- #1
Mr. Fest
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I don't quite understand the context of common tangent problems.
This is one of the problems I am trying to solve:
Prove that there is a line that is a common tangent to the parabolas y = x2 and y2 = x.
This is how I tried to solve it at least:
y2 = x --> y = [itex]\pm[/itex][itex]\sqrt{x}[/itex]
CASE I:
y = [itex]\sqrt{x}[/itex]
We have two derivatives:
One is: 2x
The other is: [itex]\frac{1}{2}[/itex]*[itex]x^{-1/2}[/itex]
Set these equal to each other and find that x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex]
--> The tangent has the slope [itex]\frac{1}{2^{1/3}}[/itex]
In this case, the tangents have the same slope for the same x-value --> It is NOT a common tangent.
CASE II:
y = -[itex]\sqrt{x}[/itex]
Here we have that the tangent of y = -[itex]\sqrt{x}[/itex] has the negative slope of [itex]\sqrt{x}[/itex] for the same x-value.
And for y = x2, the tangent has a same-value but negative slope for the negative x-value, that is y'(-x) = -y'(x)
In our case this gives us that:
The slope for y = -[itex]\sqrt{x}[/itex] at x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex] and the slope for y = x2 at x = -([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex].
Therefore, there is a line that is a common tangent for y = x2 and y2 = x
Is this, the correct way to solve this problem or would you suggest some other way that is appliable for other problems of this sorts where you have to find common tangents...
Thanks a whole lot in advance.
Mr. Fest
This is one of the problems I am trying to solve:
Prove that there is a line that is a common tangent to the parabolas y = x2 and y2 = x.
This is how I tried to solve it at least:
y2 = x --> y = [itex]\pm[/itex][itex]\sqrt{x}[/itex]
CASE I:
y = [itex]\sqrt{x}[/itex]
We have two derivatives:
One is: 2x
The other is: [itex]\frac{1}{2}[/itex]*[itex]x^{-1/2}[/itex]
Set these equal to each other and find that x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex]
--> The tangent has the slope [itex]\frac{1}{2^{1/3}}[/itex]
In this case, the tangents have the same slope for the same x-value --> It is NOT a common tangent.
CASE II:
y = -[itex]\sqrt{x}[/itex]
Here we have that the tangent of y = -[itex]\sqrt{x}[/itex] has the negative slope of [itex]\sqrt{x}[/itex] for the same x-value.
And for y = x2, the tangent has a same-value but negative slope for the negative x-value, that is y'(-x) = -y'(x)
In our case this gives us that:
The slope for y = -[itex]\sqrt{x}[/itex] at x = ([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex] and the slope for y = x2 at x = -([itex]\frac{1}{16}[/itex])[itex]^{1/3}[/itex] is -[itex]\frac{1}{2^{1/3}}[/itex].
Therefore, there is a line that is a common tangent for y = x2 and y2 = x
Is this, the correct way to solve this problem or would you suggest some other way that is appliable for other problems of this sorts where you have to find common tangents...
Thanks a whole lot in advance.
Mr. Fest
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