Problem 86 - Calculus Early Transcendentals

Question

If \(a, b, c\) are the sides of a triangle and \(A, B, C\) are the opposite angles, find \(\partial A / \partial a, \partial A / \partial b, \partial A / \partial c\) by implicit differentiation of the Law of Cosines.

Step-by-Step Solution

Answer

\(\frac{\partial A}{\partial a} = \frac{b \cos(A) - a}{ab \sin(A)}, \frac{\partial A}{\partial b} = \frac{a \cos(A) - b}{ab \sin(A)}, \frac{\partial A}{\partial c} = \frac{c}{ab \sin(A)}\)

Step 1: Recall the Law of Cosines

The Law of Cosines for a triangle with sides \(a, b, c\) opposite to angles \(A, B, C\) is given by the formula \(c^2 = a^2 + b^2 - 2ab \cdot \cos(C)\). This equation expresses the relationship between the sides and the cosine of the angle.

Step 2: Differentiate with respect to \(a\)

To find \(\frac{\partial A}{\partial a}\), differentiate both sides of the Law of Cosines with respect to \(a\). The equation becomes \(0 = 2a - 2b \cos(A) + 2ab \sin(A) \frac{\partial A}{\partial a}\), assuming \(c\) and \(b\) are constants.

Step 3: Solve for \(\frac{\partial A}{\partial a}\)

Rearrange the differentiated equation to isolate \(\frac{\partial A}{\partial a}\). \[\frac{\partial A}{\partial a} = \frac{b \cos(A) - a}{ab \sin(A)}\]

Step 4: Differentiate with respect to \(b\)

Differentiate the Law of Cosines with respect to \(b\). The equation becomes \(0 = 2b - 2a \cos(A) + 2ab \sin(A) \frac{\partial A}{\partial b}\). Again, consider \(a\) and \(c\) as constants.

Step 5: Solve for \(\frac{\partial A}{\partial b}\)

Solve the resulting equation for \(\frac{\partial A}{\partial b}\). \[\frac{\partial A}{\partial b} = \frac{a \cos(A) - b}{ab \sin(A)}\]

Step 6: Differentiate with respect to \(c\)

Differentiate the Law of Cosines with respect to \(c\). The equation becomes \(2c = 2ab \sin(A) \frac{\partial A}{\partial c}\), considering \(a\) and \(b\) constants.

Step 7: Solve for \(\frac{\partial A}{\partial c}\)

Rearrange to isolate \(\frac{\partial A}{\partial c}\). \[\frac{\partial A}{\partial c} = \frac{c}{ab \sin(A)}\]

Key Concepts

Law of Cosines

The Law of Cosines is a fundamental concept in trigonometry that establishes a relationship between the sides and angles in a triangle. It can be viewed as a generalization of the Pythagorean theorem, applicable to any triangle, not just right-angled ones. The formula is stated as:

• For a triangle with sides denoted as \(a, b, c\) and the angles opposite these sides as \(A, B, C\), the Law of Cosines is given by:
• \(c^2 = a^2 + b^2 - 2ab \cdot \cos(C)\)

The significance of this formula lies in its ability to compute unknown sides or angles, especially when certain combinations of sides and angles are known. In particular, it enables the calculation of an angle when all the side lengths are given.
By rearranging terms, we can also express angles in terms of sides, allowing further exploration via derivatives. When using implicit differentiation, the angles can be indirectly determined by differentiating with respect to different side lengths, as shown in the exercise.

Partial Derivatives

Partial derivatives are essential when dealing with functions of multiple variables. In the context of the Law of Cosines, they allow us to understand how a small change in one side of a triangle affects the angle opposite to it. Partial derivatives are denoted by symbols like \(\frac{\partial A}{\partial a}\), indicating the rate of change of the angle \(A\) with respect to side \(a\), keeping the other variables constant.
To apply partial derivatives to the Law of Cosines:

• First, choose the variable with respect to which you want to differentiate, like side \(a\).
• Differentiating the equation implicitly involves applying the chain rule and recognizing which variables are held constant.
• Solving these derived equations gives expressions for how the angle changes with alterations in respective sides, leading to useful insights in geometry and physics.

The exercise exemplifies this process, using the Law of Cosines to derive expressions such as \(\frac{\partial A}{\partial a} = \frac{b \cos(A) - a}{ab \sin(A)}\), showcasing these dynamic relationships in a triangle.

Trigonometric Functions

Trigonometric functions like sine and cosine are crucial in the context of the Law of Cosines. These functions help translate the spatial geometric properties of triangles into algebraic forms.

• "\(\cos(C)\)" in the Law of Cosines holds significant meaning as it helps relate angles to sides.
• The function \(\sin(A)\) appears when deriving partial derivatives, signifying the relationship between changes in side lengths and angles.

Understanding trigonometric identities and properties aids in the manipulation and simplification of such equations:

• For example, knowing \(\sin^2(A) + \cos^2(A) = 1\) can be useful for deriving related expressions.
• These functions are periodic, meaning they repeat their values in a predictable manner, which can influence the behavior of the equations derived from them.

Through their role in implicit differentiation and the Law of Cosines, trigonometric functions serve as a powerful tool in exploring the nuanced relationships within triangles.