Problem 33 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(y\) and \(z\) are the functions of \(x\) and if \(y^{2}+z^{2}=\lambda^{2}\), then \(y \frac{d}{d x}\left(\frac{y}{\lambda}\right)+\frac{d}{d x}\left(\frac{z^{2}}{\lambda}\right)\) is equal to (A) \(\frac{z}{\lambda} \frac{d z}{d x}\) (B) \(\frac{z}{\lambda} \frac{d x}{d z}\) (C) \(\frac{\lambda}{z} \frac{d z}{d x}\) (D) None of these

Step-by-Step Solution

Answer

(A) \(\frac{z}{\lambda} \frac{dz}{dx}\)

Step 1: Differentiate the given circle equation

Given that \(y^2 + z^2 = \lambda^2\). Differentiate both sides with respect to \(x\): \[ 2y \frac{dy}{dx} + 2z \frac{dz}{dx} = 0 \] Simplify this by dividing by 2: \[ y \frac{dy}{dx} + z \frac{dz}{dx} = 0 \]

Step 2: Derive the first expression term by term

Focus on the term \(y \frac{d}{dx} \left(\frac{y}{\lambda}\right)\). Use the product rule for differentiation: \[ y \frac{d}{dx} \left(\frac{y}{\lambda}\right) = y \left(\frac{1}{\lambda} \cdot \frac{dy}{dx}\right) = \frac{y}{\lambda} \frac{dy}{dx} \]

Step 3: Derive the second expression term by term

Now differentiate \(\frac{z^2}{\lambda}\) with respect to \(x\): \[ \frac{d}{dx}\left(\frac{z^2}{\lambda}\right) = \frac{1}{\lambda} \cdot 2z \frac{dz}{dx} = \frac{2z}{\lambda} \frac{dz}{dx} \]

Step 4: Combine and simplify expression

Combine the expression from Steps 2 and 3:\[ \frac{y}{\lambda} \frac{dy}{dx} + \frac{2z}{\lambda} \frac{dz}{dx} \]Given that \(y \frac{dy}{dx} + z \frac{dz}{dx} = 0\), replace \(y \frac{dy}{dx}\) with \(-z \frac{dz}{dx}\):\[ -\frac{z}{\lambda} \frac{dz}{dx} + \frac{2z}{\lambda} \frac{dz}{dx} = \frac{z}{\lambda} \frac{dz}{dx} \]

Step 5: Identify the correct option

Comparing with the given options, the simplified expression \(\frac{z}{\lambda} \frac{dz}{dx}\) corresponds to Option (A).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique in calculus used when differentiating equations where the dependent and independent variables are intertwined, and one is not explicitly expressed in terms of the other. For instance, when dealing with equations like a circle equation, where both variables are on the same side, implicit differentiation allows us to find the derivatives of one variable with respect to another without solving for one variable in terms of the other first.

In the given exercise, the formula \( y^2 + z^2 = \lambda^2 \) represents a circle equation, meaning both \( y \) and \( z \) depend on \( x \). To differentiate this, apply implicit differentiation:

• Differentiate each term on both sides concerning \( x \).
• Use the chain rule since both \( y \) and \( z \) are functions of \( x \).
• The derivatives are \( 2y \frac{dy}{dx} \) and \( 2z \frac{dz}{dx} \).

This process highlights how we manage terms where differentiation is not straightforward due to equations being implicit in nature.

Product Rule

The product rule is an essential part of differential calculus, especially when differentiating functions that are multiplied together. The rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product is \( u'(x)v(x) + u(x)v'(x) \).

In the exercise, the product rule helps differentiate the term \( y \frac{d}{dx} \left(\frac{y}{\lambda}\right) \):

• First, identify \( u(x) = y \) and \( v(x) = \frac{1}{\lambda} \).
• Differentiate to find \( \frac{d}{dx} \left(u(x) \times v(x)\right) = y \cdot \frac{1}{\lambda} \cdot \frac{dy}{dx} \).
• The result is \( \frac{y}{\lambda} \frac{dy}{dx} \).

Utilizing the product rule effectively aids in solving problems where functions are interconnected by multiplication.

Circle Equation

A circle equation in its standard form is \( x^2 + y^2 = r^2 \), but any rearrangement retaining this form describes a circle. In our exercise, we have \( y^2 + z^2 = \lambda^2 \), where \( \lambda \) represents the radius of a circle.

This equation describes all the points \( (y, z) \) equidistant from a central point, typically the origin. In problems involving calculus and circles, such as this one, understanding how the variables relate geometrically aids in applying differentiation techniques, especially implicit differentiation.

• The equality suggests every solution \( (y, z) \) lies on a circle with radius \( \lambda \).
• Understanding the symmetric relationship between \( y \) and \( z \) is essential when working through the problem.

Comprehending the fundamental nature of circle equations is crucial when applying differentiation to these geometric contexts.