1. ## gradient fields

$\text{find the gradient field}$ $\vec{F}=\vec{V}\varphi$
$\text{for the following functions$\varphi$}\\$
$\text{a.$\displaystyle\vec{V}(x,y)=\tan^{-1}\left(\frac{y}{x} \right)$}\\$
$\text{b.$\displaystyle\vec{V}(x,y,z)=\cos\left(\frac{y}{x} \right)$}$

ok this completely new to me

2. ## Re: gradient fields

did you learn what the gradient is?

The gradient of a scalar function $f(x,y,z)$ is known as $\nabla f$ and in cartesian coordinates $\nabla f = \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)$

3. ## Re: gradient fields

Since you have only two independent variables, x and y, you must have $\nabla f= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.
(I suspect that what you have written as " $\vec{V}f$" was intended to be $\nabla f$".)

So what are $\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial x}$ and $\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}$?

And what are $\frac{\partial cos\left(\frac{y}{x}\right)}{\partial x}$ and $\frac{\partial cos\left(\frac{y}{x}\right)}{\partial y}$?

4. ## Re: gradient fields

Originally Posted by HallsofIvy
Since you have only two independent variables, x and y, you must have $\nabla f= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.
(I suspect that what you have written as " $\vec{V}f$" was intended to be $\nabla f$".)

So what are $\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial x}$ and $\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}$?

And what are $\frac{\partial cos\left(\frac{y}{x}\right)}{\partial x}$ and $\frac{\partial cos\left(\frac{y}{x}\right)}{\partial y}$?
do you mean this?
$\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}=-\dfrac{\arctan\left(y\right)}{x^2}$

5. ## Re: gradient fields

Originally Posted by bigwave
do you mean this?
$\frac{\partial tan^{-1}\left(\frac{y}{x}\right)}{\partial y}=-\dfrac{\arctan\left(y\right)}{x^2}$
How did you arrive at this? Have you learned how to take partial derivatives?

$\dfrac{\partial \tan^{-1}\left(\dfrac{y}{x}\right)}{\partial y} = \dfrac{1}{1+\left(\dfrac{y}{x}\right)^2}\cdot \dfrac{1}{x} = \dfrac{x}{x^2+y^2}$

6. ## Re: gradient fields

no I did it thru an online derivative calculator

7. ## Re: gradient fields

Originally Posted by bigwave
no I did it thru an online derivative calculator
Wolfram|Alpha: Computational Knowledge Engine

But, honestly, if you have not learned how to take partial derivatives, this will be next to impossible for you to do on your own. It may be prudent to learn how to take partial derivatives.

8. ## Re: gradient fields

I was hoping an example of that would be done here

9. ## Re: gradient fields

Originally Posted by bigwave
I was hoping an example of that would be done here
A partial derivative with respect to a variable means treat all other variables as constants and take the derivative. So, in the example $\dfrac{\partial \tan^{-1}\left( \dfrac{y}{x} \right)}{\partial y}$, we are treating $x$ as a constant and then taking the derivative. Do you know how to take derivatives in general? Because there is an entire course on derivatives called Calculus. An example here would not help. You would need to actually learn differential calculus to be able to understand derivatives in general.

10. ## Re: gradient fields

Im in calculus III right now
I missed a lot of classes lately so this partial derivaties is new
I have another forum to go to if you don't want to show the steps
I just cant find a good example of this.

11. ## Re: gradient fields

Originally Posted by bigwave
Im in calculus III right now
I missed a lot of classes lately so this partial derivaties is new
I have another forum to go to if you don't want to show the steps
I just cant find a good example of this.
I gave you the steps. If you still do not understand what to do, you either need to take a class that will catch you up or tell us where you are getting lost. You seem to not be understanding material from Calculus I/II, which is exactly what is required to do this derivative.

From Calculus I/II:

$\dfrac{d}{dx}(\tan^{-1} u) = \dfrac{1}{1+u^2}\dfrac{du}{dx}$

I just used the Chain rule. If you don't recall the Chain rule, you may want to review Calculus I. If you don't know the derivative of the inverse trigonometric functions, you should review Calculus II. Not knowing the prerequisites to your class is the reason I said you should review how to take derivatives.