Problem 287 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } x=a(t-\sin t), y=a(1-\cos t), \text { find } \frac{d^{2} y}{d x^{2}} $$

Step-by-Step Solution

Answer

The second derivative of y with respect to x is: \(\frac{d^2y}{dx^2} = \frac{(1-\cos t)(\cos t) + \sin^2 t}{(1 - \cos t)^2(\sin t)}\).

Step 1: Find the first derivatives of x and y with respect to t

Differentiate x and y with respect to t. \(x = a(t - \sin t)\), so \(\frac{dx}{dt} = a(1 - \cos t)\) \(y = a(1 - \cos t)\), so \(\frac{dy}{dt} = a(\sin t)\)

Step 2: Find the first derivative of y with respect to x

Use the chain rule to find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) \(\frac{dy}{dx} = \frac{a(\sin t)}{a(1 - \cos t)}\)

Step 3: Find the second derivatives with respect to t

Differentiate \(\frac{dy}{dx}\) and \(\frac{dx}{dt}\) again with respect to t: \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{a\sin t}{a (1 - \cos t)}\right)\) \(\frac{d^2x}{dt^2} = \frac{d}{dt}\left(a(1 - \cos t)\right) = a\sin t\)

Step 4: Find the second derivative of y with respect to x

Use the chain rule again: \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{d(axy)}{d(atx)}\cdot\frac{d(atx)}{d(axy)}\right) \) \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\) Substitute the expressions we found before: \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{a\sin t}{a(1 - \cos t)}\right)}{a\sin t}\) Now we have to calculate the derivative of our function with respect to t: \(\frac{d}{dt}\left(\frac{a\sin t}{a(1 - \cos t)}\right) = \frac{a(1-\cos t)(\cos t) + a\sin^2 t}{a^2(1 - \cos t)^2}\) Finally, substitute this expression back into the second derivative formula: \(\frac{d^2y}{dx^2} = \frac{\frac{a(1-\cos t)(\cos t) + a\sin^2 t}{a^2(1 - \cos t)^2}}{a\sin t}\) And simplify: \(\frac{d^2y}{dx^2} = \frac{(1-\cos t)(\cos t) + \sin^2 t}{(1 - \cos t)^2(\sin t)}\)

Key Concepts

Parametric Equations

When dealing with curves in calculus, especially those which cannot be neatly expressed as functions of a single variable, parametric equations are very handy. You can describe curves by expressing the coordinates \(x\) and \(y\) as functions of a third variable, often \(t\), which is called the parameter.
For example, given:

• \( x = a(t - \sin t) \)
• \( y = a(1 - \cos t) \)

the parameter \(t\) can represent time or any other continuous variable that varies through the system. This method allows us to study complex curves and shapes beyond simple function representations. Using parametric equations, we can easily manipulate and explore the nature of curves, thereby providing a versatile tool in calculus.

Chain Rule

The chain rule is a fundamental concept in calculus, especially useful when differentiating composite functions. In simpler terms, when you have a function inside another function, you use the chain rule to differentiate it.
To apply the chain rule, differentiate the outer function and multiply it by the derivative of the inner function. It's important to remember that this rule can be applied iteratively if there are multiple layers of composition.
In our specific problem with parametric equations, the chain rule is crucial because we are essentially dealing with two functions: \(y(t)\) and \(x(t)\). When finding \(\frac{dy}{dx}\), we need to take the derivative of \(y\) with respect to \(t\), then \(x\) with respect to \(t\), and divide the two results:

• \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

which highlights the elegant use of the chain rule for parametric equations.

Second Derivative

The second derivative of a function provides information about its concavity and points of inflection. In the context of parametric equations, the second derivative \(\frac{d^2y}{dx^2}\) can be found by differentiating the first derivative of \(\frac{dy}{dx}\) with respect to \(t\) again.
Initially, we obtain \(\frac{dy}{dx}\) from the parametric functions and their derivatives concerning \(t\). The second derivative requires a further application of the chain rule:

• Start with \(\frac{dy}{dx}\) and its differentiation concerning \(t\).
• Divide the result by the derivative of \(x\) with respect to \(t\).

Mathematically, it shows us how the rate of change (slope) of the tangent line itself changes with respect to \(x\), which can be critical for understanding the behavior of a curve.

Differentiation Techniques

Differentiation is a core part of calculus, involving deriving one function from another. Several techniques can be used, and choosing the right one depends on the form of the functions. The exercise involves differentiating parametric equations, meaning we need to use:

• Basic differentiation rules, particularly for trigonometric functions like \(\sin t\) and \(\cos t\).
• The chain rule, as outlined earlier.
• Simplification skills, to transform complex expressions into a more manageable form.

With parametric differentiation, each calculation is an interplay between different differentiation rules, conducted in a sequence to derive the desired derivatives. In this specific problem, differentiating \(\frac{dy}{dx}\) and further using the derivative terms related to \(t\) demonstrates a real-world application of differentiation techniques. Selecting the appropriate differentiation method is important for simplifying and correctly solving complex calculations.