Problem 65 - Calculus Early Transcendentals

Question

Prove that the level curves of the plane \(a x+b y+c z=d\) are parallel lines in the \(x y\) -plane, provided \(a^{2}+b^{2} \neq 0\) and \(c \neq 0.\)

Step-by-Step Solution

Answer

Answer: Yes, the level curves are parallel lines in the \(xy\)-plane.

Step 1: Find the equation of the level curves

For a level curve of the plane, we keep the value of \(z\) constant. Let \(z=k,\) where \(k\) is a constant. Substitute this value in the given equation: \(ax + by + ck = d.\)

Step 2: Write the equation in terms of \(x\) and \(y\)

Now, replace \(z\) with \(k\) and rearrange the equation to isolate \(y:\) \(ax + by + c(k) = d\) \(by = d - ax - ck\) \(y = \frac{1}{b}(d - ax - ck)\)

Step 3: Determine if the lines are parallel

Observe that the equation of the level curve is a linear equation in \(x\) and \(y,\) in the form \(y = mx + n,\) where \(m=-\frac{a}{b}\) is the slope and \(n=\frac{d-bk}{b}\) is the y-intercept. To prove that the lines are parallel, they must have the same slope. In our case, the slope \(m=-\frac{a}{b}\) is constant for all level curves, regardless of the value of \(k\). Therefore, the level curves of the plane \(ax+by+cz=d\) are parallel lines in the \(xy\)-plane, given \(a^2+b^2 \neq 0\) and \(c \neq 0.\)