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anonymous

  • one year ago

The depth of water in a tank oscillates every 5 hours. The smallest depth is 0.5, and that the largest depth is 7.5. Find a formula for the depth as a function of time t measured in hours for which the depth at time t=0 is the largest depth.

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  1. anonymous
    • one year ago
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    Sorry, I have to log off now. Good luck.

  2. anonymous
    • one year ago
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    @zepdrix Can you help? :)

  3. zepdrix
    • one year ago
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    |dw:1443488320174:dw|So we've got our low and our high point.

  4. anonymous
    • one year ago
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    So the amplitude is 7.5-.5/2 which is 3.5?

  5. zepdrix
    • one year ago
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    |dw:1443488404092:dw|We start at our highest point. And you've calculated the amplitude correctly, ok good!

  6. zepdrix
    • one year ago
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    Where is the middle of our function going to be located?

  7. anonymous
    • one year ago
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    Not sure?

  8. zepdrix
    • one year ago
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    |dw:1443488549410:dw|

  9. anonymous
    • one year ago
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    3.75?

  10. zepdrix
    • one year ago
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    3.5 + 0.5 Hmm that doesn't sound right :d

  11. anonymous
    • one year ago
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    Ohhhh oops I see now, 4

  12. zepdrix
    • one year ago
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    |dw:1443488682271:dw|Ok great. And it completes one full oscillation in 5 hours. So the shape should look something like this, ya?

  13. anonymous
    • one year ago
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    Beautiful!

  14. zepdrix
    • one year ago
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    So should we model this curve with `sine` or `cosine`? What do you think? :) One will be a bit easier than the other. You have to think about your starting point.

  15. anonymous
    • one year ago
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    This is where I always have a problem. How exactly do you decide whether to use sine or cosine?

  16. zepdrix
    • one year ago
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    |dw:1443488906789:dw|

  17. anonymous
    • one year ago
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    Does sine always cross at (0,0)?

  18. zepdrix
    • one year ago
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    |dw:1443489123175:dw|So which one seems more appropriate? :)

  19. anonymous
    • one year ago
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    cosine!

  20. zepdrix
    • one year ago
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    Cosine, ok good.

  21. zepdrix
    • one year ago
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    Sine always crosses through the middle point at the start. It won't necessarily be (0,0), especially if we have any kind of vertical shift.

  22. anonymous
    • one year ago
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    Ok! Is the period 2pi/5 or did I do that incorrectly?

  23. zepdrix
    • one year ago
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    Example: \(\large\rm f(x)=\sin(x)+1\)|dw:1443489226950:dw|Yes, normally this curve would go through (0,0) to start, but it's been shift up by 1.

  24. anonymous
    • one year ago
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    Oh ok that makes so much sense

  25. zepdrix
    • one year ago
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    The period is 5 hours. The letter b that we're going to stick into our formula:\[\large\rm f(x)=A \sin(bx)+d\]Is given by \(\large\rm b=\frac{2\pi}{period}\). So if that's what you meant, then yes. Our b is going to be 2pi/5.

  26. anonymous
    • one year ago
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    yep that's what I meant, great!

  27. zepdrix
    • one year ago
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    Woops cosine :) not sine, my bad. Let's stick in what we know so far, amplitude and our b,\[\large\rm f(x)=3.5 \cos\left(\frac{2\pi}{5}x\right)+d\]

  28. anonymous
    • one year ago
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    How do we find D?

  29. zepdrix
    • one year ago
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    Ummm depends how you like to think of it I guess... I like to think of it as a vertical shift of the middle line.

  30. zepdrix
    • one year ago
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    normally that middle line is the x-axis. But now it's up at 4, ya?

  31. anonymous
    • one year ago
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    yep

  32. anonymous
    • one year ago
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    so is 4 d?

  33. zepdrix
    • one year ago
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    sounds right!\[\large\rm f(x)=3.5 \cos\left(\frac{2\pi}{5}x\right)+4\] And to be a little more accurate, it's shifting the entire function up by 4. You could calculate that shift from any point on the curve. I think it's just easier to work from the middle.

  34. zepdrix
    • one year ago
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    yay team \c:/

  35. anonymous
    • one year ago
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    YAY! Thanks so much, you're the bomb!!

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