# Have trouble starting this relatively simple derivative problem

IntegrateMe
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 :)

Homework Helper
I would first find the angles which correspond to 4:30 pm and to 5:30 pm.

IntegrateMe
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

IntegrateMe
bump!

IntegrateMe
double bump!

IntegrateMe
triple bump! anyone? having real trouble with this one...

Homework Helper
What's theta exactly ? You need 2 numbers. Then use the remaining data in your problem.

IntegrateMe
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?

IntegrateMe
Is this wrong? Sorry, I just want to be done with this problem :(

Homework Helper
Hmmm...$\theta (10.5) = \arcsin (0.5) + \frac{\pi}{2} = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$. Now assume that the rate you're given at this angle is constant throughout the hour.

IntegrateMe
Does this mean that the change in power is equal to 12 when theta moves from 10.5 to 11.5?

Homework Helper
No, it's 12 times the difference between the 2 angles.

IntegrateMe
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?

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

$$P= \int_{\theta(10.5)}^{\theta(11.5)} \frac{dP(\theta)}{d\theta} {}d\theta = 12 \cdot (\theta(11.5)-\theta(10.5))$$,

assuming the rate is kept constant between 4:30 pm and 5:30 pm.

IntegrateMe
Thank you, dextercioby, I appreciate your help greatly!