Tangent line toa curve that minimizes the area of a triangle

In summary: I've tried this a few different ways and I keep getting different answers. For example, when x=0, I get y'=-2 and when x=1, I get y'=4. I'm not sure what I'm doing wrong.In summary, the problem is finding the point on the parabola y=1-x^2 at which the tangent line cuts from the first quadrant a triangle with the smallest area. Through a closer look at the problem and running through it a lot in my mind and looking again at the other topic I think I have figured it out. Thanks to anyone who was or is working on replying (if any, lol).
  • #36
yuck i hate dealing with radicals lol

ok, i think i got it. if I am right, the extremes of the area are at P(.sqrt3 / 3 , .sqrt3 / 9) and Q( -.sqrt3 / 3 , -2.sqrt3 / 9 )
 
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  • #37
Alright, once again thanks for all your help lol, couldn't have done it without you :D. Now i can finalize these in word and put them away :)
 
  • #38
You're only value should be [tex]x=\frac {\sqrt3}{3}[/tex] and plug in it [tex]y=1-x^2[/tex] for your corresponding y-value which is your height.
 
  • #39
lmao i feel so retarded, i pluged it into the area function instead of the original >.< good thing you caught me

thx again :)
 
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  • #40
NyteBlayde said:
yuck i hate dealing with radicals lol

ok, i think i got it. if I am right, the extremes of the area are at P(.sqrt3 / 3 , .sqrt3 / 9) and Q( -.sqrt3 / 3 , -2.sqrt3 / 9 )

NyteBlayde said:
lmao i feel so retarded, i pluged it into the area function instead of the original >.< good thing you caught me

thx again :)
I'm not sure if you read post #35, but for this problem you can only take the positive x values b/c x represents the base which is a measure of the length of the triangle and it can only be positive.

Also, you're corresponding y-value should also be positive b/c you want to minimize this triangle so that it is in the 1st quadrant.

If all your values are positive, that means you probably did the question correctly!

It satisfies the physical conditions along with the xy coordinate system.
 
  • #41
I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math
 
  • #42
NyteBlayde said:
I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math
Lol, damn you're hardcore.
 
  • #43
rocophysics said:
Lol, damn you're hardcore.

damn straight :D
 
  • #44
AHHH! Lol, round 2 to fix the error?

[tex]y'=\frac{-(2x+y)}{x+2y}[/tex]

[tex]y'=0=-(2x+y)[/tex]

[tex]y=-2x[/tex]

OHHH! Big break :-] I guess I can't trust my intuition.
 
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  • #45
lol round 2 was short lived :P same damn answer lmao
 
  • #46
NyteBlayde said:
lol round 2 was short lived :P same damn answer lmao
Yeah lucked out, lol. It worked out sweet since both terms were the same sign, whew! I was like damn all that work for nothing, and plus I hate to leave things incorrect.
 
  • #47
agreed, that's why i figured when i found the error while writing it out i figured id tell you :)
 

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