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

1. mathstina

hw? i need guidance.

2. 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?

3. mathstina

is it (2,2,-2)?

4. TuringTest

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

5. mathstina

yes, how to get the P?

6. piscez.in

well the function is x2+y2-z2= 0

7. TuringTest

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

8. piscez.in

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

9. TuringTest

not the tangent, the normal vector/gradient

10. TuringTest

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

11. piscez.in

gradient and tangent are the same thing arent they

12. mathstina

ok. hw do i ans to this qn?

13. 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$\nabla z(x,y)$

14. TuringTest

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

15. TuringTest

so then what is $$\vec P-\vec P_0$$ ???

16. piscez.in

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?

17. mathstina

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

18. 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...

19. piscez.in

i think i got it thankk you @TuringTest

20. 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$\nabla g(l,m,n)\cdot(\vec P-\vec P_0)=0$

21. 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

22. mathstina

ok, i m trying to grasp.

23. 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.

24. 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...

25. mathstina

why ve to multiply by 2 for 2<l,m,n>?

26. TuringTest

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

27. mathstina

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

28. TuringTest

welcome, g'night!

29. piscez.in

@TuringTest your solution is spot on, thank you

30. 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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