Q. 38 - Calculus

Question

Find the gradient of the given functions in Exercises 37–42. 

$z={\tan }^{-1}\left(\frac{y}{x}\right)$

Step-by-Step Solution

Answer

The gradient of the given function is $\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.

Step 1. Given Information.

The given function is:

$z={\tan }^{-1}\left(\frac{y}{x}\right)$

Step 2. Calculation

The gradient of the given function is: 

$$\begin{gathered} z=f\left(x,y\right)={\tan }^{-1}\left(\frac{y}{x}\right) \\ \nabla f(x,y)=\left[f_x\left(x,y\right),f_y\left(x,y\right)\right]-------\left(1\right) \end{gathered}$$

Now find 

$$\begin{gathered} f_x=\frac{\partial f(x,y)}{\partial x}=\frac{1·\left(-{\displaystyle \frac{y}{x^2}}\right)}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}=-\frac{y}{x^2+y^2} \\ {f}_y=\frac{\partial f(x,y)}{\partial y}=\frac{1·\left({\displaystyle \frac{1}{x}}\right)}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}=\frac{x}{x^2+y^2} \end{gathered}$$

Use these above values in (1) we get 

$\nabla f(x,y)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$

Step 3. Conclusion.

The gradient of the given function is  $\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$.