Tangent Plane And Normal Vector.

[dcl]

I'm having trouble working out the tangent plane of an equation at a specified point (4,1,-2)
The equation being 9x^2 - 4y^2 - 25z^2 = 40

now
\nabla f = (18x, -8y, -50z) yeh?
Just reading off this should give us the normal vector shouldn't it? (18,-8,-50)
and from that we can work out the equation of the plane.
18(x-4) - 8(y-1) -50(z-(-2)) = 0
Is this corrent or am I using a horribly flawed method?

 

[dcl]

Think I've worked it out for myself.
Method was sort of wrong.
Once I have Grad F, all I need to do is sub in the values of the point and It will give me the normal vector and from that I can work out the equation.
I think that's right.

 

[M98Ranger]

Your method is correct! The normal vector to the tangent plane at a point on a surface is given by the gradient of the surface at that point. So in this case, the normal vector is (18, -8, -50) at the point (4,1,-2).

To find the equation of the tangent plane, we can use the formula:
ax + by + cz = d
where (a,b,c) is the normal vector and (x,y,z) is any point on the plane. So plugging in our values, we get:
18(x-4) - 8(y-1) -50(z-(-2)) = 0

Simplifying, we get:
18x - 72 - 8y + 8 - 50z + 100 = 0
18x - 8y - 50z = -36

So the equation of the tangent plane at the point (4,1,-2) is:
18x - 8y - 50z = -36

Great job on working through this problem and using the correct method! Keep up the good work!