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 \(h(x, y) \text { (see Figure } 11.17(\mathrm{A}))\) 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 $$\Delta x\left(e_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right), \text { and } \Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right).$$ 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 \(x \mathbf{b}=\) absin\theta, where \(\theta\) is the angle between vectors a and \(\mathbf{b}\). Using this fact show that the area of the parallelogram is $$a^{\prime}=\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y.$$ (c) Now assume that the partial derivatives appearing in the above equation are small and expand the squareroot function using \(\sqrt{1+x^{2}}=1+4 x^{2},\) 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

The relative change of area \(\Delta a/a\) for the deformed \(L \times L\) square due to given deformation according to the height function \(h(x, y)\) is \(\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right]\)

Step-by-step solution

Derive the Side Vectors

The deformation of a small square can be represented by two vectors. The tilting due to the height function causes the square to be deformed into a parallelogram. The small sides \(\Delta x\) and \(\Delta y\) don't just move in the x, y dimensions (on the base plane), but also 'upwards' along the z axis as per \(h(x,y)\). This representation results in the vectors: \[\Delta x\left(\mathbf{e}_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right) \text{ and }\Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right)\]

Use Vector Product to Determine Area

The area of a parallelogram defined by vectors a and b can be found using the absolute value of their vector product \(|a \times b|\). So, the area \(a^{\prime}\) of the new parallelogram is given by: \[a^{\prime}=\left|\Delta x\left(\mathbf{e}_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right) \times \Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right)\right|\] which becomes: \[a^{\prime}=\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y\] after simplifying and using \(\mathbf{e}_{x} \times \mathbf{e}_{y} = \mathbf{e}_{z}\) and \(\| \mathbf{e}_{x} \| = \| \mathbf{e}_{y} \| = 1\).

Approximate the Area Change and Derive the Relative Change of Area

If the partial derivatives are small, the square root operation can be expanded using the given approximation. Applying this gives the area change \(\Delta a = a' - a\) where \(a = \Delta x \Delta y\), which becomes: \[\Delta a = a^{\prime} - a = \frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right] \Delta x \Delta y\]. The relative change of area \(\Delta a/a\), thus can be obtained by dividing both sides by \(\Delta x \times \Delta y\) yielding: \[\frac{\Delta a}{a} = \frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right]\]

Key concepts

Membrane Area Calculation

The study of membranes in cell biology involves understanding how these structures change under various conditions. Membrane area calculation is critical when examining the effects of deformation on these delicate biological systems. By dividing the area into smaller segments, it's possible to calculate the change in the total area when deformation occurs. To do this, we consider each small square pre-deformation and its transformation into a parallelogram as a result of the altering forces.

Imagine you have a sheet representing a cell membrane that's been gently pushed from underneath. The local rise or dip changes the flat sheet into a surface with varying height. Here, basic geometry comes into play—tiny squares on the original sheet no longer stay perfect squares. Instead, their sides lengthen or tilt, covering a different amount of space. This is where the vectors, as defined in the exercise, are crucial in representing those tilted sides. By understanding vector properties, students can calculate the new areas of these parallelograms and thus quantify the membrane's deformation.

Height Function in Biology

The height function is a mathematical representation used to describe the three-dimensional shape of biological membranes or surfaces. This function, designated as \(h(x, y)\), provides information about how high a point is above a reference plane, essentially giving us the 'shape' of the biological structure in the z-direction. In the context of cell biology, this function is significant because it models deformations of cellular membranes which can be due to a variety of biological processes like osmotic pressure or mechanical forces.

In practical applications, the height function enables researchers to predict how certain areas of a cell might behave under stress or during normal physiological activities. For instance, consider a red blood cell passing through a narrow capillary; the height function could assist in understanding how its membrane changes shape to fit through the tight space.

Vector Product in Cell Biology

Understanding Vector Product through Cell Membrane Distortion

One of the fundamental concepts in physics and engineering, the vector product, finds a unique application in cell biology. When applied to membrane deformation, calculating the vector product of the sides of the small deformed squares gives the area of the parallelogram, which is essential for determining the change in membrane area due to deformation.

The vector product (also known as the cross product) results in a vector that is perpendicular to the plane formed by the two original vectors, with a magnitude that is proportional to the area subsumed by that plane. For a biology student, visualizing the cell membrane's deformation with the cross product can help grasp how two-dimensional surface area changes when extended into three dimensions.

Partial Derivatives in Physical Biology

Partial derivatives play a vital role when we wish to comprehend how a function changes in response to variations in its variables, particularly for functions with more than one variable, like the height function \(h(x, y)\) used in biology. These derivatives allow us to see how the membrane's deformation in one direction is independent of its deformation in the perpendicular direction.

In the calculation of the membrane's area change, partial derivatives help us quantify the slope of the membrane surface in the x and y directions. Biologically, understanding these slopes can reveal the nature of the cell membrane's response when subjected to varying conditions. For example, the way a cell membrane curves and bulges when it absorbs water or when it's pressed by neighboring cells can be analyzed mathematically through these partial derivatives.