Tangent line toa curve that minimizes the area of a triangle

[NyteBlayde]

Sorry if the title is a bit vague :/

The Problem: Find 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.


Relevant Equations: y = 1-x^2 ; y' = -2x ; A= 1/2bh



I'm basically stuck near square one, I found this site through a google search of the above problem and found a similar topic here, but it wasn't quite the same (or at least I didn't see how to relate my problem to it) so I'd like to ask someone to help me out here :)

What I have so far is that the derivative of the tangent line to the curve is -2x. Where should I go from here?

 

[NyteBlayde]

Aha

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). All I had to do was change A= 1/2 bh to A= 1/2 xy and I know that any point on the parabola is P(p,1-p^2) so I was able to go from here :) Always makes me feel good when I figure things out :D

 

[rocomath]

Ah! I was trying for you on/off :p GJ! At least now I've gained another way of approaching random problems.

 

[NyteBlayde]

hmmm, i may be stuck again, if i find A(p) by setting it equal to 1/2 * (p^2+1)/2p * (p^2+1) i get that A(p) = p^4+2p^2+2/4p and by differentiating this to find its critical numbers to find a min, i get -4p^2+40p+8/4p^4. Now, in order for this to be 0, -4p^2+40p+8 has to be 0. I've tried to use the quadratic formula for this but I keep getting weird roots that don't even make it 0. Am I doing something wrong? sorry for the early cancellation lol :/ :stumped:

I'll keep trying but any help is appreciated :)

 

[rocomath]

I'm not sure what you did, but this is how I'm starting out.

A=\frac 1 2 xy
y=1-x^2

A(x)=\frac 1 2 x(1-x^2)
A(x)=\frac 1 2 (x-x^3)

A'(x)=\frac 1 2 (1-3x^2)

Setting it equal to 0, and then plugging in the values found into my y-eq.

 

[NyteBlayde]

Wow i completely overlooked that :/ I used a wrong reference point, man i did a lot of extra worthless work :/ at least it was kind of fun and interesting lol Thanks a ton I'll see if I can finish with that, I'm working on a couple other problems now, 3 to be exact and I think I'm doing ok, only 1 is giving me problems right now :/ maybe you can help :)

it appears simple : find highest and lowest points of curve x^2+xy+y^2=12

By implicit differentiation i get that y' = 2x+y / 2y+x . Is that right? if so I don't know what i can do to move on with the problem. I suppose I have to make it zero somehow ?? its always the simple concepts that get me...

 

[rocomath]

Lol, at least you got both of us started on that problem. I was lost on what to do :-]

I think you forgot the product rule for your middle term, xy. You weren't given any points?

Well by just plugging in values, the points are P(2,2) b/c that equals 12.

 

[NyteBlayde]

no the exact wording is: "Find the highest and lowest points on the curve x^2+xy+y^2=12"

I did D(x) (x^2 + xy + y^2) = D(x) (12)

= 2x + y +xy' + 2yy' = 0

the y + xy' is a result of the product rule (unless i messed up)

=2yy'+xy' = 2x+y

=y'(2y+x) = 2x+y

= y' = 2x+y / 2y+x

 

[rocomath]

no the exact wording is: "Find the highest and lowest points on the curve x^2+xy+y^2=12"

I did D(x) (x^2 + xy + y^2) = D(x) (12)

= 2x + y +xy' + 2yy' = 0

the y + xy' is a result of the product rule (unless i messed up)

=2yy'+xy' = 2x+y

=y'(2y+x) = 2x+y

= y' = 2x+y / 2y+x

Ah yes you're right, I messed up sorry.

 

[NyteBlayde]

heh np, I've done worse :P

 

[rocomath]

Well the way I went about this is, that when y=0 for f', then that tells me that my f is at a max/min, so y'=\frac{2x+y}{2y+x}

set y=0, that gives me y'=2, which is the same as P(2,2) from which our original equation equals 12.

 

[NyteBlayde]

ah, didn't think of that. what about the other extrema? set x=0 ? that would give me y=1/2 no?

