Problem 41 - Calculus Single Variable

Question

In a model for optimizing the angle of release \(\theta\) of a basketball shot, suppose that \(a\) and \(b\) are positive constants. Let \(\theta_{0}\) be the value of \(\theta\) in the interval \((\arctan (b / a),\) \(\pi / 2\) ) for which $$ f(\theta)=\left(a \sin (\theta) \cos (\theta)-b \cos ^{2}(\theta)\right)^{-1} $$ is minimized. What is \(\tan \left(\theta_{0}\right) ?\)

Step-by-Step Solution

Answer

\( \tan(\theta_0) = -\frac{a}{\sqrt{a^2 + b^2} + b} \)

Step 1: Understanding the Function

The function to minimize is \( f(\theta) = \left(a \sin (\theta) \cos (\theta) - b \cos^2(\theta)\right)^{-1} \). As \(f(\theta)\) is minimized when the expression in the denominator is maximized, we focus on maximizing \( g(\theta) = a \sin (\theta) \cos (\theta) - b \cos^2(\theta) \).

Step 2: Rewrite Using Trigonometric Identities

Use the double angle identity \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\). Rewrite \(g(\theta)\) as: \[ g(\theta) = \frac{a}{2}\sin(2\theta) - b\cos^2(\theta) \]. To express \(\cos^2(\theta)\) in terms of \(2\theta\), recall \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \).

Step 3: Substitute and Simplify

Substitute for \(\cos^2(\theta)\) in \(g(\theta)\): \[ g(\theta) = \frac{a}{2} \sin(2\theta) - b \left( \frac{1 + \cos(2\theta)}{2} \right) \] which simplifies to \[ g(\theta) = \frac{a}{2} \sin(2\theta) - \frac{b}{2} - \frac{b}{2} \cos(2\theta). \]

Step 4: Differentiate and Find Critical Points

Differentiate \(g(\theta)\) with respect to \(\theta\). The derivative is \[ g'(\theta) = a\cos(2\theta) + b\sin(2\theta). \] Set \(g'(\theta) = 0\) to find critical points: \[ a\cos(2\theta) = -b\sin(2\theta). \] Divide through by \(\cos(2\theta)\), assuming \(\cos(2\theta) eq 0\), to get: \[ a = -b\tan(2\theta). \] Thus, \[\tan(2\theta) = -\frac{a}{b}.\]

Step 5: Solve for \(\tan(\theta_0)\)

The expression \(\tan(2\theta) = -\frac{a}{b}\) implies that \(2\theta_0 = \arctan\left(-\frac{a}{b}\right) + k\pi\) for some integer \(k\). Since \( \theta_0 \) is in \((\arctan(b/a), \pi/2)\), choose \(k = 0\). Thus, we get \( \theta_0 = \frac{1}{2}\arctan\left(-\frac{a}{b}\right) \). Consequently, \[ \tan(\theta_0) = \tan\left(\frac{1}{2}\arctan\left(-\frac{a}{b}\right)\right). \]

Step 6: Use the Tangent Half-angle Identity

Apply the tangent half-angle identity: \( \tan(x/2) = \frac{\sin(x)}{1 + \cos(x)} \) for \( x = \arctan(-a/b) \). Notice \( \tan(x) = -a/b \), so \[ \sin(x) = \frac{-a}{\sqrt{a^2 + b^2}}, \quad \cos(x) = \frac{b}{\sqrt{a^2 + b^2}}. \] Thus, \[ \tan(\theta_0) = \frac{\sin(x)}{1 + \cos(x)} = \frac{-a}{\sqrt{a^2 + b^2} + b}. \]

Step 7: Conclusion: Result for \( \tan(\theta_0) \)

The value of \( \tan(\theta_0) \) is \(-\frac{a}{\sqrt{a^2 + b^2} + b}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics that help simplify expressions involving angles and functions like sine, cosine, and tangent. These identities establish relationships between different trigonometric functions and are crucial in solving problems, especially in calculus optimization.

One important identity used in this exercise is the double angle identity. The double angle identity for sine is given by: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \). This identity helps rewrite expressions involving trigonometric functions in a more convenient form for differentiation. Additionally, another useful identity is \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \), which lets us express \( \cos^2(\theta) \) in terms of \( 2\theta \).

These identities make it easier to manage complex trigonometric expressions and are invaluable for deriving solutions, as seen in this problem where they help maximize the function \( g(\theta) \). By applying these identities, we transform the problem into a calculus one where we can effectively find critical points.

Derivative and Critical Points

In calculus, the derivative is a measure of how a function changes as its input changes. This concept is central to finding critical points, which are potential points for local maximums, minimums, or inflection points of a function. In optimization problems, critical points help us determine where functions like \( f(\theta) \) reach their minimum or maximum values.

In the given exercise, we focus on the function \( g(\theta) \) instead of \( f(\theta) \) because minimizing \( f(\theta) \) is equivalent to maximizing its denominator \( g(\theta) \). To identify the critical points of \( g(\theta) \), we find its derivative: \( g'(\theta) = a\cos(2\theta) + b\sin(2\theta) \). To find these points, we set the derivative equal to zero: \( g'(\theta) = 0 \).

The solution gives \( a\cos(2\theta) = -b\sin(2\theta) \), leading us to the equation \( \tan(2\theta) = -\frac{a}{b} \). Solving this equation allows us to find the angles \( \theta \) that optimize our original function, demonstrating how derivatives and critical points are used to solve calculus optimization problems.

Angle of Release

The angle of release, represented by \( \theta \), is a crucial element when considering projectile motion problems, like optimizing a basketball shot. The idea is to determine an angle that will either maximize or minimize a particular function, depending on the context.

In this exercise, the goal is to find \( \theta_0 \), the angle which minimizes the function \( f(\theta) \). As shown in the solution steps, this involves setting the derivative \( g'(\theta) \) to zero, thereby determining the optimal angle \( \theta_0 \). From the expression \( \tan(2\theta) = -\frac{a}{b} \), and considering the constraint that \( \theta_0 \) lies within \( (\arctan(b/a), \pi/2) \), we simplify to find: \( \theta_0 = \frac{1}{2}\arctan\left(-\frac{a}{b}\right) \).

Understanding the angle of release in this way allows us to use mathematical techniques to solve real-world problems, ensuring the optimal release angle for achieving the desired outcome.