Problem 43 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(t)=t \sqrt{4-t^{2}}, \quad[-1,2]$$

Step-by-Step Solution

Answer

Absolute maximum is 2 at \(t = \sqrt{2}\), absolute minimum is \(-\sqrt{3}\) at \(t = -1\).

Step 1: Understand the Problem

We are asked to find the absolute maximum and absolute minimum values of a function on a given interval. The function provided is \( f(t) = t \sqrt{4-t^{2}} \) and the interval is \([-1, 2]\).

Step 2: Find Critical Points

To find critical points, we need to compute the derivative of \( f(t) \), set it equal to zero, and solve for \( t \). First, compute the derivative using the product rule: \[ f'(t) = \sqrt{4-t^{2}} + t \cdot \frac{-2t}{2\sqrt{4-t^{2}}} = \frac{4 - 2t^2}{\sqrt{4-t^2}}. \] Set \( f'(t) = 0 \): \[ 4 - 2t^2 = 0 \Rightarrow 2t^2 = 4 \Rightarrow t^2 = 2 \Rightarrow t = \pm\sqrt{2}. \] Only \( t = \sqrt{2} \) is within the interval \([-1, 2]\).

Step 3: Evaluate at Critical Points and Endpoints

Evaluate \( f(t) \) at critical points and endpoints of the interval \([-1, 2]\). For \( t = -1 \): \[ f(-1) = -1 \sqrt{4-(-1)^{2}} = -1 \sqrt{3}. \] For \( t = \sqrt{2} \): \[ f(\sqrt{2}) = \sqrt{2} \sqrt{4-(\sqrt{2})^2} = \sqrt{2} \sqrt{2} = 2. \]For \( t = 2 \): \[ f(2) = 2 \sqrt{4-2^{2}} = 0. \]

Step 4: Determine Absolute Maximum and Minimum

By comparing the computed values: - \( f(-1) = -\sqrt{3} \) - \( f(\sqrt{2}) = 2 \) - \( f(2) = 0 \)The absolute maximum is 2 at \( t = \sqrt{2} \), and the absolute minimum is \( -\sqrt{3} \) at \( t = -1 \).

Key Concepts

Critical Points

Critical points of a function are points where the derivative is zero or undefined. They are important because they can indicate potential locations of local maxima or minima. To find them, we first need to calculate the derivative of the function. Then, after simplifying the expression, set the derivative equal to zero to solve for the specific values of the variable. In this problem, the function we are dealing with is defined as: \[ f(t) = t \sqrt{4-t^2} \] To find the critical points, we start by calculating the derivative: \[ f'(t) = \sqrt{4-t^2} + t \left( \frac{-2t}{2\sqrt{4-t^2}} \right) = \frac{4 - 2t^2}{\sqrt{4-t^2}} \] By setting this derivative equal to zero, we solve for \( t \): \[ \frac{4 - 2t^2}{\sqrt{4-t^2}} = 0 \] This simplifies to: \[ 2t^2 = 4 \quad \Rightarrow \quad t^2 = 2 \quad \Rightarrow \quad t = \pm\sqrt{2} \] However, we are provided with an interval \( [-1, 2] \), which means \( t = \sqrt{2} \) is within range, and \( t = -\sqrt{2} \) is outside. Thus, the critical point in the interval is \( t = \sqrt{2} \).

Evaluation of Endpoints

In this kind of problem, evaluating the endpoints of the interval is just as important as considering the critical points. Sometimes, absolute maxima or minima occur at these endpoints, rather than at critical points. The given interval is \( [-1, 2] \), so we must evaluate \( f(t) \) at these values.1. For \( t = -1 \): - Substitute \( t = -1 \) into the function: - \( f(-1) = -1 \sqrt{4 - (-1)^2} = -\sqrt{3} \).2. For \( t = 2 \): - Substitute \( t = 2 \) into the function: - \( f(2) = 2 \sqrt{4 - 2^2} = 0 \).By evaluating the endpoints and comparing them with the value at the critical point, we can determine the absolute extrema. This ensures we don't miss any potential maxima or minima at these boundary points.

Derivative Calculation

Calculating the derivative is a key step in finding critical points and understanding the behavior of a function. Depending on the function, this can involve a range of differentiation techniques. For the function \( f(t) = t \sqrt{4-t^2} \), we apply the product rule of differentiation. The product rule states that \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). Here, let \( u(t) = t \) and \( v(t) = \sqrt{4-t^2} \). - First, calculate \( u'(t) = 1 \).- For \( v(t) \), use the chain rule to differentiate: - \( v'(t) = \frac{-2t}{2\sqrt{4-t^2}} = \frac{-t}{\sqrt{4-t^2}} \).Substitute these into the product rule: \[ f'(t) = u'(t)v(t) + u(t)v'(t) = \sqrt{4-t^2} + t \left(-\frac{t}{\sqrt{4-t^2}}\right) = \frac{4 - 2t^2}{\sqrt{4-t^2}} \]This equation shows the derivative of \( f(t) \) and is key to identifying potential critical points where the slope is zero.