The following excerpt is taken from Multiscale modeling of red blood cell mechanics and blood flow in malaria. Bruce Caswell et al. PLoS Computational Biology. 7.12 (Dec. 2011).

Membrane macroscopic properties
Extension of the linear analysis of [19] for a regular hexagonal network allows us to uniquely relate the model parameters and the network macroscopic elastic properties (shear, area-compression, and Young's moduli), see [13, 17] for details. The derived shear modulus of the membrane is given by

μ0  =     √	3  KBT	4plmx0  (  x0	2(1 − x0)3   −   1	4(1 − x0)2   +   1	4  )  +    √	3  kp(n + 1)	(4l0m+1)  (7)

where l0 is the equilibrium spring length and x0  =  l0/lm. The area-compression K and Young's moduli are equal to 2μ0 + ka + kd  and  4Kμ0/(K + μ0), respectively.

Reading further, you learn that all quantities involved are positive. Assume that x0  =  1/2 and kp  =  0.

Which of the following statements are always true?

(i) The graph of μ0 as a function of T is linear.
(ii) Viewing μ0 as a function of p, we can say that μ0 is proportional to T.
(iii) The graph of μ0 as a function of T has a slope of   5 √	3  KB	8plm  .

Let's break down the equation for the shear modulus \( \mu_0 \) and analyze each statement step by step.

The given formula for \( \mu_0 \) is:

\[
\mu_0 = \frac{\sqrt{3} k_B T}{4 p l_m x_0} \left( \frac{x_0^2}{(1 - x_0)^3} - \frac{1}{4(1 - x_0)^2} + \frac{1}{4} \right) + \frac{\sqrt{3} k_p (n+1)}{4 l_0^{m+1}}
\]

We are told to assume:
- \( x_0 = \frac{1}{2} \),
- \( k_p = 0 \).

### Step 1: Substitution of \( x_0 = 1/2 \)
When \( x_0 = \frac{1}{2} \), let's simplify the term inside the parentheses:

\[
\frac{x_0^2}{(1 - x_0)^3} = \frac{\left(\frac{1}{2}\right)^2}{\left(1 - \frac{1}{2}\right)^3} = \frac{\frac{1}{4}}{\frac{1}{8}} = 2
\]
\[
\frac{1}{4(1 - x_0)^2} = \frac{1}{4\left(1 - \frac{1}{2}\right)^2} = \frac{1}{4 \times \frac{1}{4}} = 1
\]
\[
\frac{1}{4} = 0.25
\]

Substituting these into the equation for \( \mu_0 \), we get:

\[
\mu_0 = \frac{\sqrt{3} k_B T}{4 p l_m \times \frac{1}{2}} \left( 2 - 1 + 0.25 \right)
\]
\[
\mu_0 = \frac{\sqrt{3} k_B T}{2 p l_m} \times 1.25
\]
\[
\mu_0 = \frac{5 \sqrt{3} k_B T}{8 p l_m}
\]

### Step 2: Analyze the statements

#### (i) The graph of \( \mu_0 \) as a function of \( T \) is linear.

From the simplified formula, \( \mu_0 = \frac{5 \sqrt{3} k_B T}{8 p l_m} \), it is clear that \( \mu_0 \) is directly proportional to \( T \). Since this is a linear equation in \( T \), **statement (i) is true**.

#### (ii) Viewing \( \mu_0 \) as a function of \( p \), we can say that \( \mu_0 \) is proportional to \( T \).

In the equation, \( \mu_0 = \frac{5 \sqrt{3} k_B T}{8 p l_m} \), \( \mu_0 \) is proportional to \( T \) and inversely proportional to \( p \). Therefore, **statement (ii) is true**.

#### (iii) The graph of \( \mu_0 \) as a function of \( T \) has a slope of \( \frac{5 \sqrt{3} k_B}{8 p l_m} \).

The equation for \( \mu_0 \) is \( \mu_0 = \frac{5 \sqrt{3} k_B T}{8 p l_m} \), which is in the form of \( \mu_0 = mT \), where the slope \( m = \frac{5 \sqrt{3} k_B}{8 p l_m} \). This matches the slope given in the statement, so **statement (iii) is true**.

### Conclusion:
All three statements (i), (ii), and (iii) are **true**.

---

Would you like further details on any part of this explanation? 

Here are some related questions to deepen your understanding:

1. What is the physical significance of the shear modulus \( \mu_0 \) in this context?
2. How would the expression for \( \mu_0 \) change if \( x_0 \) were different from \( \frac{1}{2} \)?
3. What impact does the parameter \(

Explore the linear relationship between the shear modulus μ0 and temperature (T) in red blood cell mechanics, as modeled by the formula μ0 = 5√3 KBT / (8plm). This page covers concepts of elasticity, proportional relationships, and linear equations, suitable for college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation