- #1

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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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- Thread starter IntegrateMe
- Start date

- #1

- 217

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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 :)

- #2

- 13,172

- 741

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

- #3

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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

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bump!

- #5

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double bump!!

- #6

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

- #7

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

- #8

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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

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

- #10

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- 741

- #11

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

- #12

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

- #13

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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

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[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

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Thank you, dextercioby, I appreciate your help greatly!

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