Have trouble starting this relatively simple derivative problem

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In summary, the angle between the sun and a solar panel changes throughout the day, so the power output of the panel changes as well.
  • #1
IntegrateMe
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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 :)
 
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  • #2
I would first find the angles which correspond to 4:30 pm and to 5:30 pm.
 
  • #3
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
 
  • #4
bump!
 
  • #5
double bump!
 
  • #6
triple bump! anyone? having real trouble with this one...
 
  • #7
What's theta exactly ? You need 2 numbers. Then use the remaining data in your problem.
 
  • #8
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?
 
  • #9
Is this wrong? Sorry, I just want to be done with this problem :(
 
  • #10
Hmmm...[itex] \theta (10.5) = \arcsin (0.5) + \frac{\pi}{2} = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} [/itex]. Now assume that the rate you're given at this angle is constant throughout the hour.
 
  • #11
Does this mean that the change in power is equal to 12 when theta moves from 10.5 to 11.5?
 
  • #12
No, it's 12 times the difference between the 2 angles.
 
  • #13
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?
 
  • #14
From a calculus perspective, let P denote the power output you need. Then

[tex] P= \int_{\theta(10.5)}^{\theta(11.5)} \frac{dP(\theta)}{d\theta} {}d\theta = 12 \cdot (\theta(11.5)-\theta(10.5)) [/tex],

assuming the rate is kept constant between 4:30 pm and 5:30 pm.
 
  • #15
Thank you, dextercioby, I appreciate your help greatly!
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is essentially the slope of the tangent line at that point.

2. Why is starting a derivative problem difficult?

Starting a derivative problem can be difficult because it requires a strong understanding of mathematical concepts such as limits, derivatives, and rules of differentiation. It also involves a lot of practice and familiarity with various types of functions.

3. What are some common mistakes when starting a derivative problem?

Some common mistakes when starting a derivative problem include forgetting to apply the chain rule, not simplifying expressions, or mixing up rules of differentiation. It is important to be careful and double check each step to avoid making these mistakes.

4. How can I improve my ability to start derivative problems?

The best way to improve your ability to start derivative problems is through practice. Make sure to review and understand the fundamental concepts and rules, and then try solving various types of problems. You can also seek help from a tutor or online resources for additional guidance.

5. What are some tips for starting a derivative problem?

Some tips for starting a derivative problem include breaking down the problem into smaller steps, carefully applying the rules of differentiation, and simplifying expressions as much as possible. It can also be helpful to draw a graph of the function to visualize the problem and better understand what the derivative represents.

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