Prove

**esum0209** · October 10th 2010, 07:17 AM

Prove that every line normal to the cone with equation z=sqrt(x^2+y^2) intersects the z-axis.

**tonio** · October 10th 2010, 09:01 AM

Originally Posted by esum0209

Prove that every line normal to the cone with equation z=sqrt(x^2+y^2) intersects the z-axis.

Define $F(x,y,z):=\sqrt{x^2+y^2}-z\Longrightarrow$ the normal line to this surface at any given point $(x_0,y_0,z_0)$ is given by the parametric equations

$x(t):=x_0+\frac{x_0}{\sqrt{x_0^2+y_0^2}}t\,,\,\,y( t)=y_0+\frac{y_0}{\sqrt{x_0^2+y_0^2}}t\,,\,\,z(t)= z_0-t\,,\,\,t\in\mathbb{R}$ .

Now prove that there exists $t_1\in\mathbb{R}\,\,s.t.\,\, x(t_1)=y(t_1)=0$ ...

Tonio

**esum0209** · October 10th 2010, 10:48 AM

Originally Posted by tonio

Now prove that there exists $t_1\in\mathbb{R}\,\,s.t.\,\, x(t_1)=y(t_1)=0$ ...

Tonio

Sorry, but I don't know to prove this... Can u help me a bit more?

Thanks in advance

**Plato** · October 10th 2010, 11:28 AM

What happens if $t = \dfrac{{ - 1}}<br /> {{\sqrt {x_0 ^2 + y_0 ^2 } }}?$

**tonio** · October 10th 2010, 11:28 AM

Originally Posted by esum0209

Sorry, but I don't know to prove this... Can u help me a bit more?

Thanks in advance

No. If you're dealing with partial derivatives you must be able to cope with this: show that the solution to $x(t_1)=0$ is the same as the solution to $y(t_1)=0$ . This is HS stuff.

Tonio

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