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mathstina
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Show that every plane that is tangent to the cone x^2+y^2=z^2 passes through the origin.
 one year ago
 one year ago
mathstina Group Title
Show that every plane that is tangent to the cone x^2+y^2=z^2 passes through the origin.
 one year ago
 one year ago

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mathstina Group TitleBest ResponseYou've already chosen the best response.0
hw? i need guidance.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
let \[g(x,y,z)=x^2+y^2z^2=0\]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?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
is it (2,2,2)?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
\[\nabla g(x,y,z)=\langle2x,2y,2z\rangle\]
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
yes, how to get the P?
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
well the function is x2+y2z2= 0
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
oops\[\nabla g(x,y,z)=\langle2x,2y,2z\rangle\]
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
im guessing the tangent at any point would be 2x, 2y,2z
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
not the tangent, the normal vector/gradient
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
P=(x,y,z) is a general point on the plane
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
gradient and tangent are the same thing arent they
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
ok. hw do i ans to this qn?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
no because the gradient is normal to the surface in this case it would be the tangent vector to the surface if it were\[\nabla z(x,y)\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
sorry, I messed up, let \(\vec P_0=\langle l,m,n\rangle\)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
so then what is \(\vec P\vec P_0\) ???
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
P= <2,2,2> ??
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
i think i got it thankk you @TuringTest
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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\[\nabla g(l,m,n)\cdot(\vec P\vec P_0)=0\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
\[\nabla g(l,m,n)\cdot(\vec P\vec P_0)=0\]\[2\langle l,m,n\rangle\cdot\langle xl,ym,zn\rangle=2l(xl)+2m(ym)2n(xn)=0\]at the origin we have (x,y,z)=(0,0,0) so\[2l^22m^2+2n^2=0\implies l^2+m^2n^2=0\]and because of the equation of the cone, we know that whatever point on it we pick we will have x^2+y^2z^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
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
ok, i m trying to grasp.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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...
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
why ve to multiply by 2 for 2<l,m,n>?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
\[g(x,y,z)=x^2+y^2z^2\implies\nabla g(x,y,z)=\langle 2x,2y,2z\rangle=2\langle x,y,z\rangle\]
 one year ago

mathstina Group TitleBest ResponseYou've already chosen the best response.0
ok. thanks. time for me to sleep. good bye.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
welcome, g'night!
 one year ago

piscez.in Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest your solution is spot on, thank you
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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\[z^2=x^2+y^2\implies g(x,y,z)=x^2+y^2z^2=0\]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.\[\nabla g(l,m,n)\cdot(\vec P\vec P_0)=0\]\[2\langle l,m,n\rangle\cdot\langle xl,ym,zn\rangle=0\]\[2l(xl)+2m(ym)2n(zn)=0\]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\[2l^22m^2+2n^2=0\]\[l^2+m^2n^2=0\]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.
 one year ago
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