Question

Find the partial derivatives of the function.

\(u(r,\theta ) = sin(rcos\theta )\)

Short answer

Finding the first partial derivatives of the function

\(u(r,\theta ) = \sin (r\cos \theta )\)

Step-by-step solution

Let us consider \(\mathop {lim}\limits_{(x,y,z) \to (0,0,0)} \frac{{yz}}{{{x^2} + 4{y^2} + 9{z^4}}}\)

Now verify that the limit exists or not. Let the function,

The function \(f(x,y,z)\) approaches \((0,0,0)\) along the \(x\)-axis:

Then \(y = 0,z = 0\) gives \(f(x,y,z) = \frac{{(0)(0)}}{{{x^2} + 4{{(0)}^2} + 9{{(0)}^4}}}\). Use \(f(x,y,z) = \frac{{yz}}{{{x^2} + 4{y^2} + 9{z^4}}}\)

\( = \frac{0}{{{x^2}}}\)for all \(x \ne 0\).

Thus, \(f(x,y,z) \to 0\) along the lines \(y = 0\) and \(x = 0\).

The function \(f(x,y,z)\) approaches \((0,0,0)\) along the \(y\)-axis:

Then \(x = 0,z = 0\)gives \(f(x,y,z) = \frac{{(y)(0)}}{{{{(0)}^2} + 4{{(y)}^2} + 9{{(0)}^4}}}\). Use \(f(x,y,z) = \frac{{yz}}{{{x^2} + 4{y^2} + 9{z^4}}}\)

\( = \frac{0}{{4{y^2}}}\)for all \(y \ne 0\).

Thus, \(f(x,y,z) \to 0\) along the lines \(x = 0\) and \(z = 0\).

The function \(f(x,y,z)\) approaches \((0,0,0)\)along the line \(x = y = z\):

\(f(x,y,z) = \frac{{(x)(x)}}{{{x^2} + 4{{(x)}^2} + 9{{(x)}^2}}}\). Use \(f(x,y,z) = \frac{{yz}}{{{x^2} + 4{y^2} + 9{z^4}}}\)

\( = \frac{{{x^2}}}{{{x^2} + 4{x^2} + 9{x^2}}}\)

\(\begin{array}{l}\\ = \frac{{{x^2}}}{{14{x^2}}}\end{array}\)

Thus, \(f(x,y,z) \to \frac{1}{{14}}\) along the lines \(x = y = z\). Since the function \(f\) has two different limits along two different lines.

Therefore, the limit \(\mathop {\lim }\limits_{(x,y,z) \to (0,0,0)} \frac{{yz}}{{{x^2} + 4{y^2} + 9{z^4}}}\) does not exist.