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Dodo1 Group Title

Dear, Math-talent! The slope of the tangent line to the graph of the function y=3x^3 at the point (381) is limx=>3 (3x^3−81)/(x-3.) By trying values of x near 3, find the slope of the tangent line

  • one year ago
  • one year ago

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  1. ZeHanz Group Title
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    Take you calculator and try x=2.9, 2.99, 2.999, to get a good idea where this is going to...

    • one year ago
  2. Dodo1 Group Title
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    ok,Thank you. I have a calculator with me.. but why x=2.9??

    • one year ago
  3. Dodo1 Group Title
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    oh i get it beacuse its near 3....

    • one year ago
  4. ZeHanz Group Title
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    Well, you want to know the limit for x --> 3. You are not supposed to do all kinds of math trickery to do this, but just try a few values near 3. So 2.9, or 3.1 and then even closer: 2.99 or 3.01, or any other value, as long as it is "near" 3...

    • one year ago
  5. Dodo1 Group Title
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    Once I gt all answers from equation. what shall I do?

    • one year ago
  6. ZeHanz Group Title
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    Can you guess what the number is that you are getting closer and closer to?

    • one year ago
  7. Dodo1 Group Title
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    hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!

    • one year ago
  8. Dodo1 Group Title
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    Ok, So I got 78.33 for x=2.9 and 80.7303 for x=2.99... I guess it gets closer to 81?

    • one year ago
  9. Dodo1 Group Title
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    cuze at the point (3, 81)

    • one year ago
  10. ZeHanz Group Title
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    It does, so it's a safe bet 81 is the slope of the tangent line!

    • one year ago
  11. ZeHanz Group Title
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    (3,81) begin the point on the graph is in itself not of importance. By coincidence, the y-coordinate is also 81.

    • one year ago
  12. Dodo1 Group Title
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    Omg, thats prety cool! thank you very much :) are you faimilar with Tagent Velocity:

    • one year ago
  13. ZeHanz Group Title
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    You mean the velocity of a moving point along a curve is the tangent vector? Or is it a book? (havent't got a clue here..)

    • one year ago
  14. Dodo1 Group Title
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    yes thats the one

    • one year ago
  15. ZeHanz Group Title
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    I could be familiar with the stuff in the book ;)

    • one year ago
  16. Dodo1 Group Title
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    cool. The displacement (in meters) of a particle moving in a straight line is given by s=3t^3 where t is measured in seconds. 1)Find the average velocity of the particle over the time interval [7,10]. 2)Find the (instantaneous) velocity of the particle when t=7

    • one year ago
  17. ZeHanz Group Title
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    Do you know the difference between average velocity and instantaneous velocity?

    • one year ago
  18. Dodo1 Group Title
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    mmm No. I have no idea. av velo =displacement/ time elapsed?

    • one year ago
  19. ZeHanz Group Title
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    It is, so that one is easy. Just calculate (s(10-s(7))/(10-7).

    • one year ago
  20. ZeHanz Group Title
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    For instanteneous velocity, you'll have to do the limit-stuff as in the first problem. Can you write down the limit?

    • one year ago
  21. Dodo1 Group Title
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    so s(10)-s(7)/3? for the fist one?

    • one year ago
  22. ZeHanz Group Title
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    Yes, because s(10-s(7) is the distance you've travelled.

    • one year ago
  23. Dodo1 Group Title
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    and jusy plug s=3t^3 in s?

    • one year ago
  24. ZeHanz Group Title
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    Don't get what you're saying there...

    • one year ago
  25. ZeHanz Group Title
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    s(10)=3*10³=3000, s(7)=3*7³=1029 (3000-1029)/3=657 m/s (you may not be using meters, but feet...)

    • one year ago
  26. Dodo1 Group Title
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    Oh! I see.

    • one year ago
  27. Dodo1 Group Title
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    For instanteneous velocity, I have to do limit thing? x=>7?

    • one year ago
  28. Dodo1 Group Title
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    limt S(t)-S(a)/T-a?

    • one year ago
  29. ZeHanz Group Title
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    Write down the limit:\[v(7)=\lim_{t \rightarrow 7}\frac{ s(t)-s(7) }{ t-7 }\] and try numbers near 7.

    • one year ago
  30. Dodo1 Group Title
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    OK, I tired to 6.99. s(6,99)=(6.99)^3*3=1024.59 S(7)= 1029 1024.59-1029/1029-7?

    • one year ago
  31. ZeHanz Group Title
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    You have to divide by t-7, which is 6.99-7=-0.01

    • one year ago
  32. Dodo1 Group Title
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    i got -7.630!

    • one year ago
  33. Dodo1 Group Title
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    No I got 441

    • one year ago
  34. ZeHanz Group Title
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    441 is good. Just use brackets...

    • one year ago
  35. Dodo1 Group Title
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    what shall I do next?

    • one year ago
  36. ZeHanz Group Title
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    Take a drink!

    • one year ago
  37. Dodo1 Group Title
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    s(6.9999)?

    • one year ago
  38. Dodo1 Group Title
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    haha!

    • one year ago
  39. ZeHanz Group Title
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    Oh, now I get it ;) If you set t=6.999 you'll be even closer to 441, so that will be the answer

    • one year ago
  40. Dodo1 Group Title
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    Oh! So limit is how you get closer to x or function?

    • one year ago
  41. ZeHanz Group Title
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    Yes, it is the value of a formula as you get closer and closer to a certain number. Only, because of the special kind of formula, it is not possible to try that number directly. As in the formula above: if you want the speed at t=7, you can't just put t=7 and get the right answer. If you try it, you'll get 0/0, which is undefined.

    • one year ago
  42. ZeHanz Group Title
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    It is possible however, to do more advanced calculations with the formula to get the limit. It is called differentiation... Once you know how to do that, you don't need to try all these cumbersome numbers "near" to a certain value!

    • one year ago
  43. Dodo1 Group Title
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    Wow, thank you so much for your detailed explaination. It is really fun doing math with you!

    • one year ago
  44. ZeHanz Group Title
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    Thanks for the compliment! I really appreciate it.

    • one year ago
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