Problem 205 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Given } x=\sin ^{-1} t, y=\sqrt{1-t^{2}}, \text { find }\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} $$

Step-by-Step Solution

Answer

The value of \(\frac{dy}{dx}\) at \(t = \frac{1}{2}\) is \(-\frac{1}{2}\).

Step 1: Differentiate y with respect to t

We will first find the derivative of \(y\) with respect to \(t\). The function is \(y = \sqrt{1 - t^2}\), so we have: \(y = (1-t^2)^{\frac{1}{2}}\) Now, differentiate it with respect to t: \(\frac{dy}{dt} = \frac{1}{2}(1-t^2)^{-\frac{1}{2}} (-2t)\) Simplify the expression: \(\frac{dy}{dt} = \frac{-t}{\sqrt{1-t^2}}\)

Step 2: Differentiate x with respect to t

Next, we need to differentiate \(x\) with respect to \(t\). The function is \(x = \sin^{-1}t\), so we have: \(x = \sin^{-1}(t)\) Now differentiate it with respect to t: \(\frac{dx}{dt} = \frac{1}{\sqrt{1 - t^2}}\)

Step 3: Find dt/dx

Next, we need to find \(\frac{dt}{dx}\). We can do this by taking the reciprocal of \(\frac{dx}{dt}\): \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}\) Plug in the value we found in step 2: \(\frac{dt}{dx} = \sqrt{1 - t^2}\)

Step 4: Apply the Chain Rule

Now, we can apply the chain rule to find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\) Plug in the values we found in step 1 and step 3: \(\frac{dy}{dx} = \frac{-t}{\sqrt{1-t^2}} \cdot \sqrt{1 - t^2}\) Simplify the expression: \(\frac{dy}{dx} = -t\)

Step 5: Find the value of dy/dx at t = 1/2

Finally, we will find the value of \(\frac{dy}{dx}\) when \(t = \frac{1}{2}\). Plug in this value for \(t\): \(\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}} = -\frac{1}{2}\) Therefore, the value of \(\frac{dy}{dx}\) at \(t = \frac{1}{2}\) is \(-\frac{1}{2}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus for finding the derivative of a composite function. If you have two functions, say \(f(x)\) and \(g(x)\), where \(f\) is a function of \(g\), and \(g\) itself is a function of \(x\), the derivative of the composite function is found using the chain rule formula:

• \(\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\)

To apply the chain rule, you differentiate the outer function, \(f\), with respect to its inside function, \(g\), and then multiply it by the derivative of the inside function, \(g\), with respect to \(x\).
In the given exercise, we see this in action when we differentiate \(y\) with respect to \(x\) by taking the derivative \(\frac{dy}{dt}\) and multiplying it by the derivative \(\frac{dt}{dx}\). This is how we find \(\frac{dy}{dx}\) using the chain rule, interpolating between the functions \(y(t)\) and \(x(t)\). This technique is essential for solving problems involving compositions of functions.

Inverse Trigonometric Functions

Inverse trigonometric functions give us the angle whose trigonometric function is a given number. They are essential when working with angles, especially when dealing with arcs and sectors in geometry.
In this exercise, \(x = \sin^{-1}(t)\) denotes the inverse sine function, also known as arcsin. This function finds the angle \(x\) for which sine is equal to \(t\).
When differentiating inverse trigonometric functions, they have specific derivatives:

• The derivative of \(\sin^{-1}(t)\) with respect to \(t\) is \(\frac{1}{\sqrt{1 - t^2}}\), which reflects how the rate of change of the angle \(x\) varies as \(t\) changes.

This derivative is crucial for solving the exercise, as it allows us to relate changes in \(t\) to changes in \(x\) and eventually solve for \(\frac{dy}{dx}\) at a specific \(t\). Understanding these derivatives is key in problems involving trigonometric relationships and changes.

Differentiation with Respect to a Parameter

Differentiation with respect to a parameter involves finding the rate of change of one variable when the other variable is changing with some parameter—which in this context, is \(t\).
Given the functions \(x(t)\) and \(y(t)\), the whole task is to find \(\frac{dy}{dx}\), which is essentially how \(y\) changes with respect to \(x\).

• This is done by finding \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).

The process involves a back-and-forth between differentiating each of the functions with respect to the parameter and then linking these rates of change, using the chain rule, to derive \(\frac{dy}{dx}\).
Differentiating with respect to a parameter is a powerful tool, especially useful when there are two variables depending on a third one. It allows us to express relationships and rates of change between two variables indirectly through an intermediate parameter. This concept is widely used in physics and engineering where parameters describe time, angle, or any other shared variable across different functions.