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 :)

 

[dextercioby]

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...

 

[dextercioby]

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 :(

 

[dextercioby]

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?

 

[dextercioby]

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?

 

[dextercioby]

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!