anonymous
  • anonymous
tangent plane of (x^2/a^2)+ (y^2/b^2)+(z^2/c^2)=1 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[xx ^{*} /a ^{2} + yy^*/b^2 + zz^*/c^2 =1 ????\]
anonymous
  • anonymous
tangent plane at what point?
anonymous
  • anonymous
at a general point x^* y^* z^*

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anonymous
  • 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
  • anonymous
you need to take derivative of the parabola to get the tangent
anonymous
  • anonymous
a parabola? btw its cal 2 and vector algebra
anonymous
  • anonymous
x y and z are independent of each other.
anonymous
  • anonymous
scrap that last post.
anonymous
  • anonymous
rsaad, you there?
anonymous
  • anonymous
yea
anonymous
  • 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
  • anonymous
So...
anonymous
  • anonymous
i did calcu;at the equation of tangent plane.
anonymous
  • anonymous
\[f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0\]
anonymous
  • anonymous
Oh...what's wrong then?
anonymous
  • anonymous
the problem i s i am getting the volume in negative.. =(
anonymous
  • anonymous
\[\nabla f=(\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2})\]
anonymous
  • anonymous
Volume? You don't need volume here.
anonymous
  • anonymous
i took the triple product. should i attach my cal. ?
anonymous
  • anonymous
Wait and see what I get, then you can compare.
anonymous
  • anonymous
i need a volume of a tetrahedron which is formed when x=0,y=0,z=0 and with thi stangent planbe.
anonymous
  • 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
  • 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
  • anonymous
yes and on simplifying that equation you get:\[xx _{o}^{}/a^2 +yy _{o}^{}/b^2 +zz _{o}^{}/c^2 =1\]
anonymous
  • anonymous
then i used the tripple product to get vol. but its negative
anonymous
  • 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
  • anonymous
1 Attachment
anonymous
  • anonymous
hey why did you take -x,-y,-z? how did you know its negative?
anonymous
  • anonymous
?
anonymous
  • anonymous
as in i took x y z . that is the positive axis. why did u take the negative ones?
anonymous
  • anonymous
I haven't taken negative axes.
anonymous
  • 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
  • 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
  • anonymous
The triple product is only going to find the volume of a parallelopiped.
anonymous
  • 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
  • anonymous
Okay, I guessed right.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
whats wrong with my calculations =(
anonymous
  • 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
  • anonymous
the attachment is not opening here.
anonymous
  • anonymous
it finally opened.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
its hw question. anyway i'll further brood on it.
anonymous
  • anonymous
thank u!
anonymous
  • anonymous
ok. if i get a chance, i'll come back to it :)
anonymous
  • anonymous
rsaad, if you're there, I have the answer for you.
anonymous
  • anonymous
It's a long problem...
anonymous
  • anonymous
There are a couple of ways of doing this (e.g. Lagrange multipliers), but substitution works fine too.
anonymous
  • anonymous
I'll upload the first sheet where I've found the appropriate determinant using the coordinates I found before:
anonymous
  • anonymous
1 Attachment
anonymous
  • 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
  • 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
  • anonymous
lol, loki, you've got a neat writing
anonymous
  • anonymous
You may solve for x^2 in the first to find\[x^2=\frac{(b^2-y^2)a^2}{2b^2}\]
anonymous
  • anonymous
hehe, thanks :)
anonymous
  • anonymous
^_^ np
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
And, um, give us a fan point if you haven't already ;p
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
okay.
anonymous
  • anonymous
You're not convinced...
anonymous
  • 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
  • anonymous
hang on...brb
anonymous
  • 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
  • anonymous
For 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
  • anonymous
but like i said, i checked internet earlier and there it was mentioned d=000
anonymous
  • anonymous
Hang on, this is easy to deal with. I want to talk about your other question.
anonymous
  • anonymous
ok just tell me if we have found the 3 vertices, we simply subtract each of them to get 3 eqs?
anonymous
  • anonymous
ok.
anonymous
  • 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
  • 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
  • anonymous
O......!!! now i see! =) thanks!!!
anonymous
  • anonymous
good :)
anonymous
  • 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
  • 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
  • anonymous
ahan. let me see
anonymous
  • 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
  • anonymous
A very old university... ;)
anonymous
  • anonymous
IV league?
anonymous
  • anonymous
Were all your questions answered?
anonymous
  • anonymous
well yes. i'd posted some problems earlier and u solved them and so i'd fan-ed you already =)
anonymous
  • anonymous
Ah...good...you caught me just in time, too...I was about to log out -.- Happy mathing :D
anonymous
  • anonymous
=D lucky me! hey come on which uni r u from? i m curious! caltech?
anonymous
  • anonymous
no... :)
anonymous
  • anonymous
then? iv league?
anonymous
  • anonymous
I'm a citizen of the UK
anonymous
  • anonymous
oxford? cambridge?
anonymous
  • anonymous
Well, it's possibly one of the two ;)
anonymous
  • anonymous
aha! no doubt you are really good at math! =P ok thanks a lot!
anonymous
  • anonymous
no probs :D

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