## anonymous one year ago The position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

1. anonymous

Can I just plug in 2 for t?

2. anonymous

It says: "by finding derivative"!! Take derivative first

3. SolomonZelman

Ok, derivative of any function f(x), (which is denoted by f'(x) ) is a [derivative-]function that is the slope of f(x). And this way when you plug in a value x=a into the f'(x), you will then get the instantaneous slope (at the point on the f(x) where x=a)

4. anonymous

ok. I understand

5. SolomonZelman

I lost my connection for a second, apologize,

6. anonymous

thats fin e

7. anonymous

whatever. I don't care I'm trying to get help rn @David27

8. SolomonZelman

The slope/derivative of a function does what? It tells you how fast a function moves. The velocity also tells you how fast a car moves (at a certain point on its position).

9. SolomonZelman

This way, it is logical (not just because the teacher said so), that the velocity is slope of position. If velocity is slope (slope=derivative, remember) of the position, then you have to find the derivative of the postion function to find velocity

10. SolomonZelman

You need the derivative of the positon function, to find the slope-function (the derivative that generates the slope of the s(t) at any given point x=a), and then you need to find the slope at a particular value of x=a, that is x=2. (so plug in x=2 into the s'(t) )

11. SolomonZelman

and hope it is clear that s'(t) is the exact v(t). Saying the the derivative of the position function , IS THE velocity function.

12. SolomonZelman

am I typing too much? Sorry if that is the case, I can reduce wordiness

13. anonymous

It's fine. I'm trying to stay with you xD

14. SolomonZelman

Ok, so all you need is s'(2)

15. SolomonZelman

s(t) = -2 - 6t s'(t)=?

16. anonymous

That's what I need help finding id the derivative. I understand all of the stuff you posted before but this is the part that has been troubling me for ages...

17. SolomonZelman

Oh, you just apply the rules. I can post them here in colors if you would like..

18. SolomonZelman

I am going to use a d/dx notation to denote that I am taking the derivative and put [] around the peace the derivative of which I am taking.

19. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d}{dx} \left[x^\color{red}{n}\right]=\color{red}{n}x^{\color{red}{n}-1} }$$ this is for any function where variable x is raised to a constant power of n. ---------------------------------------------------------------

20. SolomonZelman

Now, lets say you got a constant (a coefficient) in front of the x, what then? $$\large\color{black}{ \displaystyle \frac{d}{dx} \left[\color{blue}{c}\cdot x^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dx} \left[ x^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}x^{\color{red}{n}-1} }$$ ^ ^ ^ ^ we can take the constant c here we do the same as in the previous out, to the front like this. rule, excpet that we just multiply the whole things times this constant c. ----------------------------------------------------------------

21. SolomonZelman

d/dx , again, is just a notation that tells us: we are taking the derivative, and treat x as the variable (n and c, are just constants).

22. SolomonZelman

is this still super abstruse, or is it starting to make sense?

23. anonymous

starting to make sense.

24. SolomonZelman

Good. You would agree with me, if I say that the same rule applies for taking a derivative with respect to any variable. (For example, say, that instead of x, we had a "t")

25. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{c}\cdot t^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dt} \left[ t^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}t^{\color{red}{n}-1} }$$

26. SolomonZelman

like this here, d/dt - is a notation that tells us that we treat t is a variable. and not, the same rules applies, except that we are using a different variable (not x, but t)

27. SolomonZelman

Do you know what a derivative of a constant? What is a derivative of a 2? derivative of a 4? derivative of a constant c (with respect to x)?

28. SolomonZelman

whole those i listed just now are =0.

29. SolomonZelman

Why, because the derivative is the slope, and the slope of a line y=3, y=6, or y=c, all these lines have a slope of 0 (at any point). Therefore, when you differentiate (take a derivative) of a constant), you get 0.

30. SolomonZelman

So lets get to our function s(t)

31. anonymous

Ok. into the good stuff

32. SolomonZelman

$$\large\color{black}{ \displaystyle s(t)=\underline{-2}~~\underline{\color{blue}{-6}\cdot t^\color{red}{1}}}$$

33. SolomonZelman

do you recongize your function s(t) ?

