Q. 39 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given Information.

The given function is:

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

Step 2. Calculation.

The gradient of the given function is:

$$\begin{gathered} z=f\left(x,y\right)=\sqrt{x^2+y^2} \\ \nabla f(x,y)=\left[f_x\left(x,y\right), f_y\left(x,y\right)\right]-------\left(1\right) \end{gathered}$$ 

Now find 

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

Use these above values in (1) we get 

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

Step 3. Conclusion.

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