# Thread: Find the times when you're 65 feet off the ground.

1. ## Find the times when you're 65 feet off the ground.

So, there's a Ferris wheel.
I have to find out the times when you are 65 feet off the ground within the first four minutes, with t being the time in minutes, with this equation:
$\displaystyle h(t)=50-42sin(\frac{\pi}{2}t)$

I also had to solve it graphically, which was easy, so I know there are two points in time at which you are 65 feet off the ground(something like ~2.23 and ~3.77 minutes). What I'm having trouble with is solving it algebraically.

This is my thought process:
$\displaystyle 50-42sin(\frac{\pi}{2}t) = 65$
$\displaystyle -42sin(\frac{\pi}{2}t) = 15$
$\displaystyle sin(\frac{\pi}{2}t) = -\frac{15}{42}$
$\displaystyle arcsin(sin(\frac{\pi}{2}t)) = arcsin(-\frac{15}{42})$
$\displaystyle \frac{\pi}{2}t = arcsin(-\frac{15}{42})$
$\displaystyle t = \frac{arcsin(-\frac{15}{42})}{\frac{\pi}{2}}$

I'm pretty sure I'm doing this right.. but when I get to the last step, it gives me something like t = -.2324981381

I've plugged that in and it is obviously not the right answer. It gives me 65 as it should, but it doesn't correspond to the answers I get when solving graphically. Any help with what I'm doing wrong?

2. Originally Posted by qleeq

I'm pretty sure I'm doing this right.. but when I get to the last step, it gives me something like t = -.2324981381
You might want to check if your calculator is in the right mode? degree/radians. Also could this be an answer just in the worng quadrant?

3. Originally Posted by pickslides
You might want to check if your calculator is in the right mode? degree/radians. Also could this be an answer just in the worng quadrant?

That is the graph of the function. the straight line is y = 65
the curve is h(t) = 50-42sin((pi/2)t)

scale is y [0,100,10] and x [0,4,1]
as you can see, it crosses in both places. my calculator is in radians mode. the graph is only in the first quadrant

4. Something I figured out..
This number: t = -.2324981381
is significant to the answer somehow.. when finding the places where y = 65 on the graph, one of them is positive 2.2324981

edit: okay WOW, the t=-.232(...) is an answer in the second quadrant for x = -.232(...)
how do I find the other values of x?

5. $\displaystyle \arcsin\left(-\frac{15}{42}\right)$ is in quads III and IV ...

when the point is in quad III ...

$\displaystyle t = \frac{2[\pi - \arcsin\left(-\frac{15}{42}\right)]}{\pi} =\approx 2.23 \, s$

$\displaystyle t = \frac{2[\arcsin\left(-\frac{15}{42}\right) + 2\pi]}{\pi} \approx 3.77 \, s$

6. Originally Posted by skeeter
$\displaystyle \arcsin\left(-\frac{15}{42}\right)$ is in quads III and IV ...

when the point is in quad III ...

$\displaystyle t = \frac{2[\pi - \arcsin\left(-\frac{15}{42}\right)]}{\pi} =\approx 2.23 \, s$

$\displaystyle t = \frac{2[\arcsin\left(-\frac{15}{42}\right) + 2\pi]}{\pi} \approx 3.77 \, s$
thanks for that, but could you explain to me where you did $\displaystyle \pi - \arcsin\left(-\frac{15}{42}$ and the one with +2pi? I don't quite understand that..

7. ## Trigonometric functions with multiple zeros

I've been posting stuff on modelling temperatures lately, and I understand it now. I have a test coming up on Tuesday, and I need a little extra help with something.

I'm given a function, and I can set it equally to zero and solve it no problem. Then there is another zero that I need to find, and I'm having a hard time doing it. This has seemed to slip my mind.

Can anybody explain to me in a few steps how I can go beyond finding the first zero of a sinusoidal graph? Algebraically of course.

Say I have something like: $\displaystyle 50-42sin(\frac{\pi}{6}t)$ and I need to find values of t where y = 65.

I can solve this by setting the equation equal to 65, but after this I get stuck. Any help please?

8. Originally Posted by qleeq
I've been posting stuff on modelling temperatures lately, and I understand it now. I have a test coming up on Tuesday, and I need a little extra help with something.

I'm given a function, and I can set it equally to zero and solve it no problem. Then there is another zero that I need to find, and I'm having a hard time doing it. This has seemed to slip my mind.

Can anybody explain to me in a few steps how I can go beyond finding the first zero of a sinusoidal graph? Algebraically of course.

Say I have something like: $\displaystyle 50-42sin(\frac{\pi}{6}t)$ and I need to find values of t where y = 65.

I can solve this by setting the equation equal to 65, but after this I get stuck. Any help please?
I suggest that you draw the graphs of $\displaystyle \displaystyle y = 50 - 42 \sin \left(\frac{\pi}{6}t\right)$ and y = 65 on the one set of axes to see what's happening. After doing this, it should be more apparent what to do.

The values of t you want (suggested by a unit circle) satisfy $\displaystyle \displaystyle \frac{\pi}{6} t = -\sin^{-1} \left(\frac{5}{13}\right) + 2n \pi$ and $\displaystyle \displaystyle \frac{\pi}{6} t = \pi + \sin^{-1} \left(\frac{5}{13}\right)+ 2 n \pi$.

9. Originally Posted by mr fantastic
I suggest that you draw the graphs of $\displaystyle \displaystyle y = 50 - 42 \sin \left(\frac{\pi}{6}t\right)$ and y = 65 on the one set of axes to see what's happening. After doing this, it should be more apparent what to do.
Yes, I've done that, and I know that for the first four values of x there are two zeros. Someone showed me in another thread but I didn't understand what he was doing.

10. Originally Posted by qleeq
Yes, I've done that, and I know that for the first four values of x there are two zeros. Someone showed me in another thread but I didn't understand what he was doing.
Did you draw the unit circle like I said to do in my previous reply? I'm not sure what else can be said until you have put more thought into what I posted.