Problem 3 - Calculus an Applied Approach

Question

Find \(d y / d x\) \(y^{2}=1-x^{2}, 0 \leq x \leq 1\)

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(x\), \(dy/dx\), is \(-x / \sqrt{1 - x^{2}}\).

Step 1: Implicit Differentiation

Differentiate both sides of the equation \(y^{2}=1-x^{2}\) with respect to \(x\). The derivative of \(y^{2}\) would be \(2y(dy/dx)\) using the chain rule and the derivative of \(1-x^{2}\) would be \(-2x\). So the differentiated equation will be: \(2y(dy/dx) = -2x\).

Step 2: Solve for \(dy/dx\)

To solve for \(dy/dx\), you just need to divide both sides of the differentiated equation by \(2y\). So we have: \(dy/dx = -2x / 2y = -x/y\).

Step 3: Substituting \(y\) from the Given Equation

From the equation \(y^{2}=1-x^{2}\), solving for \(y\) will result to two roots which are positive and negative since \(y^2\) was squared: \(y = \sqrt{1 - x^{2}}\) and \(y = -\sqrt{1 - x^{2}}\). To ensure that \(y\) is a real number in the provided interval for \(x\), we will only choose the positive root (since the interval of \(x\) is from 0 to 1): \(y = \sqrt{1 - x^{2}}\). Substitute \(y\) into \(dy/dx = -x/y\), we get: \(dy/dx = -x / \sqrt{1 - x^{2}}\).

Key Concepts

Derivative of Implicit Functions

Understanding the derivative of implicit functions is a key aspect in calculus, particularly when the relationship between variables cannot be described as a simple function. Implicit differentiation helps us to find the derivative of one variable with respect to another in equations where the dependent variable, often denoted as 'y', cannot be isolated.

For example, in the equation given in our exercise, \(y^2 = 1 - x^2\), 'y' cannot be easily solved for. By applying implicit differentiation, we take the derivative of both sides with respect to 'x' and include \(dy/dx\) wherever 'y' appears. This step usually involves applying the chain rule. Following the differentiation, we isolate \(dy/dx\) to solve the derivative within the context of the equation, allowing us to understand how 'y' changes as 'x' changes.

Using implicit differentiation can streamline the process of finding slopes or rates of change in complex relationships, where explicit differentiation is not straightforward or even possible.

Chain Rule

The chain rule comes into play when differentiating compositions of functions—essentially, a function within a function. In the context of implicit differentiation, the chain rule is paramount because we often deal with variables that are functions of one another.

For instance, when differentiating \(y^2\) with respect to 'x', we actually treat 'y' as a function of 'x' (even though that function is not explicitly stated). According to the chain rule, the derivative of \(y^2\) is \(2y \big ( dy/dx \) ), where \(2y\) is the outer derivative (derivative of \(y^2\) with respect to 'y'), and \(dy/dx\) is the inner derivative (derivative of 'y' with respect to 'x').

The chain rule is a crucial tool in calculus as it allows us to tackle complex differential problems by breaking them down into simpler parts that are easier to manage.

Solving Derivatives

The process of solving derivatives involves finding the rate at which one quantity changes with respect to another. In our textbook example, the goal is to find \(dy/dx\), which represents the rate of change of 'y' with respect to 'x'.

Once we have applied implicit differentiation and the chain rule, we arrange the resulting equation to solve for \(dy/dx\). In the provided exercise, dividing by \(2y\) brings us to \(dy/dx = -x/y\). This simplification reveals the relationship between the rate of change of 'y' to 'x'.

It's important when solving derivatives to consider the domain of the function. In this case, we selected the positive root of 'y' to ensure it's a real number within the given interval for 'x'. This domain restriction often arises in real-world applications where certain solutions are not feasible or relevant, highlighting the practicality of this mathematical step.