Calculating Power Output Change in Solar Panel Angle and Time

[IntegrateMe]

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!

 

[anathema]

To find the change in power output between 4:30PM and 5:30PM, we need to use the chain rule again. We know that θ(t) is the angle between the sun and the panel at time t, and we are given that θ(t) = arcsin(t/7 -1) + π/2. So, at 4:30PM, t = 10.5. We can plug this into our function to find θ(10.5) = arcsin(3/14) + π/2. This gives us the angle at 4:30PM.

To find the angle at 5:30PM, we need to find θ(11.5) = arcsin(4/14) + π/2. Now, we can use the chain rule to find the change in power output between these two angles:

dP/dθ (2π/3) = P'(θ(11.5)) * θ'(11.5) - P'(θ(10.5)) * θ'(10.5)
12 = P'(π/3) * θ'(11.5) - P'(π/4) * θ'(10.5)

Now, we can solve for θ'(11.5) and θ'(10.5) using the equation we found in part (B). Once we have these values, we can plug them back into the equation above to find the change in power output between 4:30PM and 5:30PM. This will give us the change in power output in watts.