Variation statement into graphs- Right?

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  • #1
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My first post, yay i already like the atmosphere here :P anyway....

Using the formula F = kQq/R2 sketch graphs between
a. F and Q (k,q, and R are constant)

b. F and R (Q,q and k are constant)

c. Q and R (k,q and F are constant)


I think i did it correctly but I'm not quite sure.

a) I wrote that F is directly proportional to Q so i guess its Direct Variation graph.

b) Is 1 / R^2 inversely proportional? And is the graph the same as y= 1/x or should it be y=SQUAREROOT of 1/x but then it wouldn't make sense?

c) i got it so that it's R = SQUAREROOT of Q so it's direct root graph?

I hope that's correct! Can anybody double check?

----------------
 

Answers and Replies

  • #2
PeterO
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My first post, yay i already like the atmosphere here :P anyway....

Using the formula F = kQq/R2 sketch graphs between
a. F and Q (k,q, and R are constant)

b. F and R (Q,q and k are constant)

c. Q and R (k,q and F are constant)


I think i did it correctly but I'm not quite sure.

a) I wrote that F is directly proportional to Q so i guess its Direct Variation graph.

b) Is 1 / R^2 inversely proportional? And is the graph the same as y= 1/x or should it be y=SQUAREROOT of 1/x but then it wouldn't make sense?

c) i got it so that it's R = SQUAREROOT of Q so it's direct root graph?

I hope that's correct! Can anybody double check?

----------------

Your (a) sounds good

In (b) you said it was 1 / R^2, then immediately inversely proportional. What happened to the square?

(c) is a true statement, but why express R in terms of Q, when the question said Q in terms of R ?
 
  • #3
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Your (a) sounds good

In (b) you said it was 1 / R^2, then immediately inversely proportional. What happened to the square?

(c) is a true statement, but why express R in terms of Q, when the question said Q in terms of R ?
Thanks for your reply,
I suspected it could be Square.root 1/R.

and does it really make a difference if you express it in the other way?
 
  • #4
PeterO
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Thanks for your reply,
I suspected it could be Square.root 1/R.

and does it really make a difference if you express it in the other way?
I have not seen many people claim the radius of a circle varies as the square root of the Area when asked to draw a graph showing the relationship between the Area and radius of a circle.

The test of whether it makes any difference would be when you change the variables you graph to get a straight line so that you can determine the complete relationship.

And I don't think the second one is Square.root 1/R. The way you are writing them, it could be Square.root 1/F
 
  • #5
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I have not seen many people claim the radius of a circle varies as the square root of the Area when asked to draw a graph showing the relationship between the Area and radius of a circle.

The test of whether it makes any difference would be when you change the variables you graph to get a straight line so that you can determine the complete relationship.

And I don't think the second one is Square.root 1/R. The way you are writing them, it could be Square.root 1/F
I'm really confused :( I'm trying to understand it but i'm having trouble grasping the concept - F is proportional to R^2 since the R^2 came from F=kQq/R^2. So why doesn't that make it an inversely proportional graph or what is it even supposed to be- this is really frustrating :(
 
  • #6
PeterO
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I'm really confused :( I'm trying to understand it but i'm having trouble grasping the concept - F is proportional to R^2 since the R^2 came from F=kQq/R^2. So why doesn't that make it an inversely proportional graph or what is it even supposed to be- this is really frustrating :(

F is proportional to 1 / R^2 - but I think you just mis-wrote it.

if you were to graph y = 1/x you would get a graph showing inverse proportion.

But there is also y = 1/x^2 - the inverse square, y = 1/x^3 , the inverse cube etc.
 
  • #7
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Alright that makes sense, thanks for all your help!
 

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