anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Can I just plug in 2 for t?
anonymous
  • anonymous
It says: "by finding derivative"!! Take derivative first
SolomonZelman
  • 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)

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anonymous
  • anonymous
ok. I understand
SolomonZelman
  • SolomonZelman
I lost my connection for a second, apologize,
anonymous
  • anonymous
thats fin e
anonymous
  • anonymous
whatever. I don't care I'm trying to get help rn @David27
SolomonZelman
  • 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).
SolomonZelman
  • 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
SolomonZelman
  • 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) )
SolomonZelman
  • 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.
SolomonZelman
  • SolomonZelman
am I typing too much? Sorry if that is the case, I can reduce wordiness
anonymous
  • anonymous
It's fine. I'm trying to stay with you xD
SolomonZelman
  • SolomonZelman
Ok, so all you need is s'(2)
SolomonZelman
  • SolomonZelman
s(t) = -2 - 6t s'(t)=?
anonymous
  • 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...
SolomonZelman
  • SolomonZelman
Oh, you just apply the rules. I can post them here in colors if you would like..
SolomonZelman
  • 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.
SolomonZelman
  • 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. ---------------------------------------------------------------
SolomonZelman
  • 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. ----------------------------------------------------------------
SolomonZelman
  • 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).
SolomonZelman
  • SolomonZelman
is this still super abstruse, or is it starting to make sense?
anonymous
  • anonymous
starting to make sense.
SolomonZelman
  • 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")
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{c}\cdot t^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dt} \left[ t^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}t^{\color{red}{n}-1} }\)
SolomonZelman
  • 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)
SolomonZelman
  • 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)?
SolomonZelman
  • SolomonZelman
whole those i listed just now are =0.
SolomonZelman
  • 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.
SolomonZelman
  • SolomonZelman
So lets get to our function s(t)
anonymous
  • anonymous
Ok. into the good stuff
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle s(t)=\underline{-2}~~\underline{\color{blue}{-6}\cdot t^\color{red}{1}}}\)
SolomonZelman
  • SolomonZelman
do you recongize your function s(t) ?
anonymous
  • anonymous
yes I do
SolomonZelman
  • 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: \(\large\color{black}{ \displaystyle \frac{d}{dx} \left[\color{blue}{c}\cdot x^\color{red}{n}\right]=\color{blue}{c}\cdot \frac{d}{dx} \left[ x^\color{red}{n}\right]=\color{blue}{c}\cdot\color{red}{n}x^{\color{red}{n}-1} }\) ----------------------------------------
anonymous
  • anonymous
so 0 and 0
SolomonZelman
  • SolomonZelman
0 for the derivative of a -2, but the derivative of \(-6t^1\) is not 0, is it?
anonymous
  • anonymous
no
SolomonZelman
  • 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)
SolomonZelman
  • 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....
anonymous
  • anonymous
-6 x nt^1-1
anonymous
  • anonymous
and n would be 1, right
SolomonZelman
  • SolomonZelman
x is × ?
anonymous
  • anonymous
times yes
SolomonZelman
  • SolomonZelman
k, will go over some basic symbols for extra, if you want to, but for now lets proceed w/ the prob
SolomonZelman
  • SolomonZelman
So, you go -6 × 1 × t^(1-1)
SolomonZelman
  • SolomonZelman
(not go, but got**)
SolomonZelman
  • SolomonZelman
yes, n is 1 in this case, because your initial power is 1.
SolomonZelman
  • SolomonZelman
So, t^(1-1) is same as t^0, right?
anonymous
  • 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
anonymous
  • anonymous
any number*
anonymous
  • anonymous
It's 0 and -6
SolomonZelman
  • SolomonZelman
Oh, but this is very good, because you were able to proceed far to take the derivative already of both parts
SolomonZelman
  • SolomonZelman
yes, 0 and -6,
SolomonZelman
  • 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
SolomonZelman
  • SolomonZelman
again, derivative of s(t) (of the positon function) IS, THE VELOCITY. so your velocity function is v(t)=-6
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
then you can plug in -2 for t, (but you aren't really plugging in for t, are you?)
anonymous
  • anonymous
no
SolomonZelman
  • 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
SolomonZelman
  • SolomonZelman
So, the velocity at t=2 is as well equal to ?
anonymous
  • anonymous
-6
SolomonZelman
  • SolomonZelman
Yup
SolomonZelman
  • SolomonZelman
And that is your final answer.
anonymous
  • anonymous
Thanks man
SolomonZelman
  • SolomonZelman
It is not a problem at all, you are always welcome!
SolomonZelman
  • SolomonZelman
By the way,
SolomonZelman
  • 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 ×
SolomonZelman
  • SolomonZelman
I mean 0 2 1 5
SolomonZelman
  • 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)
SolomonZelman
  • SolomonZelman
but this is just for symbols, if you want to get 1 tip or 2. cu

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