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Simple equation.

  • Thread starter ritwik06
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1. Homework Statement
For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
1) y increases linearly with x
2) y increases as the square of x
3) y increases as y^1/2
4) If x increases, y can increase or decrease.


2. Homework Equations
y is directly proportional to x. as a is a constant, the first option would be right!


3. The Attempt at a Solution

Is my approach right?
 

Answers and Replies

arildno
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Not all, I'm afraid. Rewrite your equation as:
[tex](y-(\sqrt{a})x)(y+(\sqrt{a})x)=0[/tex]
Now, two questions:
1. Why is this rewriting permissible?
2. What must you conclude from it?
 
Last edited:
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Not all, I'm afraid. Rewrite your equation as:
[tex](y-(\sqrt{a})x)(y+(\sqrt{a})x)=0[/tex]
Now, two questions:
1. Why is this rewriting permissible?
2. What must you conclude from it?
Ans 1 because the difference of two squares can be factorised as suc:
a^2-b^2=(a+b)(a-b)

Ans 2. I dont know what can be concluded! Please be clear!
 
arildno
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Well, if two numbers multiplied together is zero, what can we say about those numbers?
 
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Well, if two numbers multiplied together is zero, what can we say about those numbers?
That any one of them is equal to zero or both are zero.
 
arildno
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So,
EITHER:
[tex]y=\sqrt{a}x[/tex]
OR
[tex]y=-\sqrt{a}x[/tex]

Given that knowledge, which statement is true?
 
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So,
EITHER:
[tex]y=\sqrt{a}x[/tex]
OR
[tex]y=-\sqrt{a}x[/tex]

Given that knowledge, which statement is true?
The fourth option??? :approve:
 
arildno
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Yup :smile:
 
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but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!! :approve:
 
arildno
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Three flaws:

1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
But that is a totally different issue.
 
HallsofIvy
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but sir. I agree totally with you. But if you look at it straight away, it seems that y is directly proprtional to x. hence they will have a linear relationship. How will you criticise this point? Thanks for the effort!!!! :approve:
This is really only echoing arildno but:

I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.
 
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1. Homework Statement
For the graph of equation (y^2)=a(x^2), where a is a positive constant, whic of the following is true?
1) y increases linearly with x
2) y increases as the square of x
3) y increases as y^1/2
4) If x increases, y can increase or decrease.


2. Homework Equations
y is directly proportional to x. as a is a constant, the first option would be right!


3. The Attempt at a Solution

Is my approach right?
Three flaws:

1. As for direct proportionality (i.e, ratio constant), remember that the proportionality constant in such a relation can be positive or negative, so that from mere direct proportionality, you cannot conclude that y will increase with x, it could equally well decrease with x.

2. If we assume that the y-coordinate is to be regarded as a FUNCTION of the x-coordinate (so that we limit ourselves to only a part of the permissible solution pairs of the original equation), we might have a discontinuous function relationship in that in one x-region, the function values followed the upper line, and in an adjoining x-region, the function values follow the lower line.
In this case, we see that the ratio x/y is NOT constant, that is direct proportionality do not hold between x and y

3. There is no reason why we should assume that there exist any function relating x-values to y-values. Indeed, the solution set to the equation do not permit a function construction in general.

To be sure, if we limit ourselves to just a part of the solution set, then we may, indeed, have some functional relation between the picked (x,y)-values.
But that is a totally different issue.

If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??


This is really only echoing arildno but:

I would "criticise" that point by saying "Yes, it true- but off the point." The answer did not say "linearly related", it said "increases linearly". That was arildno's point.
Whats the difference between "inearly related" and "increases linearly"?
 
HallsofIvy
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If I read the question again, it says that a is "positive constant". Are you stressing on the point that the squares of a negative number is always positive??
Yes, and so with y2= ax2 we can have either
[itex]y= \sqrt{a}x[/itex] or [itex]y= -\sqrt{a}x[/itex]


Whats the difference between "inearly related" and "increases linearly"?
In both [itex]y= \sqrt{a}x[/itex] and [itex]y= -\sqrt{a}x[/itex] y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.
 
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Yes, and so with y2= ax2 we can have either
[itex]y= \sqrt{a}x[/itex] or [itex]y= -\sqrt{a}x[/itex]



In both [itex]y= \sqrt{a}x[/itex] and [itex]y= -\sqrt{a}x[/itex] y and x are "linearly related". In the first, y "increases linearly" with x but in the second y "decreases linearly" with x.
Thanks a lot for the efforts. o:)
 

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