Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ g(x)=x^{2} \cos x \text { at } a=\pi / 2,[0, \pi] $$

Step-by-Step Solution

Answer

The linear approximation at \( a = \pi/2 \) is \( L(x) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \).

Step 1: Understand Linear Approximation

Linear approximation of a function at a given point is based on the equation of the tangent line at that point. The formula is given by \( L(x) = f(a) + f'(a)(x - a) \), where \( f(a) \) is the value of the function at \( a \) and \( f'(a) \) is the derivative evaluated at \( a \).

Step 2: Calculate the Function Value at \( a \)

Evaluate the function \( g(x) = x^2 \cos x \) at \( a = \pi/2 \).\[g(\pi/2) = \left(\frac{\pi}{2}\right)^2 \cos\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} \times 0 = 0.\]

Step 3: Derive the Function

Find the derivative of \( g(x) = x^2 \cos x \) using the product rule: \( (uv)' = u'v + uv' \).\[g'(x) = (2x) \cos x + x^2(-\sin x) \]\[g'(x) = 2x \cos x - x^2 \sin x.\]

Step 4: Calculate the Derivative at \( a \)

Evaluate the derivative \( g'(x) \) at \( a = \pi/2 \).\[g'(\pi/2) = 2\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - \left( \frac{\pi}{2} \right)^2 \sin\left(\frac{\pi}{2}\right) = 0 - \frac{\pi^2}{4}\times 1 = -\frac{\pi^2}{4}.\]

Step 5: Write the Linear Approximation

Substitute into the linear approximation formula:\[L(x) = 0 + \left(-\frac{\pi^2}{4}\right)(x - \frac{\pi}{2}) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}).\]

Step 6: Plot the Function and Linear Approximation

Graph the function \( g(x) = x^2 \cos x \) and the linear approximation \( L(x) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \) over the interval \([0, \pi]\). The linear approximation should closely match the function around \( a = \pi/2 \).

Key Concepts

Tangent Line

A tangent line to a function at a certain point is the straight line that just "touches" the function at that point, matching both its value and direction. This line gives the best linear approximation of the function near that point. For the function \( g(x) = x^2 \cos x \) at the point \( a = \pi/2 \), the tangent line is computed using the derivatives, as it indicates the slope or steepness at that specific point.

• The formula for the tangent line is \( L(x) = f(a) + f'(a)(x - a) \).
• In this exercise, \( L(x) = 0 + \left(-\frac{\pi^2}{4}\right)(x - \frac{\pi}{2}) \).
• Thus, the tangent line equation simplifies to \( -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \).

The tangent line helps us approximate the function around \( a = \pi/2 \), where both the function and the tangent touch but only for a tiny interval.

Product Rule

The product rule is essential when finding derivatives of products of two functions. For two differentiable functions \( u(x) \) and \( v(x) \), the derivative of their product \( uv \) is given by:

• The formula: \( (uv)' = u'v + uv' \).
• Applied to our exercise: For \( g(x) = x^2 \cos x \), consider \( u(x) = x^2 \) and \( v(x) = \cos x \).
• The derivative: \( u'(x) = 2x \) and \( v'(x) = -\sin x \).
• Thus, \( g'(x) = 2x \cos x + x^2(-\sin x) \), which further simplifies to \( 2x \cos x - x^2 \sin x \).

This rule helps efficiently handle the complexity of differentiating products of functions, like our \( g(x) \).

Derivative Evaluation

Evaluating a derivative at a specific point allows us to determine the slope of the tangent line to a function at that point. Once we have derived the function using appropriate rules, like the product rule, the next step is to substitute the point of interest into the derivative equation.

• For our function \( g'(x) = 2x \cos x - x^2 \sin x \), we evaluate at \( a = \pi/2 \).
• The value, \( g'(\pi/2) = 0 - \frac{\pi^2}{4} \times 1 = -\frac{\pi^2}{4} \).

This calculation reveals the slope of the tangent line at \( a = \pi/2 \). It tells us how steep the line is at that exact point, which is crucial for forming the linear approximation.

Plotting Functions

Plotting functions allows us to visually compare the original function and its linear approximation. This graphical method is particularly useful to understand how well the linear approximation aligns with the function near the chosen point.

• The function \( g(x) = x^2 \cos x \) should be plotted over \([0, \pi]\).
• The linear approximation \( L(x) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \) should also be plotted within the same interval.

By plotting, you can observe how both graphs behave and converge at \( a = \pi/2 \). The linear approximation will look like a straight line closely fitting \( g(x) \) around this point, thus illustrating a vital concept in analysis using graphs.