Find explicit function or functions corresponding from implicit

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  • #1
MrNeWBiE
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find explicit function or functions corresponding from implicit ,,,

Homework Statement



1) 4x^2+9y^2=36

2)xy^2 + (x^2-1)y-x=0

The Attempt at a Solution



1) y=√((36-4x^2)/9)==>6/3√(-4x^2)) = 2√(-4x^2))

and the book answer is ±2/3 √(9-x^2)

can someone tell me what's my mistake ?

2) well i could only expand then i stop

xy^2 + yx^2-y -x=0

how can i make the y in one side if i have y^2 and y in the same time ,,?
 
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Answers and Replies

  • #2
tiny-tim
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Hi MrNeWBiE! :smile:

(try using the X2 tag just above the Reply box :wink:)
√((36-4x^2)/9)==>6/3√(-4x^2))

uhh? what happened to the 36 ? :confused:
xy^2 + yx^2-y -x=0

Treat it as an ordinary quadratic equation in y :wink:
 
  • #3
MrNeWBiE
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the 36 ==> its become 6 ,,,,
 
  • #4
Martin Rattigan
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You have taken out a factor of 6 from the square root, corresponding to the 36 inside the square root. This would need to divide the whole numerator inside the root and it doesn't divide the [itex]4x^2[/itex] term. Also the 36 inside the root has effectively disappeared after "==>".

Try

[itex]y=\sqrt{(36-4x^2)/9}=\sqrt{4(9-x^2)/9}=\frac{2}{3}\sqrt{9-x^2}[/itex]

Notice that if you arranged the original equation as:

[itex]9y^2+(4x^2-36)=0[/itex]

Then setting [itex]A=9[/itex], [itex]B=0[/itex] and [itex]C=4x^2-36[/itex] we have

[itex]Ay^2+By+C=0[/itex] when you can say

[itex]y=\frac{-B\pm\sqrt{B^2-4AC}}{2A}=\frac{\pm\sqrt{-4(9(4x^2-36))}}{18}[/itex]

which gives the same answer.

You may like to think how you could use this in your second question.

In fact you won't get a function in either case because the [itex]\pm[/itex] sign means you will get two possible answers for [itex]y[/itex] for each value of [itex]x[/itex], contradicting the definition of a function which only allows one.
 
  • #5
MrNeWBiE
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i will try with the 2nd and i will show you what i will got ,,,,
 
  • #6
MrNeWBiE
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in the begain ,,,,
yx^2-y ,,, how can i got B ?
 
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  • #7
Martin Rattigan
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It's already written as a quadratic equation in y, which should give you a formula for y in terms of x. (Notice that I arranged it as a quadratic equation in y in my example.) If you start of as you have done you will get a quadratic equation in x, which will give a formula for x in terms of y. You could do either, but I would guess the book will do the former. This would also be easier because you should then be able to just read the value of B from the formula originally given.
 
  • #8
MrNeWBiE
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aha

so i should factor ? ,,,,

the answer will be y= -x ,, y= 1/x ?
 
  • #9
Martin Rattigan
330
3


Well it wasn't what I was suggesting, but it would be neater to factorize the left hand side. Then you can say if the product of the two factors is 0, one of the factors must be zero.

That gives you two alternative expressions for y on terms of x.
 

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