I need help with some simple instantaneous change
"Find the instantaneous rate of change of y with respect to x at the specified value of x0
y= 1/x^2; x0=1; x1=2 "
I understand that you use (f(x1)-f(x0))/(x1-x0) which goes to ((1/x1^2)-1)/(x1-1) but when I try and simplify this I cannot get an answer.
Thanks for the help in advance!

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Check your Algebra...
Write it out clearly.
\[\frac{ \frac{ 1 }{ 2^{2} }-\frac{ 1 }{ 1^{1} } }{ 2-1 }\]

- anonymous

Well I have to use x1 in place of the actual x1 number. That question you wrote is part A) to my question and it's -3/4

- anonymous

that's kind of a weird question.. I am playing with my white atm.... just a sec..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

So the way it shows me in the book the answer would look something like this ((1/x^2)-1)/(x-1) which reduces to -(x-1)/x^2 and it tells me to take the limit of x to 1 which makes it 0 which isn't correct

- anonymous

It was answered on yahoo answers to be like -2 or something, I just can't figure out how they got there or if it's even right

- anonymous

still working it out.. simplifying sucks, I understand what you mean..
So you are saying use the x0 = 1 but just use x as the other as in the equation I said before?

- anonymous

Yes that's basically what I understand. I figured out the first 3 of these problems with no problem, but I got to this one and spent a long time trying to figure it out. I could get it simplified to and answer but it wasn't right. So I had no idea where I was messing up. Here's the online thread with this exact question, but no steps on how to get there.http://answers.yahoo.com/question/index?qid=20090211070219AA1iPfX

- anonymous

And I need help on part B, part A was what you explained earlier, and part c I got 2x instead of whatever they got.

- anonymous

oh.. haha.. yeah.. it's right on Yahoo...
This one:
y = 1/x^2; X0 = 1; X1 = 2
a) Find the average rate of change of y with respect to x over the interval [X0, X1]
Average rate of change = (y(x1) - y(x0) ) /(x1 - x0)
for x1 = 2 and x0 =1
average rate of change = (y(2) - y(1) )/(2 -1) =(1/4 -1)/1 = - 3/4
b) Find the instantaneous rate of change of y with respect to x at the specified value of X0.
dy/dx = (-2) 1/x^3
dy/dx(1) = -2
c) Find the instantaneous rate of change of y with respect to x at an arbitrary value of X0.
dy/dx( x0) = -2/(x0^3)
------------------
Lets talk about this... in this window..
Now.. what grade/class you in?

- anonymous

I'm a junior in high school but I am taking a college Calculus 1 class

- anonymous

and you are studying what in this class atm? limits?

- anonymous

No my idiot professor skipped over limits and said we would go back to them, whihc makes no sense to me. But I have already learned limits so it's whatever. We are currently learning "Tangent lines and rates of change" kind of a pre-derivative thingy

- anonymous

using power rule yet?

- anonymous

Sorry? Not sure what that is, so I'm guessing not. I've only been to 2 classes because school started last monday, I have class mwf and wednesday class got cancelled. So we aren't very far at all

- anonymous

ok.. I know where you are then.. Just wanted to ask so I don't tell you the wrong thing...
http://www.sosmath.com/calculus/diff/der00/img14.gif

- anonymous

Yes that looks like what we're doing

- anonymous

This is something to check out later... http://www.sosmath.com/calculus/diff/der00/der00.html
Also google Pauls Online Notes....
This is extra info for you... I know what it is like having a crap teacher.. so this should help..
-----------------
Now.. Lets look at the problem at hand..
First.. Write the problem in the form as the first image link I posted..

- anonymous

So... what does it mean... at x_0.. give me a sec, I am going to try and use the equation writing tools :/

- anonymous

The equation look a little different from how it does in my book. http://www.sosmath.com/calculus/diff/der00/img18.gif it looks like this but with liit of x to x0

- anonymous

\[\lim \Delta t \rightarrow 0 \frac{ (\frac{ 1 }{x_{0} }+\Delta t)-\frac{ 1 }{ x_{0} } }{ \Delta t }\]

- anonymous

wow that was a waste of time... can you tell me what this equation means?

- anonymous

Uhm, I mean I figure you're taking the average rate of change and taking the limit so that it goes infinitely small so that you can find the instantaneous.

- anonymous

Delta t is the same as h right?

- anonymous

Yeah.. by saying that delta t, that's just a time step.. when the book used h, it just means an arbitrary step size.. in real life, we usually use time as the small step. (comes up a lot)

- anonymous

Gotcha

- anonymous

now.. the only issue then you have is the algebra.. Usually people don't get the right answer because they forget to factor something out.. that's where you can beat the zero in the denominator issue..

- anonymous

Keep in mind for the future that sometimes it doesn't work.. you can't get the 0 out of the denominator which usually means that the function is not continuous, and you can't take a derivative of that (at least at that point)

- anonymous

do you know how to the f(x+a) - f(x) part?

- anonymous

Alright, so I'm confused as to what goes into the Delta t/ H spot then.
That's what I got the 1st time around and I factored out everything. But online they had an answer so I was completely confused, seeing that I got the first three questions right.

