Question

quick clarification

Answer 1

\[y = 400-t^{2}\]

Answer 2

find the velocity when t = 3

Answer 3

using

\[\frac{ f(x+h)-f(x) }{ h }\]

Answer 4

Velocity is the slope of y (which I presume is distance)

Answer 5

Don't know what's going on with my OS

Answer 6

again, do the same
\[f(x+h) = 400 - (x+h)^2\]and so on

Answer 7

Yea laggy for me as well

Answer 8

Was wondering @agent47

if we had to say evaluate it like if these are the proper functions

\[f(t) = 400-t^{2}\]


\[f(t+h) = 400-4(t+h)^{2}\]

Answer 9

wait that's a typo the 4 in 4(t+h)^{2} shouldn't be there

Answer 10

velocity is equal to dt/dx so its basically just applying the derivative to the given equation then substituting t= 3 into the derived equation to produce the velocity at t=3

Answer 11

Using the difference formula like here would just be substituting any x in your equation for the specific F(something)

Answer 12

I presume that they haven't done differentiation yet - this is the limit definition of a derivative.

Answer 13

\[\lim_{h \rightarrow 0} \frac{ 400-(t+h)^{2}-400-t^{2} }{ h }\]

Answer 14

Well in the form its in its the difference formula unless you drive h to 0

Answer 15

and that changes things...

Answer 16

yep will be doing that soon, so far working with the definition of a derivative

Answer 17

400 + t^2, you forgot to distribute the - in

Answer 18

\[\lim_{h \rightarrow 0} \frac{ 400-(t+h)^{2}-400+t^{2} }{ h }\], Now expand and simplify, and take the limit, you will end up with an equation in terms of t in the end

Answer 19

You mean like this right?

\[\frac{ 400-t^{2}-2th-t^{2}-400+t^{2} }{ h }\]

Answer 20

\[\frac{ 400-t^{2}-2th-h^{2}-400+t^{2} }{ h }\]

Answer 21

argh!! that should be a h

Answer 22

Yes, now just sub in your t=3 and take the limit as h goes to 0

Answer 23

\[\frac{ -2th-h^{2} }{ h } = \frac{ -h(2t+h) }{ h } = |_{h=0}-(2t+h) = -6 \frac{ m }{ s }\]

Answer 24

Okay I think I got the hang of this now

Answer 25

Yup, so the car is actually travelling backwards at 6m/s

Answer 26

It seems that the limit is the derivative

Answer 27

Yes, that is the limit definition of the slope of the tangent line, which also happens to be the first derivative.

Answer 28

I mean the equation of the tangent line, not the actual slope.

Answer 29

Slope comes in when you're asked to do something like at t=3 or 4 or whatever

Answer 30

so it gives us an equation for the tangent line?

Answer 31

The limit isn't always necessarily the derivative though. Only when driving the h to 0 in this case of the difference equation is what produces an equation to find the slope of a tangent line at any point.

Answer 32

so in this case it gives you the equation for the tangent line that can be used to find the slope at any point. but in this case it's just a constant

Answer 33

just curious why isn't it always equal to the derivative @sweetburger

Answer 34

You will probably see h be rewritten to \[\Delta x\] eventually in the limit definition which stands for a change in x

Answer 35

well just simply the (limx-->2 of y=x) is not the derivative. Your just still finding the limit of when x approaches 2. It doesn't produce a derivative.

Answer 36

I think I am making this more convoluted and you probably already understand this so sorry...

Answer 37

the limit actually is the derivative when your final result does not ask you to substitute for the variable at a given point, so in this case:\[\lim_{h \rightarrow 0} \frac{ 400-(t+h)^{2}-400+t^{2} }{ h }\]=\[\lim_{h \rightarrow 0} \frac{ 400-(t^2+2th+h^2)-400+t^{2} }{ h }\]=\[\lim_{h \rightarrow 0} \frac{ 400-t^2-2th-h^2-400+t^2 }{ h }=\lim_{h \rightarrow 0} \frac{h(-2t-h)}{h}=-2t=f'(t)\]

Answer 38

The slope of the tangent line at a given point is merely the derivative evaluated at that point, in our case:\[f'(3)=-2*3=-6\]

Answer 39

Think of it in a sense.. what was the instantaneous velocity at a given time versus what function would i use to evaluate the velocity for a given time?

