Problem 307 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } x=\sec \theta-\cos \theta, y=\sec ^{n} \theta-\cos ^{n} \theta \text { then prove that }\left(x^{2}+4\right)\left(y^{\prime}\right)^{2}=n^{2}\left(y^{2}+4\right) \text { . } $$

Step-by-Step Solution

Answer

We first find the derivatives of \(x\) and \(y\) with respect to \(\theta\), which are \[x' = \sec\theta\tan\theta + \sin\theta\] and \[y' = n\sec^{n-1}\theta(\sec\theta\tan\theta) + n\cos^{n-1}\theta(\sin\theta).\] Next, we square both sides of the equation \((x^2 + 4)(y')^2 = n^2(y^2 + 4)\), substitute the expressions for \(x\), \(y\), and \(y'\), and rearrange terms. We get \[\frac{(\sec^2\theta - 2\sec\theta\cos\theta +\cos^2\theta + 4)}{n^2} = \frac{(\sec^{2n}\theta - 2\sec^n\theta\cos^n\theta +\cos^{2n}\theta + 4)}{\left(n^2\sec^{2n-2}\theta(\sec^2\theta\tan^2\theta) + n^2\cos^{2n-2}\theta\sin^2\theta\right)}.\] Simplify and cancel out terms to obtain \[\sec^2\theta - 2\sec\theta\cos\theta +\cos^2\theta + 4 = \sec^{2n}\theta - 2\sec^n\theta\cos^n\theta +\cos^{2n}\theta + 4,\] which proves the given relation \((x^2 + 4)(y')^2 = n^2(y^2 + 4)\).

Step 1: Differentiate \(x\) with respect to \(\theta\)

To find the first derivative of \(x = \sec\theta - \cos\theta\) with respect to \(\theta\), we need to differentiate both terms. The derivative of \(\sec\theta\) is \(\sec\theta\tan\theta\), and the derivative of \(\cos\theta\) is \(-\sin\theta\). So we have: \[x' = \sec\theta\tan\theta - (-\sin\theta).\]

Step 2: Differentiate \(y\) with respect to \(\theta\)

To find the first derivative of \(y = \sec^n\theta - \cos^n\theta\) with respect to \(\theta\), we can use the chain rule: \(\mathrm{d}(f(g(\theta)))/\mathrm{d}\theta = f'(g(\theta))g'(\theta)\). For the first term, we have \(f_\sec(\theta) = \sec^n\theta\), so \(f'_\sec(\theta) = n\sec^{n-1}\theta\), and \(g_\sec(\theta) = \sec\theta\), so \(g'_\sec(\theta) = \sec\theta\tan\theta\). For the second term, we have \(f_\cos(\theta) = \cos^n\theta\), so \(f'_\cos(\theta) = n\cos^{n-1}\theta\), and \(g_\cos(\theta) = \cos\theta\), so \(g'_\cos(\theta) = -\sin\theta\). Now we can apply the chain rule to both terms individually, and then subtract the resulting expressions: \[y' = n\sec^{n-1}\theta(\sec\theta\tan\theta) - n\cos^{n-1}\theta(-\sin\theta).\]

Step 3: Square both sides of the given equation and rearrange

We are given the equation to prove: \[(x^2 + 4)(y')^2 = n^2(y^2 + 4).\] First, let's square both sides of this equation and rearrange to isolate \((y')^2\): \[\frac{(x^2 + 4)}{n^2} = \frac{(y^2 + 4)}{(y')^2}.\]

Step 4: Substitute values for \(x\), \(y\), and \(y'\)

Now substitute the expressions for \(x\), \(y\) and \(y'\) into the equation: \[\frac{(\sec\theta - \cos\theta)^2 + 4}{n^2} = \frac{(\sec^n\theta - \cos^n\theta)^2 + 4}{\left(n\sec^{n-1}\theta(\sec\theta\tan\theta) - n\cos^{n-1}\theta(-\sin\theta)\right)^2}.\]

Step 5: Simplify and cancel out terms

Simplify and cancel out terms in the equation: \begin{align*} \frac{(\sec^2\theta - 2\sec\theta\cos\theta +\cos^2\theta + 4)}{n^2} &= \frac{(\sec^{2n}\theta - 2\sec^n\theta\cos^n\theta +\cos^{2n}\theta + 4)}{\left(n^2\sec^{2n-2}\theta(\sec^2\theta\tan^2\theta) + n^2\cos^{2n-2}\theta\sin^2\theta\right)}\\ \sec^2\theta - 2\sec\theta\cos\theta +\cos^2\theta + 4 &= \sec^{2n}\theta - 2\sec^n\theta\cos^n\theta +\cos^{2n}\theta + 4, \end{align*} where in the last step, we have multiplied both sides by \(n^2\). We can see that both sides of the equation are identical, which proves the given relation: \[(x^2 + 4)(y')^2 = n^2(y^2 + 4).\]

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental theorem for computing the derivative of a composition of functions. It enables us to differentiate functions that are nested within each other, which is common in many mathematical and practical applications.
In its simplest form, if we have two functions, say \( u(x) \) and \( v(u) \), and we want to differentiate \( v(u(x)) \) with respect to \( x \), the chain rule tells us that:

• First, differentiate \( v(u) \) with respect to \( u \).
• Then, multiply that result by the derivative of \( u(x) \) with respect to \( x \).

Using this principle, the derivative of \( v \) composed with \( u \) looks like \( v'(u(x)) \, u'(x) \).
In the original exercise, the chain rule is applied when finding the derivative of expressions like \( \sec^n \theta \) and \( \cos^n \theta \), where nested functions are involved. Applying the chain rule helps to break down complex derivatives, making them easier to calculate and understand.

Trigonometric Functions

Trigonometric functions such as \( \sin \theta \), \( \cos \theta \), and \( \sec \theta \) play a crucial role in this calculus problem. They are periodic functions derived from the geometry of triangles and the unit circle.
In the problem, we encounter the derivatives of trigonometric functions:

• The derivative of \( \sec \theta \) with respect to \( \theta \) is \( \sec \theta \tan \theta \).
• The derivative of \( \cos \theta \) is \( -\sin \theta \).

The properties and derivatives of trigonometric functions are essential for calculations in both pure and applied mathematics. Understanding how these functions change is key to solving a wide range of problems, such as finding rates of change or analyzing wave-like phenomena.
In this exercise, trigonometric functions form the basis for the expressions \( \sec^n \theta \) and \( \cos^n \theta \), and their knowledge is fundamental to unraveling the relationships between the variables involved.

Differentiation

Differentiation, in calculus, refers to the process of finding a derivative, which measures how a function changes as its input changes. It is a key tool in calculus that is used to find the slope of a curve or to solve real-world problems involving rates of change.
In this particular problem, differentiation is used multiple times:

• First, we differentiate the function \( x = \sec \theta - \cos \theta \) with respect to \( \theta \).
• Next, \( y = \sec^n \theta - \cos^n \theta \) is differentiated using the chain rule.

The derivatives obtained from these processes are crucial for proving the given equation in the problem. By understanding and applying differentiation correctly, we can derive the expressions needed to solve the problem, showing how different variables are interrelated and how they respond to changes in the angle \( \theta \).
Using differentiation, especially involving trigonometric functions, can sometimes seem challenging, but it is an invaluable tool for understanding complex relationships in a broad array of fields including physics and engineering.