[soroban]

Hello, for!

I have the following equation: .\(\displaystyle f(t)\:=\:50-42\sin\left(\tfrac{\pi}{2}t\right)\)

This is a model of a ferris wheel.

At what times during the first four minutes will you be 65 feet above the ground?

\(\displaystyle \text{This is what I have so far: }\;t\:=\:\frac{\arcsin(-\frac{15}{42})}{\frac{\pi}{2}}\)

\(\displaystyle \text{So we get: }\:\arcsin(-\tfrac{15}{42}) \:=\:-0.365207221\text{ radians}\)


\(\displaystyle \text{The first two positive angles are: }\begin{Bmatrix}2.776385432 \\ 3.506799875\end{Bmatrix}\:\text{ radians.}\)


\(\displaystyle \text{The first two times are: }\:t \;=\;\begin{Bmatrix}\frac{2}{\pi}(2.776385432) & \approx & 1.77 \\ \\[-3mm] \frac{2}{\pi}(3.506799875) & \approx & 2.34\end{Bmatrix}\:\text{ minutes.}\)

 

[for]

Hello, for!

I have the following equation: .\(\displaystyle f(t)\:=\:50-42\sin\left(\tfrac{\pi}{2}t\right)\)

This is a model of a ferris wheel.

At what times during the first four minutes will you be 65 feet above the ground?

\(\displaystyle \text{This is what I have so far: }\;t\:=\:\frac{\arcsin(-\frac{15}{42})}{\frac{\pi}{2}}\)

\(\displaystyle \text{So we get: }\:\arcsin(-\tfrac{15}{42}) \:=\:-0.365207221\text{ radians}\)


\(\displaystyle \text{The first two positive angles are: }\begin{Bmatrix}2.776385432 \\ 3.506799875\end{Bmatrix}\:\text{ radians.}\)


\(\displaystyle \text{The first two times are: }\:t \;=\;\begin{Bmatrix}\frac{2}{\pi}(2.776385432) & \approx & 1.77 \\ \\[-3mm] \frac{2}{\pi}(3.506799875) & \approx & 2.34\end{Bmatrix}\:\text{ minutes.}\)

Thanks.. The problem I was having was HOW do I arrive at the "first two positive" angles? Algebraically I mean, what steps are taken?

 

[mmm4444bot]

\(\displaystyle t=\frac{arcsin(-\frac{15}{42})}{\frac{\pi}{2}}\)

That's good; we can simplify.

t = -2 arcsin(5/14)/Pi


By looking at a graph, this function is only in quadrant I, and the answer I get is in QII.

That's okay. The Sine function is odd. You can use symmetry, identities, geometry (many ways) to find the two solutions for t in QI.

What your source did is good in theory, but poor in execution.

They used 2 Pi for the period of function f; that's incorrect.

The period is 4.

So, subtract t from 2, and add t to 4, instead.

t_1 = -2 arcsin(5/14)/Pi + 4

That's about 2.2325

t_2 = 2 + 2 arcsin(5/14)/Pi

That's about 3.7675

Cheers ~ Mark

 

[mmm4444bot]

\(\displaystyle \text{The first two times are: }\:t \;=\;\begin{Bmatrix}\frac{2}{\pi}(2.776385432) & \approx & 1.77 \\ \\[-3mm] \frac{2}{\pi}(3.506799875) & \approx & 2.34\end{Bmatrix}\:\text{ minutes.}\)

I don't think so.

f(1.77) = 35.2

f(2.34) = 71.4

(rounded)

MY EDIT: Fixed typograpical error on rounded value

 

[mmm4444bot]

How do I move to the next point in QI where the height(y) = 300?

f(t) never reaches 300.

The height ranges from 8 through 92.

 

[mmm4444bot]

\(\displaystyle f(x)=50-42sin(\frac{\pi}{2}t)\)

Be careful.

f(x) is a typographical error.

The given function is f(t).

 

[for]

Re:

How do I move to the next point in QI where the height(y) = 300?

f(t) never reaches 300.

The height ranges from 8 through 92.

Ugh, sorry about that. I have two separate problems, both at different heights.
What I meant was 65 feet.

 

[mmm4444bot]

So, subtract t from 2, and add t to 4, instead.

t_1 = -2 arcsin(5/14)/Pi + 4

That's about 2.2325

t_2 = 2 + 2 arcsin(5/14)/Pi

That's about 3.7675

Can you explain to me what is meant by "add t to 4" ?

I'll try.

f(t) is a sine function with period 4.

You found a solution for t, and when you use a scientific calculator to evaluate it you obtain -0.2325, yes?

There are two locations in each period of the given function where f(t) is 65.

-0.2325 is the second location within the period [-4, 0].

Because of the periodicity of the sine function, adding the period to any solution gives another solution.

I mean, adding 4 to the solution that you already obtained yields the second location where f(t) = 65 within the next period [0,4].

That is one of the two answers you seek.

You have a graph to look at, yes? Plus, we have this identity for the sine function:

sin(x) = sin(x + 2 Pi)

In your exercise, the identity applied looks like this:

f(-0.2325) = f(-0.2325 + 4)

The period in function f is not 2 Pi; it's 4.

Your source shows adding 2 Pi to your result for t. They should have added 4.

I mean, look at it.

If you can't figure this out, let me know, and I'll try adding a sketch.

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ f(t) \ = \ 50-42sin\bigg(\frac{\pi t}{2}\bigg), \ note \ period \ = \ 4 \ and \ f(0) \ = \ f(4) \ = \ 50.\)

\(\displaystyle Find \ t, \ (0 \ < \ t \ < \ 4) \ when \ f(t) \ = \ 65.\)

\(\displaystyle Hence \ 65 \ = \ 50-42sin\bigg(\frac{\pi t}{2}\bigg), \ \implies \ t \ = \ \frac{2arcsin(-5/14)}{\pi} \ \dot= \ -.232498138085.\)

\(\displaystyle Now \ for \ 0 \ < \ t \ < \ 4,\)

\(\displaystyle t \ = \ -.232498138085+4(period) \ \dot= \ 3.7675\)

\(\displaystyle and \ t \ = \ .232498138085+2(half \ a \ period) \ \dot= \ 2.2325.\)

\(\displaystyle Check: \ f(3.7675) \ = \ f(2.2325) \ = \ 65, \ see \ graph.\)

[attachment=0:23s5owtn]eee.jpg[/attachment:23s5owtn]

 

[for]

Excellent, I fully understand this now. Thanks a bunch to everyone who helped me out