- anonymous

used a instead of delta t or h, just to keep you on your toes and just remember that the name is arbitrary.... and it's easier to type

- anonymous

Hahah yeah, so still what goes there?

- anonymous

ok.. lets makes some mathemagic here....
how to do you do f(x+a) where f(x) = 1/x <--- dropped the square for simplity

- anonymous

It's just 1/(x+a)

- anonymous

nice.. and then we get 1/(x+a) - 1/x

- anonymous

follow?

- anonymous

Yes, and then that's all over a. And then we multiply top and bottom by (x+a)(x) to get rid of the fractions, right?

- anonymous

I am breaking this down into tiny steps.. :)
not being insulting....

- anonymous

Hahah it's okay

- anonymous

Oh but wait it's squared so it would be x^2+2xa+h^2

- anonymous

*a sorry used to using h

- anonymous

yup
we want to factor out the h though

- anonymous

which ever.. a or h :p

- anonymous

Yeah so you have 2x+h ? that's what I got before at least. and then h to means we are left with 2x. (That's the answer that I got for part c, not part b, part b want's something different)

- anonymous

h to 0 ***

- anonymous

*wants -.-

- anonymous

heh... nice..
now.. you are kind misreading the directions.. they are easier than you think..
You got C right? -2/x_0^3

- anonymous

kind of*

- anonymous

No, I got 2x from factoring like I said in the previous one.

- anonymous

"Yeah so you have 2x+h ? that's what I got before at least. and then h to means we are left with 2x. "

- anonymous

ah I gotcha.. the algebra still not coming right...
I know -2/x^3 is infact the right answer..
I am working on my bored trying to figure where you are going with the definition... let me work the definition.. one sec

- anonymous

are you doing it by x^(-2)?

- anonymous

Wait wait. I don't get the last message haha

- anonymous

I'm reworking trying to get -2/x^3

- anonymous

I got 2/x^2 though o.O

- anonymous

-((h + x)^2/(x^2 (h + x)^2)) + x^2/(x^2 (h + x)^2)
you getting to this step?

- anonymous

getting closer :)

- anonymous

This problem is more geared to testing your algebra skills than calculus skills.

- anonymous

No, I prefer to get rid of fractions by multiplying by a common denominator. So basically I did \[ \frac{ \frac{ 1 }{ x-a } - \frac{ 1 }{ x^2 } }{ a }\] and then multiplied by (x-a) squared (sorry forgot to put it in there) and then x^2

- anonymous

I am writing it up.. in mathematica.. it's not a lack of understanding on your part.. just algebra.. algebra takes me down sometimes.

- anonymous

D[1/x^2, x]
-(2/x^3) <--- correct answer
-------------------
(x + h)^-2 - x^-2 // Simplify
-(1/x^2) + 1/(h + x)^2
-((h + x)^2/(x^2 (h + x)^2)) + x^2/(x^2 (h + x)^2)
Numerator:
-(h + x)^2 + x^2 // Simplify
-h (h + 2 x)
Denominator:
x^2 (h + x)^2 // Expand
h^2 x^2 + 2 h x^3 + x^4
Factor out ONE!! h in the denominator and then let all h -> 0
What you get left over is - 2/x^3

- anonymous

I don't know why I wrote ONE!! but, just did..

- anonymous

In[1]:= D[1/x^2, x]
Out[1]= -(2/x^3)
Numerator:
In[7]:= -(h + x)^2 + x^2 // Simplify
Out[7]= -h (h + 2 x)
Denominator:
In[10]:= x^2 (h + x)^2 // Expand
Out[10]= h^2 x^2 + 2 h x^3 + x^4
Factor out one h in the denominator and then let all h -> 0
What you get left over is - 2/x^3

- anonymous

Okay gimme a minute lemme look over it

- anonymous

-(1/x^2) + 1/(h + x)^2
-((h + x)^2/(x^2 (h + x)^2)) + x^2/(x^2 (h + x)^2)

- anonymous

Not sure how you got from the first part to the scond part. Honestly I'm not sure what the second part even looks like haha

- anonymous

bah.. I left something out..

- anonymous

The numerator and the denominator was all supposed to be over h

- anonymous

Limit h -> 0 ((-h (h + 2 x))/(h^2 x^2 + 2 h x^3 + x^4))/h
Should be like this before you kill out the h's

- anonymous

Okay, I mean I think I got it I can do algebra, everything jus got a little mixed up probably. I got super frusterated with everything cause of my teacher. I have a lot of resources left to look at and I'm starving so thanks for all the help! One I decipher and write out everything you wrote I'll get it!

- anonymous

-((h + 2 x)/(h^2 x^2 + 2 h x^3 + x^4))
and then take the limit....

- anonymous

oh ok.. Well that's pretty much it.. At least rest a sure that once you learn this.. or so this.. you never have to do it again :P

- anonymous

Hahah yay!

- anonymous

but just to know.. that's the answer above.. once you take the limit as h -> 0 the x in the numer takes the x^4 to x^3..
Good luck with your class. Take care.. Give me a metal if my help was worth it.. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.