Partial Derivative exercise (help, please)

[DaveMat]

Directional Derivative exercise (help, please)

Hello, I'm not new in the forum, but I've just use it once, and it's been a long time, so please excuse me and correct me if the format of my question is wrong or something.

This is the exercise:

Consider the surface \(\displaystyle \, S:\, x^2 y z\, -\, z^2 x\, =\, z y\)

<sup>\(\displaystyle P\, =\, (0,\, -1,\, 0)\)

Find \(\displaystyle \, D\, \underset{u}{\rightarrow}\, z(P)\)

if: \(\displaystyle \, \overrightarrow{u}\, =\, (2,\, -3)\)

I can calculate the gradient just fine (I think):

\(\displaystyle \nabla\, f(x,\, y,\, z)\, =\, (2xyz\, -\, z^2,\, x^2z\, -\, z,\, x^2 y\, -\, 2xz\, -\, y)\)

where \(\displaystyle \, f(x,\, y,\, z)\, =\, x^2 y z\, -\, z^2 x\, -\, zy\, =\, 0\)

and I now that I need to make u a unit vector, because the Partial Derivative is:

\(\displaystyle \nabla\, f(P)\, *\, \dfrac{\overrightarrow{u}}{\Vert \, \overrightarrow{u}\, \Vert}\)

right?

Now the question is, how I'm suppose to use the dot product between a vector (x,y,z) and a vector with just (x,y) ?

Is there something that I'm missing here?

Any help, I'd appreciate it.</sup>

 

[ksdhart2]

Now the question is, how I'm suppose to use the dot product between a vector (x,y,z) and a vector with just (x,y)?

Well, in general, you can take the dot product of a two-dimensional vector with a three-dimensional vector by assigning the two dimensional vector a z-component of zero. Two vectors are equal if they have the same direction and same magnitude. Adding a z-component of 0 changes neither the direction, nor the magnitude, so it's still an equivalent vector, yes?

That said, I'm not sure that's the approach you want here. I'll admit I'm a bit confused, and I think it might help to see the exact, full text of the exercise, but it looks to me like you're being asked to find a directional derivative of z(P) in the direction of the vector u. Assuming that it's not a typo, the z(P) bit indicates to me that you'd want to convert the surface's equation into z = (a function of x and y). What happens if you try that?

 

[DaveMat]

Well, in general, you can take the dot product of a two-dimensional vector with a three-dimensional vector by assigning the two dimensional vector a z-component of zero. Two vectors are equal if they have the same direction and same magnitude. Adding a z-component of 0 changes neither the direction, nor the magnitude, so it's still an equivalent vector, yes?

That said, I'm not sure that's the approach you want here. I'll admit I'm a bit confused, and I think it might help to see the exact, full text of the exercise, but it looks to me like you're being asked to find a directional derivative of z(P) in the direction of the vector u. Assuming that it's not a typo, the z(P) bit indicates to me that you'd want to convert the surface's equation into z = (a function of x and y). What happens if you try that?

Thank you for answering, yes! you're right, it is a Directional Derivative, I knew that, I don't know why did I write Partial Derivative.