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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- anonymous

Can I just plug in 2 for t?

- anonymous

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

- 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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## More answers

- anonymous

ok. I understand

- SolomonZelman

I lost my connection for a second, apologize,

- anonymous

thats fin e

- anonymous

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

- 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

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

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

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

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

- anonymous

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

- SolomonZelman

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

- SolomonZelman

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

- 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

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

- 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

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

- SolomonZelman

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

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

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

- anonymous

starting to make sense.

- 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

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

- SolomonZelman

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

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

whole those i listed just now are =0.

- 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

So lets get to our function s(t)

- anonymous

Ok. into the good stuff

- SolomonZelman

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

- SolomonZelman

do you recongize your function s(t) ?

- anonymous

yes I do

- 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

so 0 and 0

- SolomonZelman

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

- anonymous

no

- 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

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

-6 x nt^1-1

- anonymous

and n would be 1, right

- SolomonZelman

x is × ?

- anonymous

times yes

- SolomonZelman

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

- SolomonZelman

So,
you go -6 × 1 × t^(1-1)

- SolomonZelman

(not go, but got**)

- SolomonZelman

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

- SolomonZelman

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

- 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

any number*

- anonymous

It's 0 and -6

- SolomonZelman

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

- SolomonZelman

yes, 0 and -6,

- 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

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

- anonymous

yes

- SolomonZelman

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

- anonymous

no

- 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

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

- anonymous

-6

- SolomonZelman

Yup

- SolomonZelman

And that is your final answer.

- anonymous

Thanks man

- SolomonZelman

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

- SolomonZelman

By the way,

- 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

I mean 0 2 1 5

- 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

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

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