Problem 286 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The short answer to the given question is: \(\frac{d^{2} y}{d x^{2}} = \frac{\csc^2 t}{a \sin t}\)

Step 1: Find dy/dt and dx/dt

We'll start by finding the derivatives of x and y with respect to t: \(\frac{dx}{dt} = \frac{d}{dt}(a \cos t) = -a \sin t\) \(\frac{dy}{dt} = \frac{d}{dt}(a \sin t) = a \cos t\)

Step 2: Find the first derivative dy/dx

Next, we'll determine the first derivative (dy/dx) using the chain rule: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{-a \sin t}\)

Step 3: Simplify the first derivative

Simplify the first derivative: \(\frac{dy}{dx} = - \cot t\)

Step 4: Find the second derivative d²y/dx²

Now, differentiate the first derivative with respect to x using the chain rule to find the second derivative: \(\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(- \cot t) = \frac{d}{dt} (- \cot t) \cdot \frac{dt}{dx}\)

Step 5: Calculate d/dt(-cot t) and dt/dx

Let's find the derivative of -cot(t) with respect to t and the derivative of t with respect to x: \(\frac{d}{dt}(-\cot t) = -\csc^2 t\) \(\frac{dt}{dx} = \frac{1}{(dx/dt)} = \frac{1}{(-a \sin t)}\)

Step 6: Calculate d²y/dx²

Now, we will substitute the results from Step 5 into the equation from Step 4: \(\frac{d^{2}y}{dx^{2}} = -\csc^2 t \cdot \frac{1}{(-a \sin t)}\)

Step 7: Simplify the second derivative

Finally, we can simplify the second derivative: \(\frac{d^{2}y}{dx^{2}} = \frac{\csc^2 t}{a \sin t}\) Therefore, the second derivative of y with respect to x is: \(\frac{d^{2} y}{d x^{2}} = \frac{\csc^2 t}{a \sin t}\)

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus. It allows us to find derivatives of composite functions. When dealing with parametric equations, where variables are defined in terms of a third variable (often time \(t\)), the Chain Rule becomes particularly useful. In the context of the given exercise, functions \(x\) and \(y\) depend on the parameter \(t\).

• The Chain Rule helps us relate \( \frac{dy}{dx} \) by finding \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).
• We start by taking the separate derivatives of \(x\) and \(y\) with respect to \(t\) to set up for using the Chain Rule.
• By obtaining these, we can express \( \frac{dy}{dx} \) as \( \frac{dy/dt}{dx/dt} \). This step essentially "chains" the derivative calculations together.

This concept not only eases complex computations but also provides clarity and allows step-by-step simplification, which brings us to the calculation of higher-order derivatives like the second derivative.

Second Derivative

The second derivative, as seen in the task, represents the rate of change of the rate of change. In simple words, it's the derivative of the first derivative. The exercise requires us to find \( \frac{d^2y}{dx^2} \) from the parametric form of \(x\) and \(y\).

• First, you start with finding the first derivative using the Chain Rule, which gives us \(-\cot t\).
• The next step involves differentiating \(-\cot t\) with respect to \(x\), which means solving \(\frac{d}{dx}(-\cot t)\).
• Since \(x\) and \(y\) are in terms of \(t\), we employ the Chain Rule again to find the second derivative: we need \(\frac{d}{dt}(-\cot t)\) and \(\frac{dt}{dx}\).

By substituting the appropriate derivatives, we calculated the second derivative as \(\frac{\csc^2 t}{a \sin t}\). This gives deeper insights into the curvature and behavior of the graph represented by the parametric equations.

Trigonometric Functions

Trigonometric functions play a central role in the given exercise. Let's understand what's happening with \(x = a \cos t\) and \(y = a \sin t\). These parametric equations define a circle of radius \(a\).

• \(\cos t\) and \(\sin t\) represent the x and y coordinates of a point on the circle as \(t\) varies.
• Derivatives of these trigonometric functions reveal changes in the circle's coordinates over time. For instance, differentiating \(a \cos t\) results in \(-a \sin t\), which shows how \(x\)'s rate of change depends on \(t\).
• Similarly, differentiating \(a \sin t\) yields \(a \cos t\), presenting \(y\)'s rate of change concerning \(t\).

Understanding these derivatives is key to interpreting the behavior of parametric curves, like how the slope (derivative) changes along the curve. It emphasizes the importance of trigonometric identities and derivatives in solving calculus problems.