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Dodo1

  • 2 years ago

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

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  1. ZeHanz
    • 2 years ago
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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...

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

  3. Dodo1
    • 2 years ago
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    oh i get it beacuse its near 3....

  4. ZeHanz
    • 2 years ago
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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...

  5. Dodo1
    • 2 years ago
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    Once I gt all answers from equation. what shall I do?

  6. ZeHanz
    • 2 years ago
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    Can you guess what the number is that you are getting closer and closer to?

  7. Dodo1
    • 2 years ago
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    hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!

  8. Dodo1
    • 2 years ago
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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?

  9. Dodo1
    • 2 years ago
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    cuze at the point (3, 81)

  10. ZeHanz
    • 2 years ago
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    It does, so it's a safe bet 81 is the slope of the tangent line!

  11. ZeHanz
    • 2 years ago
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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.

  12. Dodo1
    • 2 years ago
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    Omg, thats prety cool! thank you very much :) are you faimilar with Tagent Velocity:

  13. ZeHanz
    • 2 years ago
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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..)

  14. Dodo1
    • 2 years ago
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    yes thats the one

  15. ZeHanz
    • 2 years ago
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    I could be familiar with the stuff in the book ;)

  16. Dodo1
    • 2 years ago
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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

  17. ZeHanz
    • 2 years ago
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    Do you know the difference between average velocity and instantaneous velocity?

  18. Dodo1
    • 2 years ago
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    mmm No. I have no idea. av velo =displacement/ time elapsed?

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

  20. ZeHanz
    • 2 years ago
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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?

  21. Dodo1
    • 2 years ago
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    so s(10)-s(7)/3? for the fist one?

  22. ZeHanz
    • 2 years ago
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    Yes, because s(10-s(7) is the distance you've travelled.

  23. Dodo1
    • 2 years ago
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    and jusy plug s=3t^3 in s?

  24. ZeHanz
    • 2 years ago
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    Don't get what you're saying there...

  25. ZeHanz
    • 2 years ago
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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...)

  26. Dodo1
    • 2 years ago
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    Oh! I see.

  27. Dodo1
    • 2 years ago
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    For instanteneous velocity, I have to do limit thing? x=>7?

  28. Dodo1
    • 2 years ago
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    limt S(t)-S(a)/T-a?

  29. ZeHanz
    • 2 years ago
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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.

  30. Dodo1
    • 2 years ago
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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?

  31. ZeHanz
    • 2 years ago
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    You have to divide by t-7, which is 6.99-7=-0.01

  32. Dodo1
    • 2 years ago
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    i got -7.630!

  33. Dodo1
    • 2 years ago
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    No I got 441

  34. ZeHanz
    • 2 years ago
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    441 is good. Just use brackets...

  35. Dodo1
    • 2 years ago
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    what shall I do next?

  36. ZeHanz
    • 2 years ago
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    Take a drink!

  37. Dodo1
    • 2 years ago
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    s(6.9999)?

  38. Dodo1
    • 2 years ago
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    haha!

  39. ZeHanz
    • 2 years ago
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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

  40. Dodo1
    • 2 years ago
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    Oh! So limit is how you get closer to x or function?

  41. ZeHanz
    • 2 years ago
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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.

  42. ZeHanz
    • 2 years ago
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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!

  43. Dodo1
    • 2 years ago
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    Wow, thank you so much for your detailed explaination. It is really fun doing math with you!

  44. ZeHanz
    • 2 years ago
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    Thanks for the compliment! I really appreciate it.

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