Question

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!

Answer 1

Check your Algebra...
Write it out clearly.

\[\frac{ \frac{ 1 }{ 2^{2} }-\frac{ 1 }{ 1^{1} } }{ 2-1 }\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

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.

Answer 9

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)

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Lets talk about this... in this window..

Now.. what grade/class you in?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

using power rule yet?

Answer 14

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

Answer 15

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

Answer 16

Yes that looks like what we're doing

Answer 17

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..
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Now.. Lets look at the problem at hand..

First.. Write the problem in the form as the first image link I posted..

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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.

Answer 23

Delta t is the same as h right?

Answer 24

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)

Answer 25

Gotcha

Answer 26

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..

Answer 27

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)

Answer 28

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

Answer 29

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.

Answer 30

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

Answer 31

Hahah yeah, so still what goes there?

Answer 32

ok.. lets makes some mathemagic here....

how to do you do f(x+a) where f(x) = 1/x <--- dropped the square for simplity

Answer 33

It's just 1/(x+a)

Answer 34

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

Answer 35

follow?

Answer 36

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?

Answer 37

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

Answer 38

Hahah it's okay

Answer 39

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

Answer 40

*a sorry used to using h

Answer 41

yup

we want to factor out the h though

Answer 42

which ever.. a or h :p

Answer 43

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)

Answer 44

h to 0 ***

Answer 45

*wants -.-

Answer 46

heh... nice..

now.. you are kind misreading the directions.. they are easier than you think..

You got C right? -2/x_0^3

Answer 47

kind of*

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

I got 2/x^2 though o.O

Answer 55

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

you getting to this step?

Answer 56

getting closer :)

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

Okay gimme a minute lemme look over it

Answer 64

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

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

Answer 65

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

Answer 66

bah.. I left something out..

Answer 67

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

Answer 68

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

Answer 69

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!

Answer 70

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

Answer 71

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

Answer 72

Hahah yay!

Answer 73

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.. :)