Question

The height function used in the chapter can be used to compute the change in area of a membrane when it suffers some deformation. Consider the height function \(h(x, y)\) defined over an \(L \times L\) square in the \(x-y\) plane. The function \(h(x, y)\) tells us the deformed state of the \(L \times L\) square. To compute the change in area, we divide the square into small \(\Delta x \times \Delta y\) squares and consider how each of them is deformed on the surface \(h(x, y)\) (see Figure 11.17 A). The function \(h(x, y)\) maps the small square into a tilted parallelogram. (a) Show that the two sides of the parallelogram are given by the vectors \\[ \Delta x\left(\mathbf{e}_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right) \text { and } \Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right) \\] where \(\mathbf{e}_{x}, \mathbf{e}_{y},\) and \(\mathbf{e}_{z}\) are unit vectors. (b) The magnitude of the vector product of two vectors is equal to the area subsumed by them, that is, a \(x \mathbf{b}=a b \sin \theta\) where \(\theta\) is the angle between vectors a and b. Using this fact, show that the area of the parallelogram is \\[ d^{\prime}=\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y \\] (c) Now assume that the partial derivatives appearing in the above equation are small and expand the square-root function using \(\sqrt{1+x^{2}} \approx 1+\frac{1}{2} x^{2},\) to demonstrate the useful formula for the area change, \\[ \Delta a=a^{\prime}-a=\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right] \Delta x \Delta y \\] By integrating this expression over the entire domain, we can find the total area change associated with the surface. Divide both sides by \(a=\Delta x \times \Delta y\) to find a formula for the relative change of area \(\Delta a / a\).

Short answer

For the given deformation represented by function \(h(x,y)\), the sides of the deformed square (now a parallelogram) are given by vectors \(\Delta x(\mathbf{e}_{x}+\frac{\partial h}{\partial x}\mathbf{e}_{z})\) and \(\Delta y(\mathbf{e}_{y}+\frac{\partial h}{\partial y}\mathbf{e}_{z})\). The area of this parallelogram is \(\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y\). After applying approximations and simplifications, the change in area is expressed as \(\Delta a=a^{\prime}-a=\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^2+\left(\frac{\partial h}{\partial y}\right)^2\right] \Delta x \Delta y\). The relative change in area formula is \(\Delta a / a = \frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^2+\left(\frac{\partial h}{\partial y}\right)^2\right]\).

Step-by-step solution

Interpretation of Deformation

When the square is mapped onto \(h(x, y)\) by the function, it turns into a parallelogram. The sides of this parallelogram are given by vectors which are determined by the shifts in \(x\) and \(y\) (i.e. \(\Delta x\) and \(\Delta y\)) and the deformation of the surface, as measured by the partial derivatives of \(h\).

Calculation of the Parallelogram Sides

The vectors representing the sides of the deformed square are given by \(\Delta x(\mathbf{e}_{x}+\frac{\partial h}{\partial x}\mathbf{e}_{z})\) and \(\Delta y(\mathbf{e}_{y}+\frac{\partial h}{\partial y}\mathbf{e}_{z})\) where \(\mathbf{e}_{x}, \mathbf{e}_{y},\) and \(\mathbf{e}_{z}\) are unit vectors along the \(x, y,\) and \(z\) axes respectively. These vectors are computed by adding the change in dimensions, \(\Delta x\) and \(\Delta y\), to the product of the partial derivatives of \(h\) with respect to \(x\) and \(y\), respectively, and the corresponding unit vectors.

Calculation of the Parallelogram Area

The area of the parallelogram is given by the magnitude of the cross product of two sides. Given the formulas for the sides from step 2, the cross product is \(\Delta x\Delta y\) times the square root of \(1+\left(\frac{\partial h}{\partial x}\right)^2+\left(\frac{\partial h}{\partial y}\right)^2\).

Approximation of Parallelogram Area

The partial derivatives in the equation are assumed to be small. The expression under the square root is expanded using the approximation \(\sqrt{1+x^{2}} \approx 1+\frac{1}{2} x^{2}\). For small \(x\), the expression simplifies to \(\Delta a=a^{\prime}-a=\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^2+\left(\frac{\partial h}{\partial y}\right)^2\right] \Delta x \Delta y\). Here, \(\Delta a\) is the change in area, and \(a\) and \(a'\) are the original and deformed areas, respectively.

Deriving the Formula for Relative Change of Area

To find the formula for the relative change of area, both sides of the equation are divided by \(a=\Delta x\times\Delta y\), which is the original area. By doing this, the expression simplifies to \(\Delta a/a = \frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^2+\left(\frac{\partial h}{\partial y}\right)^2\right]\).

Key concepts

Partial Derivatives

Partial derivatives are a cornerstone of multivariable calculus, playing a crucial role in many fields, including physics and engineering. When we speak about partial derivatives, we refer to the rate at which a function changes as one variable changes, while keeping all other variables constant.

In the exercise provided, the function ​\( h(x, y) \) represents the height of a membrane at any point ​(x, y) on a square grid. The partial derivatives ​\( \frac{\partial h}{\partial x} \) and ​\( \frac{\partial h}{\partial y} \) give us the slope of the function in the x-direction and y-direction, respectively. These slopes tell us how the membrane's height changes as we move in either the x or y direction independently.

To improve understanding, one might visualize the surface of a trampoline — the height changes more steeply where the partial derivatives are larger. If the trampoline (membrane) is uniformly flat, the partial derivatives at any point would be zero, as there is no change in height regardless of the direction of movement.

Parallelogram Area Calculation

The area of a parallelogram can be calculated as the base times the height, but when we deal with vectors, the cross product provides a powerful alternative. This is especially useful when we parameterize the base and height in terms of vectors.

In our exercise, the base and height of the parallelogram are represented by vector expressions. These expressions account for both the original dimensions of a square section of the membrane and the deformation imparted by the surface represented by \( h(x, y) \). The deformation converts the original square area into a parallelogram as the surface stretches and contracts.

Visualizing this can be enhanced by imagining a grid painted on a rubber sheet. If you stretch a section of the rubber sheet, the initially square grids now appear as parallelograms. For an optimal learning strategy, one could graphically illustrate how a small square distorts into a parallelogram due to deformation, which can make the calculation of the new area more intuitive to students.

Vector Product

The vector product, also known as the cross product, is a fundamental operation in vector calculus. When you take the cross product of two vectors, the result is a vector that is perpendicular to both of the original vectors, and its magnitude equals the area of the parallelogram that they span.

In the context of our problem, the vector product is used to find the area of the deformed parallelogram that represents a tiny section of the membrane after deformation. The intuition here is that the magnitude of the product not only represents the area of overlap but also includes the effect of the angle between the vectors — as the vectors become less perpendicular due to the deformation, the area they span changes.

For students seeking to deepen their conceptual understanding, exploring the properties of the cross product with interactive 3D models could be beneficial. Observing how the area changes with different vector orientations can reinforce the abstract concepts involved in calculating the area of a parallelogram using vector products.