Question

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

Answer 1

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

Answer 2

tangent plane at what point?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

a parabola? btw its cal 2 and vector algebra

Answer 7

x y and z are independent of each other.

Answer 8

scrap that last post.

Answer 9

rsaad, you there?

Answer 10

yea

Answer 11

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.

Answer 12

So...

Answer 13

i did calcu;at the equation of tangent plane.

Answer 14

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

Answer 15

Oh...what's wrong then?

Answer 16

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

Answer 17

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

Answer 18

Volume? You don't need volume here.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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,\[\frac{x_0}{a^2}(x-x_0)+\frac{y_0}{b^2}(y-y_0)+\frac{z_0}{c^2}(z-z_0)=0\]

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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,\[(\frac{a^2}{x_0},0,0)\]\[(0,\frac{b^2}{y_0},0)\]\[(0,0,\frac{c^2}{z_0})\]

Answer 27

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

Answer 28

?

Answer 29

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

Answer 30

I haven't taken negative axes.

Answer 31

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?

Answer 32

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.

Answer 33

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

Answer 34

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

Answer 35

Okay, I guessed right.

Answer 36

whats wrong with my calculations =(

Answer 37

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.

Answer 38

the attachment is not opening here.

Answer 39

it finally opened.

Answer 40

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

Answer 41

thank u!

Answer 42

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

Answer 43

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

Answer 44

It's a long problem...

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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\[\frac{\partial V}{\partial y}=0\]

Answer 49

lol, loki, you've got a neat writing

Answer 50

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

Answer 51

hehe, thanks :)

Answer 52

^_^ np

Answer 53

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,\[x^2=\frac{a^2}{3}\]

Answer 54

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

Answer 55

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

Answer 56

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\[f_{xx}f_{yy}-f_{xy}^2>0\]

Answer 57

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

Answer 58

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?

Answer 59

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.

Answer 60

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.

Answer 61

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

Answer 62

okay.

Answer 63

You're not convinced...

Answer 64

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.

Answer 65

hang on...brb

Answer 66

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?

Answer 67

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.

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

ok.

Answer 72

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?"

Answer 73

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.

Answer 74

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

Answer 75

good :)

Answer 76

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'.

Answer 77

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.

Answer 78

ahan. let me see

Answer 79

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!

Answer 80

A very old university... ;)

Answer 81

IV league?

Answer 82

Were all your questions answered?

Answer 83

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

Answer 84

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

Happy mathing :D

Answer 85

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

Answer 86

no... :)

Answer 87

then? iv league?

Answer 88

I'm a citizen of the UK

Answer 89

oxford? cambridge?

Answer 90

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

Answer 91

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

Answer 92

no probs :D