# Thread: Word Problem - Trig

1. ## Word Problem - Trig

Alright, so I'm having trouble with this question, I thought I solved it, but apparently I havent so I was wondering if anyone could tell me where I went wrong.

The average depth of water in a bay is 4m. At low tide the level is 2m. One cycle is completed every 12h and low tide last occurred at 15:00h. Calculate the next time the height of the tide will be 5.0m.

Here's what I got for the equation...
Amplitude = -2
Period = pi/6
Phase Shift = -15
Vertical Displacement = 4

therfore the equation is

When I solve for t I get 14.046h which when subbed back doesn't equal 5.

If someone can point out where I went wrong I'd appreciate it.

2. ## hi

Hi

Here is what I did, this might be wrong though.

I used the sin function instead.

$y(t) = 2 \cdot sin(\frac{\pi t}{6} - \frac{\pi}{2}) + 4$

Using amplitude 2 , and phase change so that $y(0) = 2$
which is what was described. I am here using time 15:00 as 'starting point'.

Now we need to solve: $2 \cdot sin(\frac{\pi t}{6} - \frac{\pi}{2}) + 4 = 5$

We end up at $t = 4 + 12k \; , k\in \mathbb{Z}$ , but obviously we can only have positive valus on k. $k = 0$ gives $t = 4$ , and y(4) = 5

So 19:00

3. Originally Posted by peekay
Alright, so I'm having trouble with this question, I thought I solved it, but apparently I havent so I was wondering if anyone could tell me where I went wrong.

The average depth of water in a bay is 4m. At low tide the level is 2m. One cycle is completed every 12h and low tide last occurred at 15:00h. Calculate the next time the height of the tide will be 5.0m.

Here's what I got for the equation...
Amplitude = -2
Period = pi/6
Phase Shift = -15
Vertical Displacement = 4

therfore the equation is

the phase shift of "15" is not needed. let t = 0 be 3:00 PM

When I solve for t I get 14.046h which when subbed back doesn't equal 5.

If someone can point out where I went wrong I'd appreciate it.
$h = -2\cos\left(\frac{\pi t}{6}\right) + 4$

$5 = -2\cos\left(\frac{\pi t}{6}\right) + 4$

$-\frac{1}{2} = \cos\left(\frac{\pi t}{6}\right)$

$\frac{\pi t}{6} = \frac{2\pi}{3}$

$t = 4$ or 7:00 PM

4. Why is the 15 not needed? Having a hard time understanding that

Can I put it in there or is that whats messing everything up? Cause doesn't it force the wave to have a minimum at 15:00h?

5. Originally Posted by peekay
Why is the 15 not needed? Having a hard time understanding that

Can I put it in there or is that whats messing everything up?
you can set t = 0 to be anytime you wish.

6. So if I set t=0 to be 15:00h (as in it starts at 15:00h) would my function be correct?

7. if you insist ...

$h = -2\cos\left[\frac{\pi}{6}\left(t-15\right)\right] + 4$

$5 = -2\cos\left[\frac{\pi}{6}\left(t-15\right)\right] + 4$

$-\frac{1}{2} = \cos\left[\frac{\pi}{6}\left(t-15\right)\right]$

$\frac{2\pi}{3} = \frac{\pi}{6}\left(t-15\right)$

$4 = t - 15$

$19 = t$

8. Thanks twigs and skeeter, really appreciate the help.

The answer turned out to be 19h. I did solving for t a little bit differently by letting pi/6[(t-15)] equal theta, but it turned out to be the same answer.

Thanks again for the help guys!