91Ó°ÊÓ

Level curves are curves on the Cartesian plane that represent points at which a function has a constant value. In this case, we want to find the level curves of $$z = \sqrt{y-x^2-1}$$. We will do this by choosing different values of z and plugging them into the equation, so we have an equation in terms of x and y that we can plot on the Cartesian plane.

Our function is given as: $$z = \sqrt{y-x^2-1}$$ We can rewrite this as an equation in x and y, by squaring both sides: $$(z)^{2} = (y-x^2-1)$$ Now we have an equation in x, y, and z of the form: $$z^2 + x^2 + 1 = y$$

We will choose two values for z, based on the given window of \([-5, 5] \times[-5, 5]\). Note that z cannot be negative, and it must be limited in order to fit into the window. Let's choose z=0 and z=1 as our level curve values.

1. For z=0, the equation is: $$y = x^2+1$$ This is a parabola shifted 1 unit upward. It opens upward, with a vertex at (0, 1). As x varies within \([-5,5]\), y varies within \([1,26]\). But since the window limits y to \([-5,5]\), this level curve will be truncated at the top at y=5. 2. For z=1, the equation is: $$y = (1+x^{2})^2$$ This is another parabola shifted 1 unit upward, but wider than the previous one. Its vertex is at (0, 1). As x varies within \([-5,5]\), y varies within \([1,676]\). This level curve will also be truncated at the top at y=5. After plotting these level curves, label them with their respective z-values. The level curve at $$y = x^2+1$$ corresponds to z=0, and the level curve at $$y = (1+x^{2})^2$$ corresponds to z=1.