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LaddiusMaximus

  • 4 years ago

estimate instantaneous rate of change at the point x=5 for f(x)=lnx

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  1. amistre64
    • 4 years ago
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    the derivative of lnx at x=5

  2. amistre64
    • 4 years ago
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    1/5 maybe?

  3. LaddiusMaximus
    • 4 years ago
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    That is not what it is asking. Im supposed to use the average velocity formula to figure it out and its not working

  4. amistre64
    • 4 years ago
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    the average velocity is for an average rate of change; your spose to be using the limit of that formula as h goes to 0 or some other denominator

  5. LaddiusMaximus
    • 4 years ago
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    I know that but we havent gotten that far. We are only using rates of change.

  6. amistre64
    • 4 years ago
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    instantaneous velocity is the derivatie; which can be defined as:\[\lim_{h->0}\frac{ln(x+h)-ln(x)}{h }\]

  7. amistre64
    • 4 years ago
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    then use the avg stuff with small changes

  8. amistre64
    • 4 years ago
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    ln(5)-ln(5.0001)/.0001

  9. amistre64
    • 4 years ago
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    or something akin to it

  10. LaddiusMaximus
    • 4 years ago
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    the way the book and cramster teaches it is way different. im supposed to use numbers a little more and a little less than five

  11. LaddiusMaximus
    • 4 years ago
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    like 4.9 and 5.01

  12. amistre64
    • 4 years ago
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    teh smaller the change, the more exact the estimate

  13. LaddiusMaximus
    • 4 years ago
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    (5)^3 +5-((4.9)^3 +4.9)/ 5-4.9

  14. LaddiusMaximus
    • 4 years ago
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    well it asks for estimates and all the ones I give are wrong

  15. amistre64
    • 4 years ago
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    cubed stuff?

  16. amistre64
    • 4 years ago
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    you said lnx

  17. LaddiusMaximus
    • 4 years ago
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    I end up with 68.75, 73.04,74.51,75.85

  18. LaddiusMaximus
    • 4 years ago
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    I dont know man. Ive been beating my head against the wall for the last two hours and the book is nigh on useless

  19. amistre64
    • 4 years ago
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    ln(5.00001/5)/.00001 = .2000...... approx 1/5 http://www.wolframalpha.com/input/?i=ln%285.00001%2F5%29%2F.00001

  20. LaddiusMaximus
    • 4 years ago
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    it says that the instantaneous rate of change is estimated by computing the average rate of smaller intervals

  21. amistre64
    • 4 years ago
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    thats what i did

  22. LaddiusMaximus
    • 4 years ago
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    well what am I doing then? because im lost. I dont understand how to set up the problem.

  23. amistre64
    • 4 years ago
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    ln(5.00001) is a small change from ln(5) right?

  24. amistre64
    • 4 years ago
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    in fact is .00001 small of a change

  25. LaddiusMaximus
    • 4 years ago
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    ok i follow that. but what then? how do I set up the problem to solve it? so its [5, 5.001]?

  26. amistre64
    • 4 years ago
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    yes

  27. amistre64
    • 4 years ago
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    or 4.99999 to 5 but its easier to see the change when it grows a little bigger

  28. LaddiusMaximus
    • 4 years ago
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    how do i plug it in? i see a formula for average rate of change, but it doesnt give one for instantaneous

  29. amistre64
    • 4 years ago
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    im not sure what it is youre looking at ....

  30. amistre64
    • 4 years ago
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    if its about inputing into a program, i got no idea what it wants then

  31. amistre64
    • 4 years ago
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    the instantaneous rate of change at ln(5) = 1/5 = .2

  32. amistre64
    • 4 years ago
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    so how you got those other vlaues I cant say

  33. LaddiusMaximus
    • 4 years ago
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    im sorry. the book does a very crappy job of explaining things

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