# Help me check my solutions, please!

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!

## Answers and Replies

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bump!

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!
I don't know what they mean by the change in power output. Probably it's P(5:30 PM) - P(4:30 PM), since a change is just a variation of a function. In that case you have to find P(t), integrating the expression in (A).

Sorry if this didn't help :tongue2: