Question

Find the partial derivatives of the function.

\(R(p,q) = ta{n^{ - 1}}(pq)\)

Short answer

Finding the first partial derivatives of the function

\(R(p,q) = {\tan ^{ - 1}}(pq)\)

Step-by-step solution

consider the following limit:

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}\)

The graph of the function \(z = f(x,y) = \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}\)in the window \((0,0.00001) \times (0,0.00001) \times (0.6,1.125)\) is shown below

Calculation:

From the graph, along the line \(y = x\)as \((x,y) \to (0,0)\), that is constant \(z = 1.125\)

The limit of this function is \(1.125\).

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2} + 3x(x) + 4{x^2}}}{{3{x^2} + 5{x^2}}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2} + 3{x^2} + 4{x^2}}}{{8{x^2}}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{9{x^2}}}{{8{x^2}}}\)

\( = \frac{9}{8}\)

Therefore,

Graph:

The graph of the function \(z = f(x,y) = \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}}\)in the window \((0,0.00001) \times (0.00001,0) \times (0.375,0.8)\) is shown below

Finding limit value:

From the graph, along the line \(y = - x\) as \((x,y) \to (0,0)\), that is constant \(z = 0.375\)

The limit of this function is \(0.375\)

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{2{x^2} + 3xy + 4{y^2}}}{{3{x^2} + 5{y^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{x^2} - 3x(x) + 4{x^2}}}{{3{x^2} + 5{x^2}}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{3{x^2}}}{{8{x^2}}}\)

\( = \frac{3}{8}\)

The limit of this function is not independent of the path passing through \((0,0)\) and \((x,y) \to (0,0)\).

The limit of function along different paths (\(y = x\) and\(y = - x\)) is different.

Hence, the limit of this function does not exist at the point \((x,y) = (0,0)\).