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Mitchi35

  • one year ago

I'm only 16 and I am doing some AP calculus problems... Anyone good or somewhat okay at it?

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  1. sleepyhead314
    • one year ago
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    We got plenty of people that can help you here :) I got a 5 on both the AB and BC calc tests ^_^

  2. Mitchi35
    • one year ago
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    Oh wow! Okay I'm sold. Let me just put the problem up real quick.

  3. Mitchi35
    • one year ago
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    Give the volume of the solid generated by revolving the region bounded by the graph of y = ln(x), the x-axis, the lines x = 1 and x = e, about the y-axis.

  4. anonymous
    • one year ago
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    What is the derivatives of: \[x^{-3}\] and \[ e^{2x}\] with respect to x?

  5. sleepyhead314
    • one year ago
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    haha @PlasmaFuzer I would suggest that you ask your own questions in the "Ask a question..." box to the left ^_^

  6. anonymous
    • one year ago
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    Ooo I like the question even better @Mitchi35

  7. anonymous
    • one year ago
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    @sleepyhead314 I thought he was asking for us to quiz him... I have many more where that came from, but I guess I just misunderstood what he was looking for

  8. Mitchi35
    • one year ago
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    Lol What are yo talking about? @PlasmaFuzer

  9. Mitchi35
    • one year ago
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    Oh Okay. :)

  10. sleepyhead314
    • one year ago
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    I see, it's fine xD but feel free to jump in to help anytime @PlasmaFuzer :)

  11. anonymous
    • one year ago
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    I didn't see your second post until I had already posted mine, and then I confused you for the the other person in here and not the poster XD sorry.

  12. sleepyhead314
    • one year ago
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    now, I'm guessing you Are allowed a graphing calculator for this problem? @Mitchi35

  13. anonymous
    • one year ago
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    I would @sleepyhead314 but I am afraid its been quite a while for me on these types of problems so I think I might just cheat step my way to the answer which isn't instructive if you don't understand why things work the way they do

  14. anonymous
    • one year ago
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    I.e. just integrate and rotate

  15. Mitchi35
    • one year ago
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    And I'm a she by the way.@ PlasmaFuzer. And Yes. But only I don't have one. @sleepyhead314

  16. sleepyhead314
    • one year ago
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    |dw:1444067181310:dw|

  17. anonymous
    • one year ago
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    Sorry again XD

  18. Mitchi35
    • one year ago
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    Oh wait! They have multiple choice. I"ll put them up.

  19. anonymous
    • one year ago
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    I don't want to step on any toes @sleepyhead314 but I am also curious to see if I remembered correctly can I give it a shot and if I am wrong then you can correct?

  20. sleepyhead314
    • one year ago
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    of course @PlasmaFuzer feel free to take the wheel ^_^

  21. anonymous
    • one year ago
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    I am mostly just debating between two approaches, my gut tells me you calculate the integral of the function (i.e. area under curve), then you rotate that 2pi radians about the x-axis....

  22. anonymous
    • one year ago
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    Then again I have to consider that the infinitesimal areal elements are circles of radius ln(x) and so my integrand should be ln(x)^2 (i.e. imitate the circle formula)

  23. anonymous
    • one year ago
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    times pi

  24. anonymous
    • one year ago
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    Considering further I think it should be the first, because the integral derivation of the area of a circle involves the circumference as the integrand.... therefore my first approach I believe to be correct.

  25. jigglypuff314
    • one year ago
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    try the disk method, volume = pi times the integral of the function from bound to bound squared

  26. jigglypuff314
    • one year ago
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    in this case your function is ln(x) and your bounds are from 1 to e

  27. anonymous
    • one year ago
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    Ahh yes ty jiggly Its the square of the integral not the integrand squared that makes more sense.

  28. anonymous
    • one year ago
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    You can take this one Im obviously rusty

  29. Mitchi35
    • one year ago
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    Yeah. It's fine guys. I'll just ask the teacher. Thank you very much though anyways. :)

  30. anonymous
    • one year ago
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    Oh @Mitchi35 I can help if they left.... I realized my mistake, and if they aren't there to help finish the problem I am more than able.

  31. anonymous
    • one year ago
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    I just was gonna give them first crack since I kinda screwed up.... Anyways think of the problem as assembling a bunch of disks whose area is given by pi*r^2 where the radius in this instance will be an really thing slice of the area under the curve.

  32. anonymous
    • one year ago
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    Ok @Mitchi35 I was right before I just pulled up a quick review site and it isnt the integral squard as jigglypuff suggested but it is the INTEGRAND squared prior to integration. Give me a moment and I will type up a solution.

  33. jigglypuff314
    • one year ago
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    sorry if our explanations are a little confusing @Mitchi35 I'll attach the answer I got for your question on this comment for you hopefully that might help a bit more :)

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  34. anonymous
    • one year ago
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    Oh yeah @jigglypuff314 if thats what you meant I totally agree.... I was torn between my first two approaches and thought what you said was implying a third.... but really it was just confirming the second approach.

  35. jigglypuff314
    • one year ago
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    I was simply using the approach that my BC calc teacher taught me :P

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