Problem 88 - Calculus Volume 1

Question

Confirm the approximations by using the linear approximation at \(x=0\). \(\frac{1}{\sqrt{1-x^{2}}} \approx 1\)

Step-by-Step Solution

Answer

Linear approximation confirms the expression \( \frac{1}{\sqrt{1-x^2}} \approx 1 \) at \( x = 0 \).

Step 1: Identify the Function

The function we need to approximate is given by the expression \( f(x) = \frac{1}{\sqrt{1-x^2}} \). To use linear approximation, we first need to determine the value of the function and its derivative at the point of approximation, which is \( x = 0 \).

Step 2: Calculate the Function Value at x = 0

Evaluate the function at \( x = 0 \). Since \( f(x) = \frac{1}{\sqrt{1-x^2}} \), we have:\[ f(0) = \frac{1}{\sqrt{1-0^2}} = \frac{1}{\sqrt{1}} = 1 \]

Step 3: Find the Derivative of the Function

Calculate the derivative of the function \( f(x) = \frac{1}{\sqrt{1-x^2}} \) using the chain rule:\[ f'(x) = \frac{d}{dx} (1-x^2)^{-1/2} = \frac{1}{2}(1-x^2)^{-3/2}(-2x) = \frac{x}{(1-x^2)^{3/2}} \]

Step 4: Calculate the Derivative at x = 0

Evaluate the derivative at \( x = 0 \):\[ f'(0) = \frac{0}{(1-0^2)^{3/2}} = 0 \]

Step 5: Form the Linear Approximation

The linear approximation formula is given by:\[ L(x) = f(a) + f'(a)(x-a) \] where \( a \) is the point of approximation, here \( a = 0 \). So, we have:\[ L(x) = f(0) + f'(0)(x - 0) = 1 + 0 \times x = 1 \]

Step 6: Validate the Approximation

Since the linear approximation \( L(x) = 1 \), which matches our initial approximation \( \frac{1}{\sqrt{1-x^2}} \approx 1 \), the approximation is confirmed to be correct at \( x = 0 \).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that helps us understand how a function changes at any point. It is basically the function's rate of change or slope at a specific point. When calculating the derivative, we find how the output of the function changes with respect to its input.For the given function \[ f(x) = \frac{1}{\sqrt{1-x^2}} \]we needed to find its derivative to use the linear approximation.Using the chain rule, we derived the derivative as:\[ f'(x) = \frac{x}{(1-x^2)^{3/2}} \]This gives us the rate at which our function changes with respect to \(x\). Understanding derivatives is essential as they provide the backbone for many approximations in calculus.When using derivatives:

• You analyze how slight changes in \(x\) affect \(f(x)\).
• Used in approximations to predict the function behavior near a point.
• Helps in optimizing and modeling real-world situations.

Function Evaluation

Function evaluation involves substituting the specific value of \(x\) into the function to find its output. For our linear approximation, we evaluated the function \[ f(x) = \frac{1}{\sqrt{1-x^2}} \]at \(x = 0\).After substitution, we calculated:\[ f(0) = \frac{1}{\sqrt{1-0^2}} = 1 \]This step is crucial in linear approximation because it sets the foundation for understanding the function's behavior at a specific point. It acts as a starting point for constructing the linear approximation.Key points for evaluating functions:

• Identify the exact point of interest, like \(x = 0\) here.
• Substitute this value into the function.
• Simplify the expression to get a clear numerical value.

Chain Rule

The chain rule is a powerful tool for finding derivatives, especially when dealing with composite functions. A composite function is outlined as a function nested within another function.In our exercise, the chain rule was applied to differentiate \[ f(x) = \frac{1}{\sqrt{1-x^2}} \]The process involves differentiating the outer layer first and then multiplying by the derivative of the inner function:\[ f'(x) = \frac{1}{2}(1-x^2)^{-3/2}(-2x) \]Finally, simplifying gives us:\[ f'(x) = \frac{x}{(1-x^2)^{3/2}} \]The chain rule allows us to approach complex functions one layer at a time:

• Start with the outer function and differentiate.
• Move inward, identifying and differentiating each function part.
• Multiply your results to form the complete derivative.

Taylor Expansion

Taylor expansion is a method used to approximate functions using polynomials. For a function that is smooth and differentiable, it can provide estimates around a specific point. Linear approximation is essentially the first-order Taylor expansion.In our example:\[ f(x) = \frac{1}{\sqrt{1-x^2}} \]we used the first term and the first derivative to define a linear polynomial at \(x = 0\):This expansion can be summarized as:\[ L(x) = f(a) + f'(a)(x-a) \]Using \[ a = 0, \quad f(0) = 1, \quad \text{and} \quad f'(0) = 0 \]This provides:\[ L(x) = 1 + 0 \times (x-0) = 1 \]Taylor expansion helps in:

• Establishing local approximations of complex functions.
• Simplifying expressions to make computations more feasible.
• Being a building block for further expansions beyond linear, providing more accuracy if necessary.