tangent plane of (x^2/a^2)+ (y^2/b^2)+(z^2/c^2)=1 ?

- anonymous

tangent plane of (x^2/a^2)+ (y^2/b^2)+(z^2/c^2)=1 ?

- schrodinger

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- anonymous

\[xx ^{*} /a ^{2} + yy^*/b^2 + zz^*/c^2 =1 ????\]

- anonymous

tangent plane at what point?

- anonymous

at a general point x^* y^* z^*

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## More answers

- anonymous

okay change of answer
it is x^*/a^2 +y^*/b^2 +z^*/c^2 =1 i think thats the correc tangent plane eq

- anonymous

you need to take derivative of the parabola to get the tangent

- anonymous

a parabola? btw its cal 2 and vector algebra

- anonymous

x y and z are independent of each other.

- anonymous

scrap that last post.

- anonymous

rsaad, you there?

- anonymous

yea

- anonymous

You can find the equation for the tangent plane at the point (x_0,y_0,z_0) by taking the gradient of your function, evaluating at that point (the grad. will give you a vector that is perpendicular to the plane) and sub. in those values into the equation for a plane.

- anonymous

So...

- anonymous

i did calcu;at the equation of tangent plane.

- anonymous

\[f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0\]

- anonymous

Oh...what's wrong then?

- anonymous

the problem i s i am getting the volume in negative.. =(

- anonymous

\[\nabla f=(\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})\]

- anonymous

Volume? You don't need volume here.

- anonymous

i took the triple product. should i attach my cal. ?

- anonymous

Wait and see what I get, then you can compare.

- anonymous

i need a volume of a tetrahedron which is formed when x=0,y=0,z=0 and with thi stangent planbe.

- anonymous

At the point (x_0,y_0,z_0), the normal to the tangent plane will be\[\nabla f(x_0,y_0,z_0)=(\frac{2x_0}{a^2},\frac{2y_0}{b^2},\frac{2z_0}{c^2})\]and so, at the same point, the equation of the tangent plane will be,\[\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)=0\]that is,\[\frac{x_0}{a^2}(x-x_0)+\frac{y_0}{b^2}(y-y_0)+\frac{z_0}{c^2}(z-z_0)=0\]

- anonymous

i got these coordinates \[(-a ^{2}/x , b ^{2}/y, 0), (-a ^{2}/x,0,c ^{2}/z),(-a ^{2}/x,0,0)\]

- anonymous

yes and on simplifying that equation you get:\[xx _{o}^{}/a^2 +yy _{o}^{}/b^2 +zz _{o}^{}/c^2 =1\]

- anonymous

then i used the tripple product to get vol. but its negative

- anonymous

Yeah, see, I'm thinking the plane has to form one side of the tetrahedron. Given that, it will cut the x-, y- and z-axes at the points,\[(\frac{a^2}{x_0},0,0)\]\[(0,\frac{b^2}{y_0},0)\]\[(0,0,\frac{c^2}{z_0})\]

- anonymous

##### 1 Attachment

- anonymous

hey why did you take -x,-y,-z? how did you know its negative?

- anonymous

?

- anonymous

as in i took x y z . that is the positive axis. why did u take the negative ones?

- anonymous

I haven't taken negative axes.

- anonymous

sorry my bad i was jumping to the triple product... so with the coordinates u ve mentioned, i should be taking triple product of these? or as i have done in my calculaatios in the attachment?

- anonymous

Well, it's hard for me to say because I don't have the entire question. I'm only piecing together the fact you have a tangent plane that's to form one surface of a tetrahedron, and with the other three sides lying in the x-y, x-z and x-y planes.

- anonymous

The triple product is only going to find the volume of a parallelopiped.

- anonymous

ok its find the min vol bounded by the planes, xy, yz,xz, and a plane tangent to the folowing:
x^2/a^2+(y/b)^2 + (z/c)^2 =1

- anonymous

Okay, I guessed right.

- anonymous

##### 1 Attachment

- anonymous

whats wrong with my calculations =(

- anonymous

I'd have to do the problem to see that. When do you need this? I have some things I need to take care of.

- anonymous

the attachment is not opening here.

- anonymous

it finally opened.

- anonymous

##### 1 Attachment

- anonymous

its hw question. anyway i'll further brood on it.

- anonymous

thank u!

- anonymous

ok. if i get a chance, i'll come back to it :)

- anonymous

rsaad, if you're there, I have the answer for you.

- anonymous

It's a long problem...

- anonymous

There are a couple of ways of doing this (e.g. Lagrange multipliers), but substitution works fine too.

