Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

a. Let \(y=\arctan x\) so that \(x=\tan y(x)\). Show that $$ \sec ^{2} y(x)=1+x^{2} $$ b. Use \(x=\tan y(x)\) and the chain rule, \([G(u(x))]^{\prime}=G^{\prime}(u(x)) u^{\prime}(x),\) to conclude that $$ 1=\left(\sec ^{2} y(x)\right) y^{\prime}(x) \quad \text { and } \quad y^{\prime}(x)=\frac{1}{1+x^{2}} $$ c. Use the result of b. to show that $$ \int_{0}^{1} \frac{1}{1+x^{2}} d x=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The integral \( \int_0^1 \frac{1}{1+x^2} \, dx \) evaluates to \( \frac{\pi}{4} \).

Step 1: Understand the Problem Statement for Part a

We start with the function definition \(y = \arctan(x)\) which implies that \(x = \tan(y)\). We need to show that \(\sec^2 y(x) = 1 + x^2\).

Step 2: Use Trigonometric Identity

Recall the trigonometric identity: \( \sec^2(y) = 1 + \tan^2(y) \). Since \(x = \tan(y)\), we can substitute to get \(\sec^2(y) = 1 + x^2\). This verifies part a.

Step 3: Apply the Chain Rule for Part b

Given \(x = \tan(y)\), we apply the chain rule \(\frac{d}{dx}(y) = y'(x)\) and setup the derivative \(1 = (\sec^2(y)) y'(x)\). Hence, \(y'(x) = \frac{1}{\sec^2(y)}\).

Step 4: Substitute Identity into Derivative

From step 2, we know \(\sec^2(y) = 1 + x^2\). Substitute this into the previous step to get \(y'(x) = \frac{1}{1+x^2}\). This addresses part b.

Step 5: Evaluate the Integral for Part c

We need to evaluate \( \int_0^1 \frac{1}{1+x^2} \, dx \). Recognize that \( \frac{1}{1+x^2} \) is the derivative of \( \arctan(x) \). Thus, the integral evaluates to \( \arctan(x) \) from 0 to 1.

Step 6: Compute the Definite Integral

Substitute the bounds into \( \arctan(x) \), giving \( \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \). Thus, \( \int_0^1 \frac{1}{1+x^2} \, dx = \frac{\pi}{4} \).

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between angles and sides in triangles, particularly right angles. One of the core trigonometric functions is the tangent, denoted as \( \tan(y) \), which is the ratio of the opposite side to the adjacent side in a right triangle with angle \( y \).
Another important function is the secant, \( \sec(y) \), which is the reciprocal of the cosine function (\( \cos(y) \)). The trigonometric identity \( \sec^2(y) = 1 + \tan^2(y) \) is crucial when solving problems involving the derivatives of inverse trigonometric functions. This identity simplifies to \( \sec^2(y) = 1 + x^2 \) when substituting \( x = \tan(y) \), smoothly aligning with what was shown in the original exercise to establish this trigonometric relationship.

Derivative

A derivative in calculus represents the rate at which a function is changing at any given point, often described as the function's slope or the instantaneous rate of change. In the exercise we explore how differentiating the function \( y = \arctan(x) \) leads to discovering its derivative, \( y'(x) \).
Finding the derivative of \( y = \arctan(x) \) involves taking into account the identity we proved earlier \( \sec^2(y) = 1 + x^2 \). The derivative here becomes \( y'(x) = \frac{1}{1+x^2} \), demonstrating how derivatives of inverse trigonometric functions can often resemble fractions involving known trigonometric identities. As in our example, knowing that \( \sec^2(y) = 1 + x^2 \) helps simplify the expression of the derivative.

Integral Calculus

Integral calculus is concerned with the concept of integration, which is essentially the reverse process of differentiation. It is about finding the antiderivative or the accumulated value, which gives the area under a curve of a function.
In this exercise, we evaluate the integral \( \int_{0}^{1} \frac{1}{1+x^{2}} \, dx \). This integrand, \( \frac{1}{1+x^2} \), is known to have the antiderivative \( \arctan(x) \). Evaluating the integral from 0 to 1 involves substituting the bounds into this antiderivative. This results in \( \arctan(1) - \arctan(0) \), which simplifies to \( \frac{\pi}{4} \), because \( \arctan(1) \) equals \( \frac{\pi}{4} \) and \( \arctan(0) \) equals 0. The definite integral thus represents the area under the curve \( \frac{1}{1+x^2} \) from 0 to 1 in the Cartesian plane.

Chain Rule

The chain rule is a formula for calculating the derivative of the composition of two or more functions. It is pivotal when differentiating expressions where variables are nested within other functions.
For the function \( y = \arctan(x) \) and given \( x = \tan(y) \), the chain rule allows us to differentiate as \([G(u(x))]' = G'(u(x)) \cdot u'(x)\). In this exercise, we applied the chain rule to find \( y'(x) \), initially expressed as \( 1 = (\sec^2(y))y'(x) \). The chain rule helps conclude that \( y'(x) = \frac{1}{1+x^2} \) by considering how \( \sec^2(y) = 1 + x^2 \). This showcases the power of the chain rule in connecting the derivative of composite functions with trigonometric identities and simplifying complicated expressions.