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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    I think the right answer to this is .12

  2. zepdrix
    • one year ago
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    So for linear approximation we use ummm...\[\large\rm f(x)\approx f(x_o)+\color{orangered}{f'(x_o)(x-x_o)}\]Something like that, ya? Where the orange stuff is our error. Wait.. is that the error? Hmmm thinking

  3. anonymous
    • one year ago
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    Im not sure this problem is very confusing

  4. zepdrix
    • one year ago
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    Yes it is +_+ So the error in the output \(\large\rm \Delta y \) ummm... ya ya ya, should be \(\large\rm f'(x_o)\color{green}{\Delta x}\), where Delta x is the distance between our base point x_o and x. \(\large\rm \Delta y\approx f'(x_o)\color{green}{(x-x_o)}\) So our base point is x=1. And we're drifting 0.001 away from that. \[\large\rm \Delta y\approx f'(1)(0.001)\]

  5. zepdrix
    • one year ago
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    Were you able to come up with a formula for Surface Area of a cube? :) And its derivative?

  6. anonymous
    • one year ago
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    No I was not

  7. zepdrix
    • one year ago
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    Take a cube, call the side length x. Since it's a cube, all side lengths are the same. The area of one face will be x*x, ya? So the total surface area will be the area of all 6 panels of the cube. \(\large\rm A=6x^2\)

  8. zepdrix
    • one year ago
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    Wait wait, then how were you able to come up with .12? LOL Just straight up guessing? >.< Buhaha, Cam cam, come on!!

  9. anonymous
    • one year ago
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    Haha yes it was a guess but it made sense

  10. zepdrix
    • one year ago
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    So your derivative is \(\large\rm A'=12x\). Mmm k.

  11. zepdrix
    • one year ago
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    And your error approximation for the output \(\large\rm \Delta y\) is going to be\[\large\rm =A'(x_o)\Delta x\]Our base point is x=1. And our delta is the distance between the x's, 0.001.

  12. anonymous
    • one year ago
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    Ok

  13. zepdrix
    • one year ago
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    \[\large\rm =A'(1)(0.001)=?\]

  14. anonymous
    • one year ago
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    1X10^-3

  15. zepdrix
    • one year ago
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    What? 0_o Plug 1 into A'(x) and then multiply that value by 0.001

  16. anonymous
    • one year ago
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    Oooo .12?

  17. zepdrix
    • one year ago
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    Hmm that's not what I'm getting.. lemme see..

  18. zepdrix
    • one year ago
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    \[\large\rm =\color{orangered}{A'(1)}(0.001)=\color{orangered}{12(1)}(0.001)=12(0.001)=0.012\]

  19. zepdrix
    • one year ago
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    Is that one of our options? :o Hopefully we're doing this right <.<

  20. anonymous
    • one year ago
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    Yes it is!!

  21. anonymous
    • one year ago
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    Wait no its .12

  22. zepdrix
    • one year ago
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    What do the other options look like? XD

  23. anonymous
    • one year ago
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    .02 .003 and none of the above

  24. zepdrix
    • one year ago
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    Hmm :p @ganeshie8 @dan815

  25. anonymous
    • one year ago
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    Ill just put in .12 and hope thats the answer

  26. anonymous
    • one year ago
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    Thank you for the help!

  27. zepdrix
    • one year ago
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    ya sorry :c

  28. anonymous
    • one year ago
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    Its okay thanks for the help!!

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