Question

Studying for my last final. I'm just trying to check my understanding of specific concepts. If I have a vector-valued function, say F(x,y,z)=(x^2 +e^(yz^2), sin(xy+z)) and I'm supposed to find the derivative of F, I just use the Jacobian matrix (which is a matrix of first order partial derivatives, correct?)

Answer 1

\[\vec F =\langle x^2 +e^{yz^2}, \sin(xy+z)\rangle\]So the derivative of f would just be the gradient vector , \(\nabla f\)

Answer 2

And also, where is your z-coordinate?

Answer 3

What do you mean my z-coordinate? @Jhannybean My problem just asks me to calculate the derivative of that vector valued function. My professor said the answer is:

Answer 4

I am not quite sure that method or format, but in finding derivatives, you have a defined function which is continuous, so just simply find retricepartials, \(f_x~,~ f_y~,~ f_z\) and write it as a vector.

Answer 5

\[\vec F = \langle x^2 +e^{yz^2}, \sin(xy+z)\rangle\]\[\vec F'(x,y,z) =\langle (x^2+e^{yz^2})' ~,~ (\sin(xy+z))'\rangle\]

Answer 6

you're defining derivative as the determinant of jacobian matrix
so simply find the partials of components and work the determinant

Answer 7

Can you just find the partials with respect to all 3 variables? Or does it have to be in that style...

Answer 8

i see z component is missing in your vector field
is it deliberate ?

Answer 9

I see why you cannot just find all 3 partials... it is because the variables are common to both components.

Answer 10

Yeah, that's how my professor wrote the problem.

Answer 11

Oh i just opened ur attachment, one sec..

Answer 12

@Jhannybean I understand now what you mean by the z component is missing. I guess she just wrote this problem poorly?

Answer 13

But using this method, you are able to find your partial for z regardless.

Answer 14

Overall, I just want to know if when finding the derivative of some vector-valued function, I just use the jacobian matrix and find the determinant of the that matrix?

Answer 15

then your work is easy
sinc the determinant is not defined for a non-square matrix, you just plug the partials in the matrix and you're done

Answer 16

So solving for the matrix is not needed?

Answer 17

Okay, and if it were a square matrix I would just calculate the determinant?

Answer 18

there are various kinds of derivatives of a vector valued function
depends on what derivative you're finding

Answer 19

I suppose, the answer is just a 2x3 matrix @Jhannybean

Answer 20

well it just said "the derivative"

Answer 21

Interesting. That simplifies everything!!

Answer 22

yes and the professor must have told u in the class what "derivative" means

Answer 23

https://www.youtube.com/watch?v=d8qDWRUBpYM

Answer 24

gradient, curl and divergence are other kinds of derivatives of a vector valued function

Answer 25

Otherwise you would have \[\vec F = \langle x^2 +e^{yz^2}, \sin(xy+z)\rangle\]\[f_x= \langle 2x ~,~ y\cos(xy+z)\rangle\]\[f_y = \langle z^2e^{yz^2}~,~ (1)\cos(xy+z)\rangle\]\[f_z = \langle 2yze^{yz^2}~,~\cos(xy+z)\rangle\]

Answer 26

And writing them out in that fashion is perfectly fine as well

Answer 27

Ooh, mistake.

Answer 28

\[f_y =\langle z^2e^{yz^2}~,~ x\cos(xy+z)\rangle\]

Answer 29

thanks for the help guys :)

Answer 30

your professor is refering to jacobian matrix for derivative
so you just find the jacobian matrix if he asks you for derivative in exam

determinant wont be necessary i guess