Problem 76 - Calculus Early Transcendentals

Question

Generalize Exercise 75 by considering a wave described by the function \(z=A+\sin (a x-b y)\) where \(a, b,\) and \(A\) are real numbers. a. Find the direction in which the crests and troughs of the wave are aligned. Express your answer as a unit vector in terms of \(a\) and \(b\) b. Find the surfer's direction - that is, the direction of steepest descent from a crest to a trough. Express your answer as a unit vector in terms of \(a\) and \(b\)

Step-by-Step Solution

Answer

The unit vector representing the direction of the crests and troughs is \(\frac{\left\langle b, a \right\rangle}{\sqrt{a^2 + b^2}}\), and the unit vector representing the surfer's direction is \(\frac{\left\langle -a, b \right\rangle}{\sqrt{a^2 + b^2}}\).

Step 1: Find the gradient of z

To find the gradient of the function \(z\), we need to find the partial derivatives of \(z\) with respect to \(x\) and \(y\). Given the wave function: $$ z = A + \sin(ax - by) $$ Using basic calculus, find the partial derivatives. The partial derivative of \(z\) with respect to \(x\) is: $$ \frac{\partial z}{\partial x} = a\cos(ax - by) $$ The partial derivative of \(z\) with respect to \(y\) is: $$ \frac{\partial z}{\partial y} = -b\cos(ax - by) $$ Thus, the gradient of \(z\) is given by the vector: $$ \nabla z = \left\langle a\cos(ax-by), -b\cos(ax-by) \right\rangle $$

Step 2: Find the direction of crests and troughs

The crests and troughs of the wave are aligned in the direction perpendicular to the gradient. The vector perpendicular to the gradient of \(z\) can be found by swapping the x and y components, and changing the sign of one of the components: $$ \text{Wave direction} = \left\langle b\cos(ax-by), a\cos(ax-by) \right\rangle $$ By normalizing the above vector, we obtain the unit vector representing the direction of the crests and troughs: $$ \text{Unit vector of wave direction} = \frac{\left\langle b\cos(ax-by), a\cos(ax-by) \right\rangle}{\sqrt{(b\cos(ax-by))^2 + (a\cos(ax-by))^2}} $$ $$ = \frac{\left\langle b, a \right\rangle}{\sqrt{a^2 + b^2}} $$

Step 3: Find the surfer's direction

The surfer's direction is the direction of steepest descent, which is also the negative of the gradient. Using the gradient we calculated in step 1, the negative of the gradient is: $$ -\nabla z = \left\langle -a\cos(ax-by), b\cos(ax-by) \right\rangle $$ By normalizing the above vector, we obtain the unit vector representing the surfer's direction: $$ \text{Unit vector of surfer's direction} = \frac{\left\langle -a\cos(ax-by), b\cos(ax-by) \right\rangle}{\sqrt{(-a\cos(ax-by))^2 + (b\cos(ax-by))^2}} $$ $$ = \frac{\left\langle -a, b \right\rangle}{\sqrt{a^2 + b^2}} $$ The unit vector of the surfer's direction will be: $$ \text{Unit vector of surfer's direction} = \frac{\left\langle -a, b \right\rangle}{\sqrt{a^2 + b^2}} $$ In summary, the unit vector representing the direction of the crests and troughs is \(\frac{\left\langle b, a \right\rangle}{\sqrt{a^2 + b^2}}\), and the unit vector representing the surfer's direction is \(\frac{\left\langle -a, b \right\rangle}{\sqrt{a^2 + b^2}}\).