Q. 42 - Calculus

Question

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

$f\left(x,y,z\right)={\cos }^{-1}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)$

Step-by-Step Solution

Answer

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

Step 1. Given Information.

The given function is:

$f\left(x,y,z\right)={\cos }^{-1}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)$

Step 2. Calculation.

The gradient of the given function is: 

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

Now find 

$$\begin{gathered} f_x\left(x,y,z\right)=\frac{\partial f}{\partial x}=\frac{-1\left[z{\left(-{\displaystyle \frac{1}{2}}\left(x^2+y^2+z^2\right)\right)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}·2x\right]}{\sqrt{1-{\displaystyle \frac{z^2}{x^2+y^2+z^2}}}}=\frac{zx}{x^2+y^2+z^2\sqrt{x^2+y^2}} \\ {f}_y\left(x,y,z\right)=\frac{\partial f}{\partial y}=\frac{zy}{x^2+y^2+z^2\sqrt{x^2+y^2}} \\ {f}_z\left(x,y,z\right)=\frac{\partial f}{\partial z}=\frac{-1\left[z{\left(-{\displaystyle \frac{1}{2}}\left(x^2+y^2+z^2\right)\right)}^{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}·2z+{\left(x^2+y^2+z^2\right)}^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}·1\right]}{\sqrt{1-{\displaystyle \frac{z^2}{x^2+y^2+z^2}}}}=-\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2} \end{gathered}$$

Use these above values in (1) we get 

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

Step 3. Conclusion.

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