# Thread: Tangent to a surface

1. ## Tangent to a surface

Just wondering if someone would be able to confirm if my answer is correct or where I've gone wrong?
I've never really done anything like this before so I'm not sure if I'm even remotely on the right track.

1. Find the equation of the tangent plane to the surface

at the point

Here's my solution:

Hope someone can help...Thanks

2. I did not read it the moment I saw your first line.

The gradient is not a real number! It is a vector. For example given $\displaystyle f(x,y,z) = C$ then $\displaystyle \nabla f = \partial_x f\bold{i}+ \partial_y f\bold{j}+ \partial_z f\bold{k}$ (where $\displaystyle \partial_x$ means partial derivative along $\displaystyle x$ and so on ... ). Notice this is not a number, it is a vector.

Here are the steps.
1)Given $\displaystyle f(x,y,z)=C$.
2)Find $\displaystyle \nabla f(x,y,z)$.
3)Evaluate this gradient at the specific point you are given.
4)The vector in #3 is orthogonal to the surface. Therefore, it is the normal to the tangent plane to the surface at this point. Since you know the normal vector and the point you can describe the tangent plane.

3. Thanks for that .
I'm still getting my head around the whole 3 plane thing at the moment.

Why is the gradient vector orthogonal to the surface? Isn't it by definition equal to the tangent in that particular plane?

4. Is this any better for the gradient of the function at the particular point?
I got this by differentiating each variable then replacing x,y,z with the value at the particular point.
I know it's not very well refined (having fractions within fractions) but I'll fix this up later.

5. What you have is not the equation of a plain.

What TPH said can be condensed into one formula. you have a level curve here.

recall that, for a level curve $\displaystyle F(x,y,z) = k$ for $\displaystyle k \in \mathbb{R}$, the tangent plane to the surface at a point $\displaystyle (x_0,y_0,z_0)$ is given by:

$\displaystyle F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$

where $\displaystyle F_x, Fy \mbox{ and } F_z$ and the partial derivatives of $\displaystyle F(x,y,z)$ with respect to x,y, and z respectively, at the point $\displaystyle (x_0,y_0,z_0)$

now try the problem again

6. And this equation will give the tangent in the form of (i,j,k)?

Ok I'm going to forget everything I've done so far and try to start again. Thanks for being patient with me. I know this is prob so easy for someone people it's hard to understand why I don't get it .

Sooo...

The tangent plane to a level curve is given by

Where:

x= (x^2)/25
y= (y^2)/4
z= (z^2)/4

fx = 2x/25
fy = 2y/4
fz = 2z/4

Xo = 1
Yo = (4root3)/5
Zo = (4root3)/5

I'll give it a shot and hope for the best. Thanks again!!

7. Originally Posted by Spimon
x= (x^2)/25
y= (y^2)/4
z= (z^2)/4
no. this is not so. F(x,y,z) = (x^2)/25 + (y^2)/4 + (z^2)/4

but everything else is ok. post your solution when you're done, so we can check it

8. Originally Posted by Jhevon
$\displaystyle F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$

where $\displaystyle F_x, Fy \mbox{ and } F_z$ and the partial derivatives of $\displaystyle F(x,y,z)$ with respect to x,y, and z respectively, at the point $\displaystyle (x_0,y_0,z_0)$
This was the part that confused me. I read that as 'the partial derivative multiplied by the different of the function value and the point value.
From your next post I guess it should mean 'the function value multiplied by the different of the partial derivative and the point value.

Hoooopefully this is more on track . I'm going to get this sooner or later...

9. Ok, since you have really been working hard at this, i'll do it for you.

remember what i said
Originally Posted by Jhevon
... the tangent plane to the surface at a point $\displaystyle (x_0,y_0,z_0)$ is given by:

$\displaystyle F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$

where $\displaystyle F_x, Fy \mbox{ and } F_z$ and the partial derivatives of $\displaystyle F(x,y,z)$ with respect to x,y, and z respectively, at the point $\displaystyle \bold{ \color{red}(x_0,y_0,z_0)}$
$\displaystyle F(x,y,z) = \frac {x^2}{25} + \frac {y^2}4 + \frac {z^2}4 = 1$

$\displaystyle \Rightarrow F_x = \frac 2{25}x$

$\displaystyle \Rightarrow F_x \left( 1, \frac {4 \sqrt{3}}5, \frac {4 \sqrt {3}}5\right) = \boxed{ \frac 2{25}}$

$\displaystyle \Rightarrow F_y = \frac 12 y$

$\displaystyle \Rightarrow F_y \left( 1, \frac {4 \sqrt{3}}5, \frac {4 \sqrt {3}}5\right) = \boxed{ \frac {2 \sqrt{3}}5}$

$\displaystyle \Rightarrow F_z = \frac 12 z$

$\displaystyle \Rightarrow F_z \left( 1, \frac {4 \sqrt{3}}5, \frac {4 \sqrt {3}}5\right) = \boxed{ \frac {2 \sqrt{3}}5}$

Recall that $\displaystyle (x_0,y_0,z_0) = \left( 1, \frac {4 \sqrt{3}}5, \frac {4 \sqrt {3}}5\right)$

Thus, the tangent plain is given by:

$\displaystyle F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0$

$\displaystyle \Rightarrow \boxed{ \frac 2{25}(x - 1) + \frac {2 \sqrt{3}}5 \left( y - \frac {4 \sqrt{3}}5 \right) + \frac {2 \sqrt{3}}5 \left( z - \frac {4 \sqrt{3}}5\right) = 0}$

and you can expand this if you wish and write it in the form: $\displaystyle ax + by + cz = d$, where $\displaystyle a,b,c, \mbox{ and }d$ are constants

the above is not a vector, it is a plane. i noticed you had i,j,k notation in one of your answers above, that's wrong

10. Thanks so much! I guess I'm still getting my head around this 3d calculus stuff. I'll study your solution and try to make sense of it. Thanks again