Active Calculus - Multivariable

Evaluate \(f(x,y) = 6 - \frac{x^2}{2} - y^2\) and its first partial derivatives at \((x_0,y_0)\text{;}\) that is, find \(f(1,1)\text{,}\) \(f_x(1,1)\text{,}\) and \(f_y(1,1)\text{.}\)

We want our plane to match the height of our surface at \((x_0,y_0)\) and the value \(z_0\) from Key Idea 11.5.3 will need to be the \(z\)-coordinate value where our plane will intersect the surface. What is the \(z\)-coordinate of the point where the tangent plane and the surface should intersect?

If we want our plane to have the same behavior as the surface \(z=f(x,y)\) near our input \((x_0,y_0)\text{,}\) then the plane will need to match the behavior of the traces \(x=1\) and \(y=1\) near our point of interest.

Sketch the traces of \(f(x,y) = 6 - \frac{x^2}2 - y^2\) for \(y=y_0=1\) and \(x=x_0=1\) below in Figure 11.5.4. Draw the tangent lines to the each of the traces when the input (either \(x\) or \(y\) as appropriate) is 1.

Give the slopes of the tangent lines to the traces that you drew in the previous part and write a few sentences to explain why the tilt of the tangent plane in the \(x\)-direction is given by the partial derivative \(f_x\) and the tilt of the tangent plane in the \(y\)-direction is given by the partial derivative \(f_y\text{.}\) You will likely want to talk about how each of these slopes/partial derivatives relates to the traces (of the surface and the plane) in Figure 11.5.5.