Gradient of a Vector/Function

  • Thread starter vorcil
  • Start date
  • Tags
    Gradient
In summary, the gradient of f(x,y,z) is found to be: .
  • #1
vorcil
398
0

Homework Statement



Find the gradients of the following functions:

When I say gradient, I'm not just differentiating the functions, apparently I have to do it this way (because it's in my physics book)

f(x,y,z) = x^2 + y^3 + z^4
f(x,y,z) = x^2 y^3 z^4
f(x,y,z) = e^x sin(y) ln(z)

Homework Equations



[tex]\nabla {\bf r} = \frac{\partial r}{\partial x} \hat{x} + \frac{\partial r}{\partial y} \hat{y} + \frac{\partial r}{\partial z} \hat{z} [/tex]

I'm not just differentiating the functions

where [tex] \frac{r}{|r|} [/tex] = gradient

The Attempt at a Solution



for the first question, finding the gradient of
f(x,y,z) = x^2 + y^3 + z^4

[tex]\nabla {\bf f(x,y,z) } = \frac{\partial r}{\partial x} \hat{x} + \frac{\partial r}{\partial y} \hat{y} + \frac{\partial r}{\partial z} \hat{z} [/tex]

differentiating partially,[tex]\nabla[/tex]
[tex] \nabla
(x^2 + y^3 + z^4) = 2x \hat{x} + 3y^2 \hat{y} + 4z^3 \hat{z} [/tex]

finding the magnitude of the function
[tex] \sqrt{(x^2)^2 + (y^3)^2 + (z^4)^2} = \sqrt{x^4 + y^6 + z^8} [/tex]

using the formula to find the gradient,
[tex] \frac{r}{|r|} = \frac{\nabla (x^2 + y^3 + z^4) }{\sqrt{x^4 + y^6 + z^8}} = \frac{2x \hat{x} + 3y^2 \hat{y} + 4z^3 \hat{z}}{\sqrt{x^4 + y^6 + z^8}} [/tex]

not sure if this is right or not, also I don't know if I can simplifier it further and if this satisfies as an answer,
my book has no answers in the back to check, and solutions manuals are impossible to get to students, and the only way to find out if I'm right is to either post it here, or ask my lecturer, and I feel that this is a question that most people should be able to do, so it's embarrassing for me to ask my lecturer at tutorials
 
Last edited:
Physics news on Phys.org
  • #2
[edit] fixed the latex
 
Last edited:
  • #3
i may be missing something, but why not just use the cartesian form of the gradient

[tex] \nabla = \begin{pmatrix}
\frac{\partial }{\partial x} \\
\frac{\partial }{\partial y} \\
\frac{\partial }{\partial z}
\end{pmatrix}[/tex]
 

What is the gradient of a vector?

The gradient of a vector is a mathematical operation that calculates the rate of change of a vector's magnitude and direction in a specific direction. It is represented by a vector with the direction of the steepest increase and its magnitude equal to the rate of change.

How is the gradient of a vector calculated?

The gradient of a vector is calculated by taking the partial derivatives of the vector's components with respect to each of its independent variables. The resulting vector is then represented as a sum of its components.

What is the relationship between the gradient of a vector and its direction?

The gradient of a vector is always perpendicular to the direction of the vector's steepest increase. This means that the gradient points in the direction of the vector's greatest rate of change.

Can the gradient of a vector be negative?

Yes, the gradient of a vector can be negative. A negative gradient indicates that the vector's magnitude is decreasing in a specific direction, while a positive gradient indicates an increase in magnitude in that direction.

What is the gradient of a function?

The gradient of a function is a vector that represents the direction and magnitude of the function's steepest increase at a specific point. It is calculated by taking the partial derivatives of the function with respect to each of its independent variables and combining them into a vector.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
770
  • Calculus and Beyond Homework Help
Replies
8
Views
470
  • Calculus and Beyond Homework Help
Replies
3
Views
772
  • Calculus and Beyond Homework Help
Replies
8
Views
876
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
694
  • Calculus and Beyond Homework Help
Replies
9
Views
547
  • Calculus and Beyond Homework Help
Replies
6
Views
853
  • Calculus and Beyond Homework Help
Replies
2
Views
512
  • Calculus and Beyond Homework Help
Replies
20
Views
461
Back
Top