Question

Find \(z=f(x, y)\) and use the total differential to approximate the quantity. \(\frac{1-(3.05)^{2}}{(5.95)^{2}}-\frac{1-3^{2}}{6^{2}}\)

Short answer

The approximate value of the given expression using the total differential is -0.00926.

Step-by-step solution

Rewrite the function

Firstly, the given function is \(f(x, y) = \frac{1-x^{2}}{y^{2}}\), where \(x = 3\) and \(y = 6\).

Compute Partial Derivatives

Calculate the partial derivative with respect to \(x\) and \(y\), denoted as \(f_{x}\) and \(f_{y}\) respectively. We get \(f_{x} = \frac{-2x}{y^{2}}\) and \(f_{y} = \frac{2(1-x^{2})}{y^{3}}\).

Calculate the values of \(f_{x}\) and \(f_{y}\) at \(x = 3, y = 6\)

Substitute the values of \(x\) and \(y\) into \(f_{x}\) and \(f_{y}\), we get \(f_{x} = \frac{-2*3}{6^{2}} = -\frac{1}{6}\) and \(f_{y} = \frac{2(1-3^{2})}{6^{3}} = -\frac{1}{18}\).

Determine the changes

Next, we consider a small change in values of \(x\) and \(y\) which are \(dx = 3.05 - 3 = 0.05\) and \(dy = 6 - 5.95 = 0.05\).

Compute the total differential

The total differential \(dz\) is given by \(dz = f_{x}*dx + f_{y}*dy. Plug in the calculated values, we get \(dz = -\frac{1}{6} * 0.05 - \frac{1}{18} * 0.05 = -\frac{0.05}{6} - \frac{0.05}{18} = -0.00925925\).

Get the approximation

The total differential \(dz\) can be used to approximate the change in the value of the function \(f(x, y)\) for small changes in \(x\) and \(y\). So the approximate value of given expression is \(dz = -0.00926\) (rounded to five decimal places).