Expert: Paul Klarreich - 11/30/2007

Question
Hello and good afternoon. I have three questions that I really need help with. The book I'm using does not show how to solve. I will appreciate it if can help me out with the ones you can and show me how you got to the answer step-by step. I hope to hear from you very soon.

1. Find the area of in the first quadrant bounded by the arc of a circle described by the polar equation r = 2 sin 2è + 4 cos è.


2.Find the average value of the function i= 15 [(l- e^(-1/2t)] from t=0 to t=4.

My ans.- 7.5(2 + e^-2)


3.Find the rms value of the functions i= 15 [(l- e^(-1/2t)] from t=0 to t=4.


Thank you so much for you help.

Answer
Questioner:   Cindy
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  Hello and good afternoon. I have three questions that I really need help with. The book I'm using does not show how to solve. I will appreciate it if can help me out with the ones you can and show me how you got to the answer step-by step. I hope to hear from you very soon.

1. Find the area of in the first quadrant bounded by the arc of a circle described by the polar equation r = 2 sin 2t + 4 cos t.


2.Find the average value of the function i= 15 [(l- e^(-1/2t)] from t=0 to t=4.

My ans.- 7.5(2 + e^-2)


3.Find the rms value of the functions i= 15 [(l- e^(-1/2t)] from t=0 to t=4.


Thank you so much for you help.
................................................
Hi, Cindy,
[How do you know this is the afternoon here? Actually the time stamp on your question is 3:50 AM]  

1. The area in P.C. is: [use t = theta]

{t2
|  1/2 r^2 dt
}t1

r = (2 sin 2t + 4 cos t)  << are you sure this is a circle?
r = (4 sin t cos t + 4 cos t)
r = 4 cos t (sin t + 1)

r^2 = 16 cos^2(t)(sin^2(t) + 2 sin t + 1)
Now integrate this from 0 to pi/2.

It's messy, so I cheated and used THE INTEGRATOR:

http://integrals.wolfram.com/index.jsp

which gave:

(60*x - 48*Cos[x] - 16*Cos[3*x] + 24*Sin[2*x] - 3*Sin[4*x])/6

You can do the rest.
...................................

2.Find the average value of the function i= 15 [(l- e^(-1/2t)] from t=0 to t=4.

Average value over [a,b] = Integral over [a,b] divided by (b-a)

{4
|   [(l- e^(-1/2t)] dt << put the 15 back later.
}0


= t - e^(-1/2t)/(-1/2) =

= t + 2e^(-1/2t) from  0 to 4 =

4 + 2e^(-2) - (0 + 2) =

2 + 2e^(-2) times 15.
...............................................
For an RMS value, you:

1. Square the function.
2. Average it over the interval, like #2.
3. Square-root the result.



[(l- e^(-1/2t)]^2 = 1 - 2 e^(-1/2t) + e^(-t)  << put back 15 later.

{4
| [1 - 2 e^(-1/2t) + e^(-t)] dt =
}0

t + e^(-1/2t) - e^(-t), from 0 to 4:

4 + e^(-2) - e^(-4) - (0 + 1 - 1) =

4 + e^(-2) - e^(-4)

Now divide by 4, take sqrt, finally X15.