34. anonymous

yes I do

35. SolomonZelman

Now, take the derivative of each underlined part separately. for -2 , apply what you know about the derivative of a constant. for -6•t¹ , apply the rule that I posted (also posted below): ---------------------------------------- The rule is: $$\large\color{black}{ \displaystyle \frac{d}{dx} \left[\color{blue}{c}\cdot x^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dx} \left[ x^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}x^{\color{red}{n}-1} }$$ ----------------------------------------

36. anonymous

so 0 and 0

37. SolomonZelman

0 for the derivative of a -2, but the derivative of $$-6t^1$$ is not 0, is it?

38. anonymous

no

39. SolomonZelman

$$\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{-6}\cdot t^\color{red}{1}\right]=}$$ please take a shot to proceed from here (if not i got your back)

40. SolomonZelman

apply the same rule, $$\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{-6}\cdot t^\color{red}{1}\right]=\color{blue}{-6}\cdot \frac{d}{dt} \left[ t^\color{red}{1}\right]= }$$ go ahead....

41. anonymous

-6 x nt^1-1

42. anonymous

and n would be 1, right

43. SolomonZelman

x is × ?

44. anonymous

times yes

45. SolomonZelman

k, will go over some basic symbols for extra, if you want to, but for now lets proceed w/ the prob

46. SolomonZelman

So, you go -6 × 1 × t^(1-1)

47. SolomonZelman

(not go, but got**)

48. SolomonZelman

yes, n is 1 in this case, because your initial power is 1.

49. SolomonZelman

So, t^(1-1) is same as t^0, right?

50. anonymous

I just noticed what I did wrong and number to the 0 power is 1 not 0. That's why I got 0 and 0

51. anonymous

any number*

52. anonymous

It's 0 and -6

53. SolomonZelman

Oh, but this is very good, because you were able to proceed far to take the derivative already of both parts

54. SolomonZelman

yes, 0 and -6,

55. SolomonZelman

so, we had: s(t)=-2-6t and then our derivative (or the slope, or the velocity in this case) s'(t)=0-6 so s'(t)=-6

56. SolomonZelman

again, derivative of s(t) (of the positon function) IS, THE VELOCITY. so your velocity function is v(t)=-6

57. anonymous

yes

58. SolomonZelman

then you can plug in -2 for t, (but you aren't really plugging in for t, are you?)

59. anonymous

no

60. SolomonZelman

v(t)=-6 there is no t to plug in for, your function is a line s(t)=-6 that means that it has the same output for all values of t 9and that output is -6). That means that for any t=a, you get -6

61. SolomonZelman

So, the velocity at t=2 is as well equal to ?

62. anonymous

-6

63. SolomonZelman

Yup

64. SolomonZelman

65. anonymous

Thanks man

66. SolomonZelman

It is not a problem at all, you are always welcome!

67. SolomonZelman

By the way,

68. SolomonZelman

You can click ALT click 0, 1, 2, 5 on your number pad on the right of your keyboard release ALT it should get you ×

69. SolomonZelman

I mean 0 2 1 5

70. SolomonZelman

1) Click and hold ALT 2) click the number code (using the numbers that are on the right of the keyboard, and NOT the ones below F1, F2, F3, etc., ) 3) release the ALT number code result 0 2 1 5  × 2 4 6  ÷  7  • ──────────────── among with other symbols.  2 5 1  √ 7 5 4  ≥ 7 5 5 ≤ 2 4 1  or 7 5 3 ± 2 4 7 ≈ 0 1 8 5  ¹ 2 5 3  ² 0 1 7 9 ³ ALSO, 1 6 6 ª 2 5 2 ⁿ 1 6 7 º 2 4 8 ° 0 1 9 0 ¾ 4 2 8 ¼ 1 7 1 ½ 2 2 7 π 1 5 5 ¢ 2 3 6 ∞ 1 5 9 ƒ 4 ♦ 2 5 4 ■ 2 1 9 █ 1 9 6 ─ 7 • 6 ♠ 5 ♣ 3 ♥ 13 ♪ 14 ♫ 4 8 9 Θ (there are more)

71. SolomonZelman

but this is just for symbols, if you want to get 1 tip or 2. cu