## 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

1. ZeHanz

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

2. Dodo1

ok,Thank you. I have a calculator with me.. but why x=2.9??

3. Dodo1

oh i get it beacuse its near 3....

4. ZeHanz

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

Once I gt all answers from equation. what shall I do?

6. ZeHanz

Can you guess what the number is that you are getting closer and closer to?

7. Dodo1

hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!

8. Dodo1

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

cuze at the point (3, 81)

10. ZeHanz

It does, so it's a safe bet 81 is the slope of the tangent line!

11. ZeHanz

(3,81) begin the point on the graph is in itself not of importance. By coincidence, the y-coordinate is also 81.

12. Dodo1

Omg, thats prety cool! thank you very much :) are you faimilar with Tagent　Velocity:

13. ZeHanz

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

yes thats the one

15. ZeHanz

I could be familiar with the stuff in the book ;)

16. Dodo1

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

Do you know the difference between average velocity and instantaneous velocity?

18. Dodo1

mmm No. I have no idea. av velo =displacement/ time elapsed?

19. ZeHanz

It is, so that one is easy. Just calculate (s(10-s(7))/(10-7).

20. ZeHanz

For instanteneous velocity, you'll have to do the limit-stuff as in the first problem. Can you write down the limit?

21. Dodo1

so s(10)-s(7)/3? for the fist one?

22. ZeHanz

Yes, because s(10-s(7) is the distance you've travelled.

23. Dodo1

and jusy plug s=3t^3 in s?

24. ZeHanz

Don't get what you're saying there...

25. ZeHanz

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

Oh! I see.

27. Dodo1

For instanteneous velocity, I have to do limit thing? x=>7?

28. Dodo1

limt S(t)-S(a)/T-a?

29. ZeHanz

Write down the limit:$v(7)=\lim_{t \rightarrow 7}\frac{ s(t)-s(7) }{ t-7 }$ and try numbers near 7.

30. Dodo1

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

You have to divide by t-7, which is 6.99-7=-0.01

32. Dodo1

i got -7.630!

33. Dodo1

No I got 441

34. ZeHanz

441 is good. Just use brackets...

35. Dodo1

what shall I do next?

36. ZeHanz

Take a drink!

37. Dodo1

s(6.9999)?

38. Dodo1

haha!

39. ZeHanz

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

Oh! So limit is how you get closer to x or function?

41. ZeHanz

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

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

Wow, thank you so much for your detailed explaination. It is really fun doing math with you!

44. ZeHanz

Thanks for the compliment! I really appreciate it.