Problem 64 - Biocalculus Calculus for the Life Sciences

Question

(a) Show that \(\cos \left(x^{2}\right) \geqslant \cos x\) for 0\(\leqslant x \leqslant 1\) (b) Deduce that \(\int_{0}^{\pi / 6} \cos \left(x^{2}\right) d x \geqslant \frac{1}{2}\)

Step-by-Step Solution

Answer

(a) Since \( x^2 \leqslant x \), we have \( \cos(x^2) \geqslant \cos(x) \); (b) Therefore, \( \int_{0}^{\pi/6} \cos(x^2) \, dx \geqslant \frac{1}{2} \).

Step 1: Analyze the Functions Differently at x^2 and x

We start by comparing the behavior of the functions within the given range. Since the cosine function is decreasing for the interval \[ 0 \leqslant x \leqslant 1, \]this implies that for any value of \( x \) in this range, \( x^2 \leqslant x \) because squaring values between 0 and 1 results in smaller numbers. Hence, we have \[ \cos(x^2) \geqslant \cos(x) \]since cosine decreases as the input increases when \( 0 \leqslant x \leqslant 1 \).

Step 2: Deduce the Integral Inequality

To show the inequality \[ \int_{0}^{\pi / 6} \cos(x^2) \, dx \geqslant \frac{1}{2}, \]we will use the result from the previous step. Since \( \cos(x^2) \geqslant \cos(x) \) for all \(x \in [0, \pi/6] \), we have:\[ \int_{0}^{\pi / 6} \cos(x^2) \, dx \geqslant \int_{0}^{\pi / 6} \cos(x) \, dx. \] Let's compute the simpler integral:\[ \int_{0}^{\pi/6} \cos(x) \, dx. \]The integral of \( \cos(x) \) is \( \sin(x) \). Applying the limits \( 0 \) and \( \pi/6 \) gives\[ \sin\left(\frac{\pi}{6}\right) - \sin(0) = \frac{1}{2} - 0 = \frac{1}{2}. \] Thus, \[ \int_{0}^{\pi / 6} \cos(x^2) \, dx \geqslant \frac{1}{2}. \]

Key Concepts

Understanding the Cosine Function

The cosine function, denoted as \( \cos(x) \), is a fundamental trigonometric function that helps us understand relationships in right-angled triangles and periodic phenomena. In a right triangle, \( \cos(\theta) \) represents the ratio of the adjacent side's length to the hypotenuse for a given angle \( \theta \). Cosine is particularly notable for its periodic nature:

• **Range:** The value of cosine ranges from -1 to 1.
• **Period:** The function repeats every \(2\pi\), meaning \(\cos(x + 2\pi) = \cos(x)\).
• **Symmetry:** Cosine is an even function, implying \(\cos(-x) = \cos(x)\).

These properties impact the investigation of functions like \( \cos(x^2) \) compared to \( \cos(x) \) in specific intervals. Here, since the cosine function decreases as the argument increases in the range \([0, 1]\), we know that because \(x^2 \leq x\), \( \cos(x^2) \geq \cos(x)\). This relation holds because squaring reduces the value of \(x\) when \(0 \leq x \leq 1\).

Integral Inequality Explained

An integral inequality involves comparing the integrals of two functions over a specific interval. For this problem:

• We are asked to compare \( \int_{0}^{\pi / 6} \cos(x^2) \, dx \)
• Against the integral \( \int_{0}^{\pi / 6} \cos(x) \, dx \).

Since \( \cos(x^2) \geq \cos(x) \) in the interval \([0, \pi/6]\), it follows that the area under the curve \( \cos(x^2) \) from 0 to \( \pi/6 \) is greater than or equal to the area under \( \cos(x) \) over the same limits.
If we calculate \( \int_{0}^{\pi / 6} \cos(x) \, dx \), we find it equals \(1/2\), confirming the original inequality \( \int_{0}^{\pi / 6} \cos(x^2) \, dx \geq 1/2\). This deduced relation is a simple yet profound insight into how functions behave and can be used to compare areas under their curves in calculus.

The Role of Calculus in Life Sciences

Calculus, with its tools of differentiation and integration, is essential to the life sciences. It allows researchers and practitioners to explore dynamic changes in studying living systems. Important applications include:

• **Exploring Growth Rates:** Derivatives help determine the growth rates of populations, cells, and bacteria.
• **Modeling Biological Processes:** Integral calculus is used to model the accumulation of substances, such as drug concentration in the bloodstream or nutrient absorption.
• **Analyzing Structures:** Analysis of curves and shapes through calculus aids in understanding anatomical structures like neural pathways or blood vessel formations.

The integral inequality concept from the given exercise demonstrates an application of calculus in exploring functional behaviors, providing insights into potential applications like modeling the effectiveness of a drug over time when related to concentration, or analyzing rates of reaction in biochemistry.