Tangent line toa curve that minimizes the area of a triangle

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SUMMARY

The discussion centers on finding the point on the parabola defined by the equation y = 1 - x² where the tangent line creates a triangle with the smallest area in the first quadrant. The area A of the triangle is expressed as A = 1/2 * xy, leading to the derivative A'(x) = 1/2 * (1 - 3x²). Participants explore critical points by setting A'(x) to zero, ultimately leading to the conclusion that the minimum area occurs at x = ±√3/3, with corresponding y-values calculated from the original parabola equation.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and critical points.
  • Familiarity with the properties of parabolas, particularly the equation y = 1 - x².
  • Knowledge of geometric principles related to the area of triangles.
  • Experience with implicit differentiation and solving quadratic equations.
NEXT STEPS
  • Study the application of derivatives in optimization problems, focusing on finding minima and maxima.
  • Learn about the geometric interpretation of derivatives in relation to tangent lines and areas.
  • Explore the use of the quadratic formula in solving for critical points in polynomial equations.
  • Investigate the implications of physical constraints in optimization problems, such as limiting values to positive domains.
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Students and educators in calculus, mathematicians interested in optimization problems, and anyone studying the geometric properties of functions and their derivatives.

  • #31
Lol, I made that mistake too but I realized no that can't be right since I have to move 2x to the other side :-]

But yeah, that was a good problem.
 
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  • #32
REPOST: wasnt sure if ud see it no since this is page 3 lol
thanks a ton for helping out on that one :D

back on the topic of the first question, u set 1/2(1-3x^2) = 0

so i should end up with 1/2 - 3/2x^2 = 0 and all i need to do is solve for x and then find my min right?
 
  • #33
Instead of multiplying the half through, just divide and get rid of it. Then yes just find where x=0, which is a critical point and plug it back into your original.
 
  • #34
lol this going to be kinda ugly, +/- 1/.sqrt 3 (without rationalizing)
 
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  • #35
NyteBlayde said:
lol this going to be kinda ugly, +/1 1/.sqrt 3 (without rationalizing)
Yeah that's what I got too, but remember, x=base which is representation of length so we take only the positive value.
 
  • #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 x=\frac {\sqrt3}{3} and plug in it y=1-x^2 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?

y&#039;=\frac{-(2x+y)}{x+2y}

y&#039;=0=-(2x+y)

y=-2x

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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