Show that the given curve is normal to the given surface at the given point).

Question

Show that the curve $r(t) = (t^2, t, 5t − 4)$ is normal to the surface $2x^2 +y^2 +5z^2 = 8$ at the point $(1, 1, 1)$. My thought is that find the derivatives of $r(t)$ and the gradient of the surface. Then make these two have a dot product is $0$. But I found that after my calculations. The result is not $0$. Hope someone can help with my confusion Thanks.

Answer 1

We can use this statement for definition of normal plane:

A plane P passing through a point M of a curve L and perpendicular to the tangent MT is called a normal plane to the curve L. The direction vector of the tangent $r'=x', y' z')$, is a normal vector to the plane P.The equation of a normal plane is:

$(X-x)x'+(Y-y)y'+(Z-z)z'=0$

Now we have:

$r(t)=f(x, y, z)=(x=t^2, y=t, z=5t-4)$

So the equation of normal plane on curve L is:

$(X-t^2)2t+(Y-t)\cdot 1+(Z-5t+4)5=0$

$2tX-2t^3+Y-t+5Z-25t+20=0$

Or:

$2yX-2y^3+Y-y+5Z-25y+20=0$

$\Rightarrow 2yX-2y^3+Y-26y+5Z+20=0$

At point $(x, y, z)=(1, 1, 1)$ we have:

$2X+Y+5Z-8=0$

So the direction vector (normal vector of plane) is:

$N=(2, 1, 5)$

The vector of a tangent plane on surface is:

$(2x^2+y^2+5z^2)'=4x +2y+10z=2(2x+y+5z); N'=(2, 1, 5)$

So $N=N'$

That is normal plane is tangent on surface, so the surface is perpendicular on the curve at point $1, 1, 1$.

$L: $

Answer 2

The surface is a plane in canonical form so its normal vector is $(2,1,5)$.

The derivative of the curve is $(2t,1,5)=(2,1,5)$ at $(1,1,1)$

Therefore, ...