1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help me check my solutions, please!

  1. Oct 27, 2011 #1
    The power output of a particular type of solar panel varies with the angle of the sun shining on the panel. The panel outputs P (θ) watts when the angle between the sun and the panel is θ for 0 ≤ θ ≤ π. On a typical summer day in Ann Arbor, the angle between a properly mounted panel and the 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.

    (A) Calculate dP/dt using the chain rule, and give interpretations for each part of your calculation.

    P(θ(t)). So, dP/dt = P(θ(t))' = P'(θ(t)) * θ'(t)
    P'(θ(t)) is the average rate of change of power with respect to θ.
    θ'(t) is the average rate of change of θ with respect to t.

    (B) Suppose θ(t) = arcsin(t/7 -1) + π/2. Calculate θ'(t) using the equivalent expression: sin(θ(t) - π/2) = t/7 - 1

    I just differentiated the equivalent function:

    cos(θ(t) - π/2)*θ'(t) = 1/7
    θ'(t) = 1/(7*cos(θ(t) - π/2))

    (C) Suppose dP/dθ (2π/3) = 12 and θ(t) is the function in part (B). Find the change in power output between 4:30PM and 5:30PM.

    This is where I'm having trouble. I would think the "change in power output" would simply be dP/dθ, since this represents the change in power with respect to θ, but I feel as though I'm incorrect here. Any help would be awesome. Thanks!
     
  2. jcsd
  3. Oct 27, 2011 #2
    Change in power output is dP/dt here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help me check my solutions, please!
  1. Please check my work (Replies: 0)

Loading...