# Thread: Tangent plane and normal line..

1. ## Tangent plane and normal line..

I got 2 questions that I am finding problems with, would someone help me please?

(1) Fine equation of the tangent plane
(2) Find equation of normal line to the surface
represented by the equation:

x-z = 4arctan(yz)

at the points (1+Pi, 1, 1)

Thanks.
taurus

2. Originally Posted by taurus
I got 2 questions that I am finding problems with, would someone help me please?

(1) Fine equation of the tangent plane
(2) Find equation of normal line to the surface
represented by the equation:

x-z = 4arctan(yz)

at the points (1+Pi, 1, 1)
Let $\displaystyle f(x,y,z) = x - z - 4\tan^{-1} yz$ so the surface is represented as $\displaystyle f(x,y,z) = 0$.
The normal to the surface at $\displaystyle (1+\pi , 1 , 1)$ is given by $\displaystyle \nabla f(1+\pi , 1 , 1) = a\bold{i}+b\bold{j}+c\bold{k}$. (Whatever $\displaystyle a,b,c$ are, you need to compute that).

For (1) to find the tangent plane you use the fact that is passes through the point $\displaystyle (1+\pi , 1,1)$ and its normal vector is $\displaystyle a\bold{i}+b\bold{j}+c\bold{k}$.

For (2) to find the normal line procede as above. The line contains $\displaystyle (1+\pi,1,1)$ and is aligned with vector $\displaystyle a\bold{i}+b\bold{j}+c\bold{k}$.

3. Am still a bit confused with my equation how its achieved?

Could someone show me some steps?

thanks

4. if i have f(x,y,z) = x - z - 4arctan(yz) = 0

So i need to differentiate for fx, fy and fz.

so is the following right:
fx = dx

but whats fy and fz?

Thanks

5. Originally Posted by taurus
if i have f(x,y,z) = x - z - 4arctan(yz) = 0
If $\displaystyle f(x,y,z) = x - z - 4\tan^{-1} (yz)$.

Then, $\displaystyle \frac{\partial f}{\partial x} = 1$, $\displaystyle \frac{\partial f}{\partial y} = -\frac{4z}{1+(yz)^2}$, $\displaystyle \frac{\partial f}{\partial z} = -1-\frac{4y}{1+(yz)^2}$.