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 April 18th, 2014, 06:13 PM #1 Newbie   Joined: Apr 2014 From: florida Posts: 3 Thanks: 0 How to do sinusoid problem?? At 2:00 PM, on April 6, Olive Oyl says to Popeye, "Oh dear, the tide is completely in and the water is 15 meters deep." "Yes, Olive." Popeye replies At 8:00 PM on the same day, Olive says to Popeye, "Look, dear, the tide is completely out and the water is only 11 meters deep." "Yes, Olive" Popeye yawns "Popeye, Popeye, let's write the equation that describes this sinusoidal motion of the tides!" "Yes, Olive, do that. Start the timing back at noon today, OK?" Popeye says, knowing that choosing this time will keep her occupied longer than if he had chosen 2:00 PM. A short while later, Olive Oyl whispers to Popeye, "I've got it, Popeye." Popeye grimaces and hurriedly scribbles several questions on the back of a spinach label, hands them to Olive, and goes back to sleep. 1. Sketch the graph 2. What was Olive's equation? 3. What will be the depth of the water tonight at 10:00 PM? 4. What will b the depth of the water tomorrow morning at 7:00 AM? 5. When is the first low tide tomorrow? 6. What's the earliest time on April 7th that the water will be 12.7 meters deep? Last edited by skipjack; June 19th, 2014 at 05:20 AM.
 April 18th, 2014, 06:15 PM #2 Newbie   Joined: Apr 2014 From: florida Posts: 3 Thanks: 0 I'll pay someone to do it, just pm me. Honestly, I'm desperate. Last edited by skipjack; June 19th, 2014 at 05:29 AM.
April 19th, 2014, 07:07 AM   #3
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From the given information, $(2,15), \;\ (8,11)$

Use these two points to find a sine wave equation that models the tide.

Use $y=a\cdot \sin(bx+c)+d$.

Knowing what the various pieces of this represent allows you to pick at each piece.

'a' is the amplitude. It is determined by $\frac{\text{high}-\text{low}}{2}$

or $a=\frac{15-11}{2}=2$

Let's say we go over a 24 hour period from noon one day until noon the next.

From the given info, the graph crosses the midline at 5 hours and 11 hours.

Since the graph of the 'regular' sine function has high and low points of -2 and 2, this means we have to shift the graph up to 11 and 15.
Thus $d=13$

So, we have this much so far:

$y=2\sin(bx+c)+13$

For $(2,15)$:

$15=2\sin(2b+c)+13$

For $(8,11)$:

$y=2\sin(8b+c)+13$

Solving these two equations gives:

$b=\frac{-\pi}{6}$ and $c=\frac{5\pi}{6}$

Thus, we have:

$y=2\sin(\frac{-\pi}{6}t+\frac{5\pi}{6})+13$

Here's the graph to see if the values given work out. They appear to.

Apparently, at noon when the graph begins, the tide is 14 meters.
Attached Images
 tide.jpg (13.2 KB, 0 views)

 April 19th, 2014, 01:37 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 The tide varies from the mean depth of 13m by 2cos((t - 2)π/6)m, where t is the number of hours from noon (and π is pi). Assuming you can graph that, does it help?

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