Chapter 3: Problem 38
Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ g(x)=x^{2} \cos x \text { at } a=\pi / 2,[0, \pi] $$
Short Answer
Expert verified
The linear approximation at \( a = \pi/2 \) is \( L(x) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \).
Step by step solution
01
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 \( a \).
02
Calculate the Function Value at \( a \)
Evaluate the function \( g(x) = x^2 \cos x \) at \( a = \pi/2 \).\[g(\pi/2) = \left(\frac{\pi}{2}\right)^2 \cos\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} \times 0 = 0.\]
03
Derive the Function
Find the derivative of \( g(x) = x^2 \cos x \) using the product rule: \( (uv)' = u'v + uv' \).\[g'(x) = (2x) \cos x + x^2(-\sin x) \]\[g'(x) = 2x \cos x - x^2 \sin x.\]
04
Calculate the Derivative at \( a \)
Evaluate the derivative \( g'(x) \) at \( a = \pi/2 \).\[g'(\pi/2) = 2\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - \left( \frac{\pi}{2} \right)^2 \sin\left(\frac{\pi}{2}\right) = 0 - \frac{\pi^2}{4}\times 1 = -\frac{\pi^2}{4}.\]
05
Write the Linear Approximation
Substitute into the linear approximation formula:\[L(x) = 0 + \left(-\frac{\pi^2}{4}\right)(x - \frac{\pi}{2}) = -\frac{\pi^2}{4}(x - \frac{\pi}{2}).\]
06
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 \( a = \pi/2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( -\frac{\pi^2}{4}(x - \frac{\pi}{2}) \).
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 \( 2x \cos x - x^2 \sin x \).
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, \( g'(\pi/2) = 0 - \frac{\pi^2}{4} \times 1 = -\frac{\pi^2}{4} \).
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.