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?