## anonymous 3 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. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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. anonymous

cuze at the point (3, 81)

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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. anonymous

yes thats the one

15. anonymous

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

16. anonymous

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. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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. anonymous

Oh! I see.

27. anonymous

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

28. anonymous

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

29. anonymous

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

30. anonymous

OK, I tired to 6.99. s(6,99)=(6.99)^3*3=1024.59 S(7)= 1029 1024.59-1029/1029-7?

31. anonymous

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

32. anonymous

i got -7.630!

33. anonymous

No I got 441

34. anonymous

441 is good. Just use brackets...

35. anonymous

what shall I do next?

36. anonymous

Take a drink!

37. anonymous

s(6.9999)?

38. anonymous

haha!

39. anonymous

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. anonymous

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

41. anonymous

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. anonymous

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. anonymous

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

44. anonymous

Thanks for the compliment! I really appreciate it.