Question

1. for the following functions, sketch a contour map with at least five level curves, with evenly spaced differences of z - values.

(a) z = x² + y².
(b) z = √(x² + y²).
(c) z = 3x - y.

Explanation:

Step1: Recall the concept of level - curves

Level - curves of a function $z = f(x,y)$ are given by $f(x,y)=k$, where $k$ is a constant.

Step2: For part (a) $z=x^{2}+y^{2}$

Set $x^{2}+y^{2}=k$. These are circles centered at the origin with radius $r = \sqrt{k}$ for $k\geq0$. When $k = 0$, the level - curve is just the point $(0,0)$. For $k = 1$, $x^{2}+y^{2}=1$ (a unit circle), for $k = 4$, $x^{2}+y^{2}=4$ (a circle of radius 2), for $k=9$, $x^{2}+y^{2}=9$ (a circle of radius 3), for $k = 16$, $x^{2}+y^{2}=16$ (a circle of radius 4).

Step3: For part (b) $z=\sqrt{x^{2}+y^{2}}$

Set $\sqrt{x^{2}+y^{2}}=k$. Then $x^{2}+y^{2}=k^{2}$ ($k\geq0$). These are also circles centered at the origin with radius $r = k$. When $k = 0$, the level - curve is the point $(0,0)$. For $k = 1$, $x^{2}+y^{2}=1$; for $k = 2$, $x^{2}+y^{2}=4$; for $k=3$, $x^{2}+y^{2}=9$; for $k = 4$, $x^{2}+y^{2}=16$.

Step4: For part (c) $z = 3x-y$

Set $3x-y=k$, which can be rewritten as $y=3x - k$. These are straight lines with slope $m = 3$ and $y$ - intercept $b=-k$. When $k = 0$, $y = 3x$; when $k = 1$, $y=3x - 1$; when $k=-1$, $y=3x + 1$; when $k = 2$, $y=3x - 2$; when $k=-2$, $y=3x+2$.

To sketch the contour maps:

• For (a) and (b), draw concentric circles centered at the origin with appropriate radii for different non - negative values of $k$.
• For (c), draw straight lines with slope 3 and different $y$ - intercepts for different values of $k$.

Answer:

Sketch concentric circles for (a) and (b) and straight lines with slope 3 for (c) as described above for different values of $k$.