# Show angle between $P-P_0$ and plane tangent to surface at $P_0$ approaches 0

Let $f(x,y) \rightarrow R$ be a function with continuous partial derivatives, and let $S$ be the surface $z=f(x,y)$ in $R^2$. Let $P_{0}=(x_0,y_0,z_0 )$ be a point in S and $P=(x,y,z)$ be some other point in S. We're asked to show that $a$, the angle between the plane tangent to S at $(x_0,y_0,z_0)$ and the vector $P-P_0$, approaches $0$ as $P\rightarrow P_0$.

I'd appreciate some help with proving this. The question is from a former exam in my multivariable calculus course.

The angle thing ticks me off here since I have no idea how to approach this. I tried proving this using dot products to derive the angle between the vectors, but calculating the limit was difficult.

Thanks!

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Hint only, since this looks like homework and you should try to do the calculations yourself:

Instead of $P-P_0$, consider the corresponding normalized unit vector $Q= \frac{P-P_0}{|P-P_0|}$, so you have not to bother about estimating it's length. Instead of the tangent plane, consider the unit normal $n$ to the tangent plane and show that $\langle n,Q\rangle$ tends to zero as $P$ approaches $P_0$.

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Thanks for your help. Like I said, I actually already tried this approach, but I'll give it another try, maybe the calculation isn't as daunting as I thought. BTW It's not homework (what's with the speculation?) - rather, preparation for an exam. –  ro44 Aug 19 '12 at 14:54
So in essence I'd like to show that $Q\cdot (f_x,f_y,-1)/||(f_x,f_y,-1)||$ tends to $0$, which is the same as showing $Q \cdot (f_x,f_y,-1)$ tends to $0$. But this doesn't seem correct... is there an error somewhere? –  ro44 Aug 19 '12 at 14:59
Actually this might be correct since $f$ is differentiable. I missed that last time I thought about this... –  ro44 Aug 19 '12 at 15:07
Orright, success. This answer didn't tell me something I didn't already know but it pointed me in the right direction and told me to "push further" so I'm accepting it. Thanks! –  ro44 Aug 19 '12 at 15:09
@ro44 It's the same thing if $|\nabla f| \ge c >0$ . –  user20266 Aug 19 '12 at 15:10

$$P-P_0=(x_0-x\,,\,y_0-y\,,\,z_0-z)$$

As a plane's normal at $\,P\,$ is $\,N:=\left(f_x(P)\,,\,f_y(P)\,,\,-1\right)\,$ , we get that the sine of the angle between the line and the plane is $$(P-P_0)\cdot N=f_x(P)(x_0-x)+f_y(P)(y_0-y)-(z_0-z)\xrightarrow [(x,y,z)\to(x_0,y_0,z_0)]{}0$$ since the partial derivatives $\,f_x\,,\,f_y\,$ are given continuous at $\,P\,$

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