 

[rocomath]

I'm not sure about that, I wish I could plug it into my calculator. LOL

 

[NyteBlayde]

lol same, quick question, from y'=2, how did you get P(2,2) I may just be overlooking that aswell. I thought i got it but now i realized i messed up.

 

[rocomath]

Ah, I just read my Calculus book and we need to do it over again. It says to set y'=0, and basically just set the numerator equal to 0.

y'=\frac{2x+y}{2y+x}

y'=0=2x+y

y=-2x

So plug that back into our original equation ...

x^2+x(-2x)+(-2x)^2=12

Set it equal to 0 and solve for x. Then plug it back into the numerator of our y'.

 

[NyteBlayde]

lol ok, from where? :)

 

[rocomath]

lol ok, from where? :)

Ok sorry, refresh! I just updated after figuring out what my book did.

 

[NyteBlayde]

so 3x^2-12=0 ?

 

[NyteBlayde]

wait wait wait duh, can't i just go x^2=4 and my roots are +/- 2 ??

 

[rocomath]

wait wait wait duh, can't i just go x^2=4 and my roots are +/- 2 ??

Yes, those are the roots and just plug back into the numerator of y'.

Are there answers and what book is this from?

 

[NyteBlayde]

Yes, those are the roots and just plug back into the numerator of y'.

Why do I need to plug them into y' ? don't i just use them as x values for the original equation and find y values and see which point is smaller?

And no I don't have answers for these problems in the student edition, it's Single Variable Calculus Fourth Edition by James Stewart (editor?/author?) oh, they're on pg 310 of that book BTW :)

 

[rocomath]

I'm not sure if this explanation will suffice, but we're trying to solve for where y'=0, so we found that it will be 0 when y=-2x. By using our original equation, we find the x values in which y is at a max and plug it back into the numerator of y' that makes it 0.

(Worst explanation ever, lol.)

 

[NyteBlayde]

neither 2 or -2 make it 0 though.

 

[rocomath]

neither 2 or -2 make it 0 though.

Hm, the thing is we set y'=0 already, so we did solve for our condition. We used it to find the max/min of our original equation.

 

[NyteBlayde]

so what do I do with +/- 2, are these the y coords for my min/max? so all i would have to do is find the x value?EDIT: errr sorry, x coords for the min/max so id have to find the y values? which would just be: max: (2,2) min(-2,-2) ?

 

[rocomath]

No x=+/-2

So go back to the numerator for y'=0=2x+y and plug it in there, and it you will get the y values for your max/min.

 

[rocomath]

so what do I do with +/- 2, are these the y coords for my min/max? so all i would have to do is find the x value?


EDIT: errr sorry, x coords for the min/max so id have to find the y values? which would just be: max: (2,2) min(-2,-2) ?

No, y'=0=2x+y

x=+/-2

y=-2x, P(2,-4) and Q(-2,4)

 

[rocomath]

neither 2 or -2 make it 0 though.

Ah, now see it satisfies our conditions that make y'=0

 

[NyteBlayde]

lol sorry, i was just doing a step twice, once on paper once in my head :P so min(2,-4) max(-2,4) correct?
EDIT: lol u beat me to it

 

[NyteBlayde]

3 down 1 togo :D, now i got to go finish that pesky area of a triangle one
thanks a ton for helping me lol, id be up much later if i was trying to go solo :/

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?

 

[rocomath]

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.

 

[NyteBlayde]

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?

 

[rocomath]

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.

 

[NyteBlayde]

lol this going to be kinda ugly, +/- 1/.sqrt 3 (without rationalizing)

 

[rocomath]

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.

 

[NyteBlayde]

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]

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 :)

 

[rocomath]

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.

 

[NyteBlayde]

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

thx again :)

 

[rocomath]

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 )

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.

 

[NyteBlayde]

I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math

 

[rocomath]

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.

 

[NyteBlayde]

Lol, damn you're hardcore.

damn straight :D

 

[rocomath]

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.

 

[NyteBlayde]

lol round 2 was short lived :P same damn answer lmao

 

[rocomath]

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.

 

[NyteBlayde]

agreed, that's why i figured when i found the error while writing it out i figured id tell you :)