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Have trouble starting this relatively simple derivative problem

  1. Oct 26, 2011 #1
    A solar panel outputs P(θ) watts when the angle between the sun and the panel is θ for 0 ≤ θ ≤ π. On any day, the angle between the panel and sun t hours after 6 a.m. is θ(t) for 0 ≤ t ≤ 14. Assume that sunrise is at 6 a.m. and sunset is 8 p.m.

    Suppose dP/dθ (2π/3) = 12 and θ(t) = arcsin(t/7 - 1) + π/2

    Find the change in power for the panel between 4:30pm-5:30pm.

    I'm not asking any of you to do the problem for me, I just need some guidance as how to start.

    Thanks everyone :)
     
  2. jcsd
  3. Oct 26, 2011 #2

    dextercioby

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    I would first find the angles which correspond to 4:30 pm and to 5:30 pm.
     
  4. Oct 26, 2011 #3
    Well, t = 10.5 would be 4:30pm, and t = 11.5 would be 5:30pm

    So, since θ(t) = arcsin(t/7 - 1) + π/2

    θ(10.5) = arcsin(10.5/7 - 1) + π/2
    and
    θ(11.5) = arcsin(11.5/7 - 1) + π/2
     
  5. Oct 26, 2011 #4
  6. Oct 26, 2011 #5
    double bump!!
     
  7. Oct 27, 2011 #6
    triple bump!!! anyone? having real trouble with this one...
     
  8. Oct 27, 2011 #7

    dextercioby

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    What's theta exactly ? You need 2 numbers. Then use the remaining data in your problem.
     
  9. Oct 27, 2011 #8
    θ(10.5) = 2.0944
    and
    θ(11.5) = 2.2690

    Now, I think I would simply need to plug these numbers into P(θ), but why do they tell me that dP/dθ (2π/3) = 12?
     
  10. Oct 27, 2011 #9
    Is this wrong? Sorry, I just want to be done with this problem :(
     
  11. Oct 27, 2011 #10

    dextercioby

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    Hmmm...[itex] \theta (10.5) = \arcsin (0.5) + \frac{\pi}{2} = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} [/itex]. Now assume that the rate you're given at this angle is constant throughout the hour.
     
  12. Oct 27, 2011 #11
    Does this mean that the change in power is equal to 12 when theta moves from 10.5 to 11.5?
     
  13. Oct 27, 2011 #12

    dextercioby

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    No, it's 12 times the difference between the 2 angles.
     
  14. Oct 27, 2011 #13
    Let me understand this intuitively...

    dP/dθ (2π/3) = 12 tells us that (since 2pi/3 = 10.5) when we move from 10.5 to 11.5, the change in power will be equal to the power at 10.5 = 2pi/3 - the power at 11.5 = 2.269 * 12

    So:

    (P(11.5) - P(10.5))(12)

    Correct?
     
  15. Oct 27, 2011 #14

    dextercioby

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    From a calculus perspective, let P denote the power output you need. Then

    [tex] P= \int_{\theta(10.5)}^{\theta(11.5)} \frac{dP(\theta)}{d\theta} {}d\theta = 12 \cdot (\theta(11.5)-\theta(10.5)) [/tex],

    assuming the rate is kept constant between 4:30 pm and 5:30 pm.
     
  16. Oct 27, 2011 #15
    Thank you, dextercioby, I appreciate your help greatly!
     
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