Problem 45 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(t)=2 \cos t+\sin 2 t, \quad[0, \pi / 2]$$

Step-by-Step Solution

Answer

The absolute maximum is \(\frac{3\sqrt{3}}{2}\), and the minimum is 0.

Step 1: Find the derivative of f(t)

The first step is to find the derivative of the given function, which is necessary for finding critical points. The function is \(f(t) = 2 \cos t + \sin 2t\). Using differentiation rules:The derivative of \(2 \cos t\) is \(-2 \sin t\).The derivative of \(\sin 2t\) is \(2 \cos 2t\) (chain rule applied: derivative of \(\sin x\) is \(\cos x\) and multiply by the derivative of \(2t\), which is 2).Thus, the derivative \(f'(t) = -2 \sin t + 2 \cos 2t\).

Step 2: Solve for critical points by setting f'(t) = 0

To find critical points where the function might have extreme values, set the derivative to zero:\(-2 \sin t + 2 \cos 2t = 0\).This simplifies to \(- \sin t + \cos 2t = 0\).We know that \(\cos 2t = 1 - 2 \sin^2 t\).Thus, \(- \sin t + 1 - 2 \sin^2 t = 0\). Solve this quadratic equation for \(\sin t\).

Step 3: Solving quadratic for \(\sin t\)

Substitute \(u = \sin t\), giving the equation:\(-u + 1 - 2u^2 = 0\).Rewrite as \(2u^2 + u - 1 = 0\).Solve this quadratic equation using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), \(c = -1\).

Step 4: Calculate the solutions using the quadratic formula

Applying the quadratic formula:\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{4} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \].This gives us solutions:\(u = \frac{2}{4} = \frac{1}{2}\)and \(u = \frac{-4}{4} = -1\) (which is not valid as\(u\) is \(\sin t\) and must be between -1 and 1).

Step 5: Determine t from sin t = 1/2

From \(\sin t = \frac{1}{2}\), determine \(t\) within the interval \([0, \frac{\pi}{2}]\). Knowing that \(\sin t = \frac{1}{2}\) at \(t = \frac{\pi}{6}\).

Step 6: Evaluate f(t) at critical points and endpoints

Evaluate the original function \(f(t) = 2 \cos t + \sin 2t\) at the critical point \(t = \frac{\pi}{6}\) and at the endpoints \(t = 0\) and \(t = \frac{\pi}{2}\):- \(f(0) = 2 \cdot 1 + \sin 0 = 2\)- \(f(\frac{\pi}{2}) = 2 \cdot 0 + \sin \pi = 0\)- \(f(\frac{\pi}{6}) = 2 \cdot \cos \frac{\pi}{6} + \sin(\frac{\pi}{3}) = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\)

Step 7: Identify the absolute maximum and minimum values

Compare the values obtained:- At \(t=0\), \(f(0) = 2\).- At \(t=\frac{\pi}{2}\), \(f(\frac{\pi}{2}) = 0\).- At \(t=\frac{\pi}{6}\), \(f(\frac{\pi}{6}) = \frac{3\sqrt{3}}{2}\).The absolute maximum value is \(\frac{3\sqrt{3}}{2}\) and the absolute minimum value is \(0\).

Key Concepts

Derivative

The derivative of a function is a fundamental tool in calculus, allowing us to understand the rate at which one quantity changes with respect to another. In simple terms, it tells us the slope of a function at any given point. When working with the function \(f(t) = 2 \cos t + \sin 2t\), calculating the derivative was the first crucial step in identifying its critical points.

• The derivative of a basic trigonometric function \(\cos t\) is \(-\sin t\).
• Applying rules of differentiation to \(\sin 2t\), using the chain rule, yields \(2 \cos 2t\) since we multiply by the derivative of the inner function \((2t)\).

By combining these, we found the derivative: \(f'(t) = -2 \sin t + 2 \cos 2t\). Derivatives are essential for many applications, especially in finding critical points, which leads us to the next concept.

Critical Points

Critical points of a function occur where the first derivative equals zero or is undefined. These points are where the function could potentially have a local maximum, minimum, or point of inflection. To find the critical points of the function \(f(t) = 2 \cos t + \sin 2t\), we set the derivative \(f'(t) = -2 \sin t + 2 \cos 2t\) equal to zero.

• Solving this equation helps us understand where the function's slope changes.
• These points are candidates for local extremities, necessary to determine absolute maximum or minimum within a closed interval.

In our case, simplifying the equation involved trigonometric identities and solving a quadratic equation which ultimately allowed us to find the angles at which these critical points occur.

Quadratic Equation

Quadratic equations appear frequently in calculus, especially when dealing with critical points. A quadratic equation is of the form \(ax^2 + bx + c = 0\). In our problem, after substituting \(u = \sin t\), we derived the equation \(2u^2 + u - 1 = 0\). This required us to use the quadratic formula
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = 1\), and \(c = -1\). Solving this equation gives us solutions for \(\sin t\).

• It is crucial to ensure the solutions make sense within the context, as not all solutions may fit the domain of the trigonometric functions.
• With the quadratic formula, we filter out solutions that don't pertain to our problem, like \(u = -1\), since this was outside the range of sine values in our interval.

Understanding quadratic equations and how to manipulate them is vital for solving such calculus problems.

Trigonometric Functions

Trigonometric functions like sine and cosine are periodic functions that describe oscillatory behavior, and they play a vital role in calculus. In the problem at hand, the function \(f(t) = 2 \cos t + \sin 2t\) involves these functions.

• The cosine function, \(\cos t\), measures the horizontal component of a unit circle at angle \(t\).
• The sine function, \(\sin t\), measures the vertical component of the same circle.
• The double-angle identity, used in simplifying \(\cos 2t = 1 - 2 \sin^2 t\), is essential for transforming equations.

These characteristics are crucial, as they allow us to manipulate and solve problems involving these types of functions by leveraging their properties and identities. Understanding these functions' behaviors aids in evaluating the function at key points and identifying maximum and minimum values.