- anonymous

I'll upload the first sheet where I've found the appropriate determinant using the coordinates I found before:

- anonymous

##### 1 Attachment

- anonymous

Now, the x, y and z are points that satisfy the equation of your surface, so we may eliminate z as\[z=c \sqrt{1-\frac{x^2}{a^2}+\frac{y^2}{b^2}}\]

- anonymous

and sub. this in for z in the volume equation. Now we have a function in x and y only. We can find the extrema by finding\[\frac{\partial V}{\partial x}=0\]and\[\frac{\partial V}{\partial y}=0\]

- anonymous

lol, loki, you've got a neat writing

- anonymous

You may solve for x^2 in the first to find\[x^2=\frac{(b^2-y^2)a^2}{2b^2}\]

- anonymous

hehe, thanks :)

- anonymous

^_^ np

- anonymous

and sub. this into the partial derivative found for y to find:\[y^2=\frac{b^2}{3}\]Substituting this back into the first expression for x^2 we have,\[x^2=\frac{a^2}{3}\]

- anonymous

To find z, we only need to realise that z is constrained by the surface, so substituting these in, we find,\[z^2=c^2(1-\frac{x^2}{a^2}-\frac{y^2}{b^2})=c^2(1-\frac{a^2/3}{a^2}-\frac{b^2/3}{b^2})=\frac{c^2}{3}\]

- anonymous

Substituting these into the expression for volume, we have\[V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}=\frac{1}{6}\frac{a^2b^2c^2}{\frac{a}{\sqrt{3}}\frac{b}{\sqrt{3}}\frac{c}{\sqrt{3}}}=\frac{\sqrt{3}}{2}abc\]

- anonymous

You have to check that these values for x and y do yield a maximum in V by using the appropriate test\[f_{xx}<0\]and\[f_{xx}f_{yy}-f_{xy}^2>0\]

- anonymous

And, um, give us a fan point if you haven't already ;p

- anonymous

thanks lokisan =) but i've a question.why did you use the surface equation when actually it is the tangent plane which is bounding the tetrahedron?

- anonymous

surface eq to eliminate z. why not the tangent plane equation?
and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

- anonymous

You have to think of the surface equation as being the set of values x, y and z are allowed to take. Remember, the x, y, and z used in the volume formula were obtained from the plane equation, and that particular point was the point on the surface where the tangent plane was generated.

- anonymous

I dropped the notation,\[V=\frac{1}{6}\frac{a^2b^2c^2}{x_0y_0z_0}\]for\[V=\frac{1}{6}\frac{a^2b^2c^2}{xyz}\]

- anonymous

okay.

- anonymous

You're not convinced...

- anonymous

what about my 2nd q:and also, why in calculating the detrminent, u used a-b, b-c,c-d where d was 0,0,0. on one website, that i consukted the formula was d-a,d-b,d-c... i am confused.

- anonymous

hang on...brb

- anonymous

well yes. cuz when i plotted the tetrahedron, the sid was flanked by tangent PLANE. so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?

- anonymous

For the determinant, I used the formula (1/6)Â·det(aâˆ’b, bâˆ’c, câˆ’d), where a, b, c and d are the vertices. This one makes no specific mention of assuming a vertex (e.g. d) to be the origin.

- anonymous

but like i said, i checked internet earlier and there it was mentioned d=000

- anonymous

Hang on, this is easy to deal with. I want to talk about your other question.

- anonymous

ok just tell me if we have found the 3 vertices, we simply subtract each of them to get 3 eqs?

- anonymous

ok.

- anonymous

You asked, "so why not set of points satisfied by that tangent plane? i mean if we look at the surface as a whole, wont that be misleading?"

- anonymous

Okay...the volume is determined by the tangent plane, and the tangent plane is determined by the surface. You're being asked to find the specific coordinates (i.e. point on surface which generates the tangent plane) that leads to a tangent plane that gives you the maximal volume. That's why the x, y and z must satisfy the surface equation, because we're generating tangent planes for various points (x,y,z) on the surface. We can't choose arbitrary x, y AND z, since these won't necessarily be on the surface, and therefore, won't give us a tangent plane to the surface. We can choose x and y, but the surface equation will dictate the allowable z we may take.

- anonymous

O......!!! now i see! =) thanks!!!

- anonymous

good :)

- anonymous

Now, as for the equation for the volume...I used a formula from one of my books, and I've also seen it on wikipedia:
http://en.wikipedia.org/wiki/Tetrahedron
Look under 'volume'.

- anonymous

You can derive this for yourself if you need convincing by understanding that the volume of any pyramid is 1/3 x base x perpendicular height.

- anonymous

ahan. let me see

- anonymous

right. so then it follows what you have done / thanks a lot! btw which uni r u from? i m curious 'cuz u r able to answer all my questions!

- anonymous

A very old university... ;)

- anonymous

IV league?

- anonymous

Were all your questions answered?

- anonymous

well yes. i'd posted some problems earlier and u solved them and so i'd fan-ed you already =)

- anonymous

Ah...good...you caught me just in time, too...I was about to log out -.-
Happy mathing :D

- anonymous

=D
lucky me! hey come on which uni r u from? i m curious! caltech?

- anonymous

no... :)

- anonymous

then? iv league?

- anonymous

I'm a citizen of the UK

- anonymous

oxford? cambridge?

- anonymous

Well, it's possibly one of the two ;)

- anonymous

aha! no doubt you are really good at math! =P ok thanks a lot!

- anonymous

no probs :D

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