Vector calculus problem

To use LaTeX on here, enclose your LaTeX code in [noparse]${stuff}$[/noparse].

e.g. [noparse]$\dfrac{-b \pm \sqrt{b^2-4ac}}{2}$[/noparse] will give $\dfrac{-b \pm \sqrt{b^2-4ac}}{2}$


Even if you can't use LaTeX, have you not heard of paragraphs? Full stops? Capitalisation?

Sorry to snipe on presentation, but what you've written is incredibly difficult to read.

As far as the actual question, it's kind of unclear what you've done so far. What are your thoughts on finding the tangent plane?
For the last bit, I have no idea what vector you mean by $1/3$ so it's hard to comment, but unless the rate of change was 0 I think it's unlikely to help you in terms of finding a vector where the directional derivative is zero.

Sorry about this I didn't really think to check on the latex used on this website I just assumed it would be like others I had also completely forgot to make paragraphs hopefully it is now easier to read
Anyway to find the tangent plane to the surface I calculated nabla phi at the point p then worked out a normal to the surface at the same point p but I'm unsure what to do next, do I need to use the dot product with the normal and another vector say

And for the last bit I mean I calculated a directional derivative which was 1/3 at the point p

So you have a vector n normal to the plane, you now just need to find the appropriate constant d so that the plane r.n = d passes through the given point on the surface.

This is indeed how to approach this.



and to do this bit, note that:

a) for any vector $\vec{v}$ lying in the plane, $\vec{v} \cdot \nabla \phi = 0$
b) the general position vector of a point in the plane is $(x,y,z)$
c) the point with position vector $(2,1,\pi/2)$ is also in the plane

Hi thanks for your response in the question it is asking to find the tangent plane to the surface phi=4 so would it be nabla Phi so would it be each term of the nabla phi evaluated at the point so the equation of the tangent would be dphi/dx(X-xo)+dphi/dy(y-yo) +dphi/dz(z-z0)=4 or 0? Then for where X0 and y0 z0 would be coordinates

Sorry, I can't really think about this now, but I've just noticed something that looks a bit odd: the point $(2,1,\pi/2)$ doesn't lie on the surface $\phi=4$ - have you copied the question correctly?

[Aside: please use latex - I rarely respond to questions that don't use it when it's needed]

Hi thanks for your response and yeah I'm 100% sure that I copied the question correctly does the point not lie on the Phi(X,y,z)=4?

Well, when x,y,z have those values, what does phi equal? Is it 4?