Problem 47 - Single Variable Calculus

Question

Strain on Vertebrae The strain (percentage of compression) on the lumbar vertebral disks in an adult human as a function of the \(\operatorname{load} x\) (in kilograms) is given by $$ f(x)=7.2956 \ln \left(0.0645012 x^{0.95}+1\right) $$ What is the rate of change of the strain with respect to the load when the load is \(100 \mathrm{~kg}\) ? When the load is \(500 \mathrm{~kg}\) ? Source: Benedek and Villars, Physics with Illustrative Examples from Medicine and Biology.

Step-by-Step Solution

Answer

The rate of change of the strain with respect to the load when the load is 100 kg is approximately \(0.23933 \%/\text{kg}\), and when the load is 500 kg, it is approximately \(0.057165 \%/\text{kg}\).

Step 1: Find the derivative of the function f(x)

We are given the function: $$ f(x) = 7.2956 \ln \left(0.0645012 x^{0.95} + 1\right) $$ To find the rate of change of the strain with respect to the load, we need to find the derivative of this function with respect to x. We can use the chain rule of differentiation: $$ \dfrac{d}{dx} \ln (u(x)) = \dfrac{u'(x)}{u(x)} $$ So, let \(u(x) = 0.0645012 x^{0.95} + 1\). Then, $$ u'(x) = 0.0645012 \times 0.95 x^{0.95 - 1} $$ Now we can find \(\dfrac{d}{dx} f(x)\): $$ f'(x) = 7.2956 \times \dfrac{u'(x)}{u(x)} = 7.2956 \times \dfrac{0.0645012 \times 0.95 x^{0.95 - 1}}{0.0645012 x^{0.95} + 1} $$ Now that we have the derivative, we can move on to evaluate it at the given loads (100 kg and 500 kg).

Step 2: Evaluate the derivative at x = 100 kg

Let's find the rate of change of the strain with respect to the load when the load is 100 kg. We need to plug in x = 100 into our derivative: $$ f'(100) = 7.2956 \times \dfrac{0.0645012 \times 0.95 \times 100^{0.95 - 1}}{0.0645012 \times 100^{0.95} + 1} $$ Calculating the numerical values, we get: $$ f'(100) \approx 0.23933 $$ So, the rate of change of the strain with respect to the load when the load is 100 kg is approximately 0.23933 %/kg.

Step 3: Evaluate the derivative at x = 500 kg

Lastly, let's find the rate of change of the strain with respect to the load when the load is 500 kg. We need to plug in x = 500 into our derivative: $$ f'(500) = 7.2956 \times \dfrac{0.0645012 \times 0.95 \times 500^{0.95 - 1}}{0.0645012 \times 500^{0.95} + 1} $$ Calculating the numerical values, we get: $$ f'(500) \approx 0.057165 $$ So, the rate of change of the strain with respect to the load when the load is 500 kg is approximately 0.057165 %/kg.

Key Concepts

Derivative of a Logarithmic Function

Understanding the derivative of a logarithmic function is crucial when dealing with growth patterns, decay, and many phenomena in physics and engineering, such as strain measurement in biological tissues. In our case, the function representing strain incorporates a natural logarithm, which is written in mathematical syntax as \(\ln(u)\).

To differentiate a function like \(f(x) = \ln(u(x))\), we apply the rule \(\frac{d}{dx}\ln(u(x)) = \frac{u'(x)}{u(x)}\). It's essential to remember that \(u(x)\) must be a differentiable function of \(x\), and \(u'(x)\) represents the derivative of \(u(x)\) with respect to \(x\).

Chain Rule of Differentiation

When we come across a composite function like \(\ln(0.0645012 x^{0.95} + 1)\), the chain rule of differentiation comes into play. This rule allows us to find the derivative of a function that is composed of other functions—referred to as the outer and inner functions.

The chain rule states that if you have a composite function \(f(g(x))\), then the derivative \(f'(x)\) is \(g'(x)\cdot f'(g(x))\). In simpler terms, it's the derivative of the outer function, multiplied by the derivative of the inner function. So, by identifying \(u(x)\) as our inner function in the strain formula, we can use the chain rule to find the rate of change effectively.

Real-World Applications of Calculus

Calculus is not just an abstract mathematical tool; it has countless real-world applications across various fields, including physics, engineering, economics, and medicine. In the context of physics and biology, calculus helps us measure how physical quantities such as strain on the lumbar vertebral disks change under loads.

For example, the derivative \(f'(x)\) we found gives us the rate of change of strain per unit change in load. Knowing this rate of change can be vital for designing medical equipment, understanding human biomechanics, and even preempting injury by understanding the limits of strain that human tissues can safely withstand under different conditions.