Answer 40

I think we were arguing two different points.

Answer 41

nonetheless what @Agent47 said is a great explanation on this

Answer 42

The derivative gives you the the slope of the tangent line, but you still have to find the intercept.

Answer 43

Okay so in this problem the limit would give you the instantaneous velocity,

but the derivative gives you a function that would be used to find the velocity at any point in time.

Answer 44

No in this case you still had to plug in t=3 into what came out of the limit

Answer 45

The derivative of position with respect to time gives you velocity.

Answer 46

Limit still just gave you a function, it's just that you substituted t=3 while you were calculating that limit - you could have also kept it in terms of t until the very end, which is the idea behind this. But usually substituting earlier makes the equation simpler.

Answer 47

okay so we did get an equation, which was the derivative i.e. velocity it's just that we evaluated it at t = 3

Answer 48

yes

Answer 49

Another great way to think of this, as @sweetburger mentioned is in terms of change of x. How does my distance change as time goes on, so like how did my distance change from hour 1 to hour 2, or second 1 to second 2, etc.. So the delta gets smaller and smaller which then ends up being a delta at a given snapshot of time. And that change actually depends on your speed.

Answer 50

And by delta I mean \[\Delta t\]

Answer 51

So not in terms of x, but in terms of t (sorry)

Answer 52

To me, I think of derivative as \[
\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}
\]I mean, I don't read into it much beyond that. I understand that \(\Delta y/\Delta x\) is just a ratio between the change in variables.

Answer 53

Thanks for the explanation everyone!

Answer 54

In fact, I don't think of derivative as necessarily something you do to functions, I think of it as something you do to relations.

Answer 55

what do you mean by that @wio?

Answer 56

For example \[
y^2 = x^2+2
\]You can't use a function to represent \(y\) or \(x\) in terms of each other, but you can differentiate this equivalence relation.

Answer 57

Implicit differentiation (don't worry about it if you haven't gotten there yet) with respect to \(x\) gives you:\[
2y\frac{dy}{dx}=2x
\]

Answer 58

apply the operator d/dx to both sides so you take the derivative of each side with respect to x

Answer 59

However, there does not exist a function such that \(y=f(x)\) or \(x=f(y)\).
There just exists an equivalence relation between the two.

Answer 60

Couldn't we just rewrite this? as

\[y^{2} = x^{2}+2; y = \sqrt{x^{2}+2}; y = x+\sqrt{2}~? \]

Answer 61

Well, technically: \[
y=\sqrt{x^2+1}\quad \text{ or }\quad y=-\sqrt{x^2+1}
\]

Answer 62

This is why it isn't a function. There are two different \(y\) values for each \(x\).

Answer 63

Also, unfortunately \(\sqrt{x^2+1} \) does not always equal \(x+\sqrt 2\).
1. Square roots are exponents, which don't distribute over addition.
2. \(\sqrt{x^2}=|x|\).

Answer 64

These are the finer details that used to bite me in the butt as well, though

Answer 65

yep I tried plugging in a number lol the simple way just to see. that's not a function.

Answer 66

You can graph it, even though it isn't a function. It's a hyperbola.

Answer 67

so i'm guessing you can't figure out the derivative of this because it's not a function.. because i guess to me it seems the derivative gives the relationship between two variables. but there doesn't seem to be one it's not a function.

Answer 68

like I guess there wouldn't be a relationship because you are getting multiple values for y?

Answer 69

All functions are relations, but not all relations are function. I'm saying you can differentiate equivalence relations.

Answer 70

Unlike functions, relations allow you to set \(x\) to a value and get multiple \(y\) values.

Answer 71

But it is true that most of the time derivatives are spoken about in the context of functions.

Answer 72

yeah, so I guess you can still differentiate relations?

Answer 73

Yes, it is called implicit differentiation.

Answer 74

interesting i'm going to re-read this again tomorrow thanks for you help everyone