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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[xx ^{*} /a ^{2} + yy^*/b^2 + zz^*/c^2 =1 ????\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tangent plane at what point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at a general point x^* y^* z^*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay change of answer it is x^*/a^2 +y^*/b^2 +z^*/c^2 =1 i think thats the correc tangent plane eq

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to take derivative of the parabola to get the tangent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a parabola? btw its cal 2 and vector algebra

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x y and z are independent of each other.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0scrap that last post.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0i did calcu;at the equation of tangent plane.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}1=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh...what's wrong then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the problem i s i am getting the volume in negative.. =(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\nabla f=(\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Volume? You don't need volume here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i took the triple product. should i attach my cal. ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait and see what I get, then you can compare.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need a volume of a tetrahedron which is formed when x=0,y=0,z=0 and with thi stangent planbe.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At 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,\[\frac{2x_0}{a^2}(xx_0)+\frac{2y_0}{b^2}(yy_0)+\frac{2z_0}{c^2}(zz_0)=0\]that is,\[\frac{x_0}{a^2}(xx_0)+\frac{y_0}{b^2}(yy_0)+\frac{z_0}{c^2}(zz_0)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got these coordinates \[(a ^{2}/x , b ^{2}/y, 0), (a ^{2}/x,0,c ^{2}/z),(a ^{2}/x,0,0)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes and on simplifying that equation you get:\[xx _{o}^{}/a^2 +yy _{o}^{}/b^2 +zz _{o}^{}/c^2 =1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then i used the tripple product to get vol. but its negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, see, I'm thinking the plane has to form one side of the tetrahedron. Given that, it will cut the x, y and zaxes 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
 5 years ago
Best ResponseYou've already chosen the best response.0hey why did you take x,y,z? how did you know its negative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as in i took x y z . that is the positive axis. why did u take the negative ones?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I haven't taken negative axes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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 xy, xz and xy planes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The triple product is only going to find the volume of a parallelopiped.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I guessed right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats wrong with my calculations =(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0the attachment is not opening here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its hw question. anyway i'll further brood on it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. if i get a chance, i'll come back to it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rsaad, if you're there, I have the answer for you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a long problem...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are a couple of ways of doing this (e.g. Lagrange multipliers), but substitution works fine too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll upload the first sheet where I've found the appropriate determinant using the coordinates I found before:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0lol, loki, you've got a neat writing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You may solve for x^2 in the first to find\[x^2=\frac{(b^2y^2)a^2}{2b^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0To 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
 5 years ago
Best ResponseYou've already chosen the best response.0Substituting 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0And, um, give us a fan point if you haven't already ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks 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
 5 years ago
Best ResponseYou've already chosen the best response.0surface eq to eliminate z. why not the tangent plane equation? and also, why in calculating the detrminent, u used ab, bc,cd where d was 0,0,0. on one website, that i consukted the formula was da,db,dc... i am confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0You're not convinced...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about my 2nd q:and also, why in calculating the detrminent, u used ab, bc,cd where d was 0,0,0. on one website, that i consukted the formula was da,db,dc... i am confused.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0For the determinant, I used the formula (1/6)·det(a−b, b−c, c−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
 5 years ago
Best ResponseYou've already chosen the best response.0but like i said, i checked internet earlier and there it was mentioned d=000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hang on, this is easy to deal with. I want to talk about your other question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok just tell me if we have found the 3 vertices, we simply subtract each of them to get 3 eqs?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...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
 5 years ago
Best ResponseYou've already chosen the best response.0O......!!! now i see! =) thanks!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0right. 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
 5 years ago
Best ResponseYou've already chosen the best response.0A very old university... ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Were all your questions answered?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well yes. i'd posted some problems earlier and u solved them and so i'd faned you already =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah...good...you caught me just in time, too...I was about to log out . Happy mathing :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0=D lucky me! hey come on which uni r u from? i m curious! caltech?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm a citizen of the UK

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it's possibly one of the two ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0aha! no doubt you are really good at math! =P ok thanks a lot!
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