gwiyomi17 Posted June 15, 2013 Share Posted June 15, 2013 During high tide the water depth in a harbour is 22 m and during low tide it is 10m(Assume a 12h cycle).Calculate the times at which the water level is at 18m during the first 24 hours.My solution:I found the cos equation: H(t)=-6cos(π/6t)+16then..Let π/6t=Θ18=-6cosΘ+1618-16=-6cosΘΘ=1.230959417Let Θ=π/6t π/6t=1.230959417Then I don't know what's next.... Link to comment Share on other sites More sharing options...

Amaton Posted June 15, 2013 Share Posted June 15, 2013 You're assuming that the tide is at low when the 24 hours starts. Did the problem specify what the water level should be at [math]t=0[/math]? Link to comment Share on other sites More sharing options...

gwiyomi17 Posted June 15, 2013 Author Share Posted June 15, 2013 You're assuming that the tide is at low when the 24 hours starts. Did the problem specify what the water level should be at [math]t=0[/math]? it is also in low tide ... Link to comment Share on other sites More sharing options...

Amaton Posted June 15, 2013 Share Posted June 15, 2013 Alright -- your function [math]H(t)=-6\cos(\frac{\pi}{6} t)+16[/math] (where [math]t[/math] is in hours) seems like a good model. Mathematically speaking, we'll want to restrict the domain to [math]0\le t \le 24[/math] since we're only dealing with the first 24 hours of tide. If you don't make the restriction, then solving for [math]t[/math] will give solutions outside the time frame the problem wants. We'll keep this in mind throughout the problem. You started right by substituting [math]H(t)=18[/math]. So now we have [math]18=-6\cos(\frac{\pi}{6} t)+16[/math]. You can make another substitution if you want, but that might complicate the problem. We're just trying to find solutions for [math]t[/math], by getting it by itself. At least try to get it down to where we have the [math]\cos(\frac{\pi}{6} t)[/math] part alone on one side. See if you can continue from there. Link to comment Share on other sites More sharing options...

gwiyomi17 Posted June 15, 2013 Author Share Posted June 15, 2013 Alright -- your function [math]H(t)=-6\cos(\frac{\pi}{6} t)+16[/math] (where [math]t[/math] is in hours) seems like a good model. Mathematically speaking, we'll want to restrict the domain to [math]0\le t \le 24[/math] since we're only dealing with the first 24 hours of tide. If you don't make the restriction, then solving for [math]t[/math] will give solutions outside the time frame the problem wants. We'll keep this in mind throughout the problem. You started right by substituting [math]H(t)=18[/math]. So now we have [math]18=-6\cos(\frac{\pi}{6} t)+16[/math]. You can make another substitution if you want, but that might complicate the problem. We're just trying to find solutions for [math]t[/math], by getting it by itself. At least try to get it down to where we have the [math]\cos(\frac{\pi}{6} t)[/math] part alone on one side. See if you can continue from there. Thats what I did (above), I calculated for the "t" but it says on the problem "Calculate the times at which the water level is at 18m during the first 24 hours." so it means that there's more solution... but i don't know how to find the others. Link to comment Share on other sites More sharing options...

Amaton Posted June 15, 2013 Share Posted June 15, 2013 (edited) Thats what I did (above), I calculated for the "t" but it says on the problem "Calculate the times at which the water level is at 18m during the first 24 hours." so it means that there's more solution... but i don't know how to find the others. There are 4 solutions. Look at a graph of the function. Where the line crosses our cosine function is where the solutions are. Once you find the first, call it [math]t_a[/math], you can find the other three. Your calculation seems correct. I'll give you the second solution: [math]12-t_a[/math]. Do you understand how this works out? EDIT: Better image. Edited June 15, 2013 by Amaton Link to comment Share on other sites More sharing options...

gwiyomi17 Posted June 15, 2013 Author Share Posted June 15, 2013 (edited) There are 4 solutions. Look at a graph of the function. Capture.PNG Where the line crosses our cosine function is where the solutions are. Once you find the first, call it [math]t_a[/math], you can find the other three. Your calculation seems correct. I'll give you the second solution: [math]12-t_a[/math]. Do you understand how this works out? EDIT: Better image. so the next solution would be [math]12+t_a[/math] and [math]24-t_a[/math]?????? then i'll just substitute... thanks! Edited June 15, 2013 by gwiyomi17 Link to comment Share on other sites More sharing options...

daniton Posted June 16, 2013 Share Posted June 16, 2013 Sorry for the interruption but once you find the value of the angle in one full cycle (12hr.) the rest can be found just by adding the period(12hr.) since you're asked for 24hrs. Link to comment Share on other sites More sharing options...

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