Question

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.

Answer 1

Can I just plug in 2 for t?

Answer 2

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

Answer 3

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)

Answer 4

ok. I understand

Answer 5

I lost my connection for a second, apologize,

Answer 6

thats fin e

Answer 7

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

Answer 8

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).

Answer 9

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

Answer 10

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) )

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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...

Answer 17

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

Answer 18

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.

Answer 19

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[x^\color{red}{n}\right]=\color{red}{n}x^{\color{red}{n}-1} }\)

this is for any function where variable x is raised to a constant power of n.
---------------------------------------------------------------

Answer 20

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.
----------------------------------------------------------------

Answer 21

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).

Answer 22

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

Answer 23

starting to make sense.

Answer 24

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")

Answer 25

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

Answer 26

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)

Answer 27

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)?

Answer 28

whole those i listed just now are =0.

Answer 29

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.

Answer 30

So lets get to our function s(t)

Answer 31

Ok. into the good stuff

Answer 32

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

Answer 33

do you recongize your function s(t) ?

Answer 34

yes I do

Answer 35

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:

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

Answer 36

so 0 and 0

Answer 37

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

Answer 38

no

Answer 39

\(\large\color{black}{ \displaystyle \frac{d}{dt} \left[\color{blue}{-6}\cdot t^\color{red}{1}\right]=}\)

please take a shot to proceed from here
(if not i got your back)

Answer 40

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....

Answer 41

-6 x nt^1-1

Answer 42

and n would be 1, right

Answer 43

x is × ?

Answer 44

times yes

Answer 45

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

Answer 46

So,
you go -6 × 1 × t^(1-1)

Answer 47

(not go, but got**)

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

any number*

Answer 52

It's 0 and -6

Answer 53

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

Answer 54

yes, 0 and -6,

Answer 55

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

Answer 56

again, derivative of s(t) (of the positon function)

IS, THE VELOCITY.
so your velocity function is v(t)=-6

Answer 57

yes

Answer 58

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

Answer 59

no

Answer 60

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

Answer 61

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

Answer 62

-6

Answer 63

Yup

Answer 64

And that is your final answer.

Answer 65

Thanks man

Answer 66

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

Answer 67

By the way,

Answer 68

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 ×

Answer 69

I mean 0 2 1 5

Answer 70

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)

Answer 71

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