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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.

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  1. anonymous
    • one year ago
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    Can I just plug in 2 for t?

  2. anonymous
    • one year ago
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    It says: "by finding derivative"!! Take derivative first

  3. SolomonZelman
    • one year ago
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    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
    • one year ago
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    ok. I understand

  5. SolomonZelman
    • one year ago
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    I lost my connection for a second, apologize,

  6. anonymous
    • one year ago
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    thats fin e

  7. anonymous
    • one year ago
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    whatever. I don't care I'm trying to get help rn @David27

  8. SolomonZelman
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    am I typing too much? Sorry if that is the case, I can reduce wordiness

  13. anonymous
    • one year ago
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    It's fine. I'm trying to stay with you xD

  14. SolomonZelman
    • one year ago
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    Ok, so all you need is s'(2)

  15. SolomonZelman
    • one year ago
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    s(t) = -2 - 6t s'(t)=?

  16. anonymous
    • one year ago
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    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
    • one year ago
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    Oh, you just apply the rules. I can post them here in colors if you would like..

  18. SolomonZelman
    • one year ago
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    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
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{d}{dx} \left[x^\color{red}{n}\right]=\color{red}{n}x^{\color{red}{n}-1} }\) this is for any function where variable x is raised to a constant power of n. ---------------------------------------------------------------

  20. SolomonZelman
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    is this still super abstruse, or is it starting to make sense?

  23. anonymous
    • one year ago
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    starting to make sense.

  24. SolomonZelman
    • one year ago
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    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
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{c}\cdot t^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dt} \left[ t^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}t^{\color{red}{n}-1} }\)

  26. SolomonZelman
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    whole those i listed just now are =0.

  29. SolomonZelman
    • one year ago
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    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
    • one year ago
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    So lets get to our function s(t)

  31. anonymous
    • one year ago
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    Ok. into the good stuff

  32. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle s(t)=\underline{-2}~~\underline{\color{blue}{-6}\cdot t^\color{red}{1}}}\)

  33. SolomonZelman
    • one year ago
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    do you recongize your function s(t) ?

  34. anonymous
    • one year ago
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    yes I do

  35. SolomonZelman
    • one year ago
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    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
    • one year ago
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    so 0 and 0

  37. SolomonZelman
    • one year ago
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    0 for the derivative of a -2, but the derivative of \(-6t^1\) is not 0, is it?

  38. anonymous
    • one year ago
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    no

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

  40. SolomonZelman
    • one year ago
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    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
    • one year ago
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    -6 x nt^1-1

  42. anonymous
    • one year ago
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    and n would be 1, right

  43. SolomonZelman
    • one year ago
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    x is × ?

  44. anonymous
    • one year ago
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    times yes

  45. SolomonZelman
    • one year ago
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    k, will go over some basic symbols for extra, if you want to, but for now lets proceed w/ the prob

  46. SolomonZelman
    • one year ago
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    So, you go -6 × 1 × t^(1-1)

  47. SolomonZelman
    • one year ago
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    (not go, but got**)

  48. SolomonZelman
    • one year ago
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    yes, n is 1 in this case, because your initial power is 1.

  49. SolomonZelman
    • one year ago
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    So, t^(1-1) is same as t^0, right?

  50. anonymous
    • one year ago
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    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
    • one year ago
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    any number*

  52. anonymous
    • one year ago
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    It's 0 and -6

  53. SolomonZelman
    • one year ago
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    Oh, but this is very good, because you were able to proceed far to take the derivative already of both parts

  54. SolomonZelman
    • one year ago
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    yes, 0 and -6,

  55. SolomonZelman
    • one year ago
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    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
    • one year ago
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    again, derivative of s(t) (of the positon function) IS, THE VELOCITY. so your velocity function is v(t)=-6

  57. anonymous
    • one year ago
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    yes

  58. SolomonZelman
    • one year ago
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    then you can plug in -2 for t, (but you aren't really plugging in for t, are you?)

  59. anonymous
    • one year ago
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    no

  60. SolomonZelman
    • one year ago
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    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
    • one year ago
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    So, the velocity at t=2 is as well equal to ?

  62. anonymous
    • one year ago
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    -6

  63. SolomonZelman
    • one year ago
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    Yup

  64. SolomonZelman
    • one year ago
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    And that is your final answer.

  65. anonymous
    • one year ago
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    Thanks man

  66. SolomonZelman
    • one year ago
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    It is not a problem at all, you are always welcome!

  67. SolomonZelman
    • one year ago
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    By the way,

  68. SolomonZelman
    • one year ago
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    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
    • one year ago
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    I mean 0 2 1 5

  70. SolomonZelman
    • one year ago
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    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
    • one year ago
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    but this is just for symbols, if you want to get 1 tip or 2. cu

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