Question

A contour map is given for a function \(f\). Use it to estimate \({f_x}(2,1)\)and \({f_y}(2,1) \)

Short answer

\({f_y}(2,1) \approx - 2\)

\({f_x}(2,1) \approx 3\)

Step-by-step solution

Estimating \({f_x}(2,1)\) and \({f_y}(2,1)\)

Let some function values at some points \((x,y)\):

\(f(3,1) \approx 14\), \(f(2,1) \approx 10\), \(f(1,1) \approx 8\), \(f(2,0) \approx 12\), \(f(12,2) \approx 8\)

The partial derivative of \(f\) w.r.t \(x\)

\({f_x}(2,1) = \mathop {\lim }\limits_{n \to 0} \frac{{f(2 + h) - f(2,1)}}{h}\)

Take \(h = \Delta x = 1\)

\({f_x}(2,1) = f(2 + 1,1) - f(2,1)\)

\( \approx f(3,1) - f(2,1)\)

\( \approx 14 - 10\)

\( \approx 4\)

Calculating \({f_x}(2,1)\)

Take \(h = \Delta x = - 1\)

\({f_x}(2,1) = f(2 - 1,1) - f(2,1)\)

\( \approx f(1,1) - f(2,1)\)

\( \approx \frac{{8 - 10}}{{ - 1}}\)

\( \approx 2\)

Taking average of two values:

The partial derivative of \(f\) w.r.t \(x\) at the point \((2,1)\) is calculated as follow:

\(f(2,1) \approx \frac{{4 + 2}}{2} = 3\)

Therefore, \(f(2,1) \approx 3\)

The partial derivative of \(f\) w.r.t \(y\)

Take \(h = \Delta x = 1\), \({f_y}(2,1) = f(2,2) - f(2,1) \approx \frac{{8 - 10}}{1} \approx - 2\)

The partial derivative of \(f\) w.r.t \(y\)at the point \((2,1)\)is \({f_y}(2,1) \approx \frac{{ - 2 - 2}}{2} = - 2\)

Therefore , \({f_y}(2,1) \approx - 2\)

\({f_x}(2,1) \approx 3\)