Show that every plane that is tangent to the cone x^2+y^2=z^2 passes through the origin.

- anonymous

Show that every plane that is tangent to the cone x^2+y^2=z^2 passes through the origin.

- schrodinger

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- anonymous

hw? i need guidance.

- TuringTest

let \[g(x,y,z)=x^2+y^2-z^2=0\]then the normal vector to the surface is given by\[\nabla g(x,y,z)\]further we know that the point Po=(0,0,0) is on the cone.
The equation of the tangent plan at any point x,y,z is then given by\[\nabla g\cdot(\vec P-\vec P_0)=0\]where \(\vec P=\langle x,y,z\rangle\)
pluf in the pieces and what do you get?

- anonymous

is it (2,2,-2)?

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## More answers

- TuringTest

\[\nabla g(x,y,z)=\langle2x,2y,2z\rangle\]

- anonymous

yes, how to get the P?

- anonymous

well the function is x2+y2-z2= 0

- TuringTest

oops\[\nabla g(x,y,z)=\langle2x,2y,-2z\rangle\]

- anonymous

im guessing the tangent at any point would be 2x, 2y,-2z

- TuringTest

not the tangent, the normal vector/gradient

- TuringTest

P=(x,y,z) is a general point on the plane

- anonymous

gradient and tangent are the same thing arent they

- anonymous

ok. hw do i ans to this qn?

- TuringTest

no because the gradient is normal to the surface in this case
it would be the tangent vector to the surface if it were\[\nabla z(x,y)\]

- TuringTest

sorry, I messed up, let \(\vec P_0=\langle l,m,n\rangle\)

- TuringTest

so then what is \(\vec P-\vec P_0\) ???

- anonymous

i think gradient is the rate of change in genera, a scalar quantity, and tangent by definition is the rate of change of a function and its a vector quantity, normal, on the other hand is the vector thats perpendicular to the tangent? no?

- anonymous

P= <-2,-2,2> ??

- TuringTest

for the surface \(f(x,y,z)\) the gradient \(\nabla f\) represents a vector normal to the surface
http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx
@mathstina no, there should be no pure numbers here...

- anonymous

i think i got it thankk you @TuringTest

- TuringTest

general point\[\vec P=\langle x,y,z\rangle\]specific point\[\vec P_0=\langle l,m,n\rangle\]gradient at specific point\[\nabla g(l,m,n)=2\langle l,m,n\rangle\]since \(\vec P-\vec P_0\) lies in the plane, then the gradient is perpendicular to it, so the equation for a tangent plane at a point (l,m,n) is\[\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0\]

- TuringTest

\[\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0\]\[2\langle l,m,n\rangle\cdot\langle x-l,y-m,z-n\rangle=2l(x-l)+2m(y-m)-2n(x-n)=0\]at the origin we have (x,y,z)=(0,0,0) so\[-2l^2-2m^2+2n^2=0\implies l^2+m^2-n^2=0\]and because of the equation of the cone, we know that whatever point on it we pick we will have x^2+y^2-z^2=0, so the above is always true regardless of where we draw out tangent plane (l,m, and n don't matter).
therefore every tangent plane to the cone contains the origin

- anonymous

ok, i m trying to grasp.

- TuringTest

It is a tricky one, I know. I used the concept of the plane equation heavily here. If you are not good with plane equations this is a tough thing to grasp. Let me know any specific parts you don't understand.

- TuringTest

oh sorry, I did that same typo again :P
should be\[\nabla g(l,m,n)=2\langle l,m,-n\rangle\]but the rest is correct...

- anonymous

why ve to multiply by 2 for 2?

- TuringTest

\[g(x,y,z)=x^2+y^2-z^2\implies\nabla g(x,y,z)=\langle 2x,2y,-2z\rangle=2\langle x,y,-z\rangle\]

- anonymous

ok. thanks. time for me to sleep.
good bye.

- TuringTest

welcome, g'night!

- anonymous

@TuringTest your solution is spot on, thank you

- TuringTest

Let me try to elaborate so it is more understandable for @mathstina .
To get the surface of the cone written as a function we need to get all variables on the same side\[z^2=x^2+y^2\implies g(x,y,z)=x^2+y^2-z^2=0\]Now we pick a point (actually a direction vector) on the surface of the cone \(\vec P_0=\langle l,m,n\rangle\)and find the gradient at that point\[\nabla g(x,y,z)=2\langle x,y,-z\rangle\implies \nabla g(l,m,n)=2\langle l,m,-n\rangle\]now let \(\vec P=\langle x,y,z\rangle\) be a point in the tangent plane to the curve at \(\vec P_0\). This implies that the vector \(\vec P-\vec P_0\) is in the tangent plane. Any vector in the tangent plane should be perpendicular to the normal vector (the gradient), so their dot product is zero.\[\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0\]\[2\langle l,m,-n\rangle\cdot\langle x-l,y-m,z-n\rangle=0\]\[2l(x-l)+2m(y-m)-2n(z-n)=0\]Plugging in the origin to see if it satisfies both the plane and surface equations, at \(x=y=z=0\) we see that the equation of the plane gives\[-2l^2-2m^2+2n^2=0\]\[l^2+m^2-n^2=0\]We can see that this means that these numbers are a solution to \(g(x,y,z)\) and so are on the surface of the cone. Since l, m, and n are arbitrary, this means that every tangent plane to the cone contains the origin.

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