Problem 159 - Calculus Volume 3

Question

The law of cosines can be thought of as a function of three variables. Let \(x, y,\) and \(\theta\) be two sides of any triangle where the angle \(\theta\) is the included angle between the two sides. Then, \(F(x, y, \theta)=x^{2}+y^{2}-2 x y \cos \theta\) gives the square of the third side of the triangle. Find \(\frac{\partial F}{\partial \theta}\) and \(\frac{\partial F}{\partial x}\) when \(x=2, y=3,\) and \(\theta=\frac{\pi}{6}\).

Step-by-Step Solution

Answer

\(\frac{\partial F}{\partial \theta} = 6\), \(\frac{\partial F}{\partial x} = 4 - 3\sqrt{3}\).

Step 1: Understand the Function

The function given is the law of cosines expressed as a function \( F(x, y, \theta) = x^2 + y^2 - 2xy \cos \theta \). This function computes the square of the third side of a triangle given two sides \(x\), \(y\), and the included angle \(\theta\).

Step 2: Differentiate with Respect to \(\theta\)

Start by finding the partial derivative \(\frac{\partial F}{\partial \theta}\). Since only the term \(-2xy \cos \theta\) depends on \(\theta\), differentiate\[\frac{\partial}{\partial \theta}(-2xy \cos \theta) = 2xy \sin \theta\]Thus, \[\frac{\partial F}{\partial \theta} = 2xy \sin \theta\]

Step 3: Differentiate with Respect to \(x\)

Find the partial derivative \(\frac{\partial F}{\partial x}\). The function is \(x^2 + y^2 - 2xy \cos \theta\):- Derivative of \(x^2\) is \(2x\).- Derivative of \(-2xy \cos \theta\) with respect to \(x\) is \(-2y \cos \theta\).Thus,\[\frac{\partial F}{\partial x} = 2x - 2y \cos \theta\]

Step 4: Evaluate \(\frac{\partial F}{\partial \theta}\) at Given Values

Substitute \(x = 2\), \(y = 3\), and \(\theta = \frac{\pi}{6}\) into the partial derivative with respect to \(\theta\):\[\frac{\partial F}{\partial \theta} = 2 \times 2 \times 3 \times \sin\left(\frac{\pi}{6}\right)\]\(= 12 \times \frac{1}{2}\)\(= 6\)

Step 5: Evaluate \(\frac{\partial F}{\partial x}\) at Given Values

Substitute \(x = 2\), \(y = 3\), and \(\theta = \frac{\pi}{6}\) into the partial derivative with respect to \(x\):\[\frac{\partial F}{\partial x} = 2 \times 2 - 2 \times 3 \times \cos\left(\frac{\pi}{6}\right)\]\(= 4 - 6 \times \frac{\sqrt{3}}{2}\)\(= 4 - 3\sqrt{3}\)

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as only one of its variables changes, while keeping the other variables constant. In our context, we look at how the function related to the law of cosines, \( F(x, y, \theta) = x^2 + y^2 - 2xy \cos \theta \), changes as we specifically tweak the values of \(\theta\) and \(x\). This provides us insights into how the triangle's third side length is affected by changes in angle or one side. To find a partial derivative, we treat the other variables as constants and differentiate with respect to the variable of interest.

• For \(\frac{\partial F}{\partial \theta}\), we focus on the influence of the angle, which is present in the term \(-2xy \cos \theta\). Differentiating this with respect to \(\theta\) yields \(2xy \sin \theta\), reflecting how changes in the angle, while sides \(x\) and \(y\) are held constant, affect the third side.
• Similarly, for \(\frac{\partial F}{\partial x}\), we see how variations in \(x\), while keeping \(y\) and \(\theta\) constant, change the third side's square. The derivatives of \(x^2\) and \(-2xy \cos \theta\) give us \(2x - 2y \cos \theta\).

Partial derivatives are powerful in analyzing multidimensional problems, as they dissect complex functions into simpler, one-variable equations.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. It is essential for solving problems involving physical systems, economics, and engineering where multiple variables interact. The law of cosines formula here provides an example of a multivariable function \( F(x, y, \theta) \), which depends on two side lengths \(x\) and \(y\), and the angle \(\theta\) between them.
In problems like this, multivariable calculus aids in understanding how one quantity affects another. By using partial derivatives, we can study how each individual variable influences the dependent outcome, the square of the third side, without changing the other variables.

• By evaluating \(\frac{\partial F}{\partial \theta}\), we find out precisely how changes in the angle impact the triangle, holding side lengths constant.
• Similarly, \(\frac{\partial F}{\partial x}\) shows how altering one side length affects the result when other factors don’t vary.

This approach enables tackling real-world changes step by step, ensuring a detailed understanding of each factor individually.

Trigonometric Functions

Trigonometric functions like sine and cosine are integral when dealing with angles in geometry and calculus. They describe relationships in triangles and are crucial for expressing the law of cosines. In the formula \( F(x, y, \theta) = x^2 + y^2 - 2xy \cos \theta \), cosine plays a key role by representing the effect of the angle on the length of the triangle’s third side.

• The term \(-2xy \cos \theta\) shows how cosine adjusts the impact of the angle between sides \(x\) and \(y\), reflecting its direct contribution to the length of the third side. This is because \(\cos \theta\) interacts with the sides multiplicatively.
• When differentiating with respect to \(\theta\), the derivative \(\frac{\partial}{\partial \theta}(-2xy \cos \theta)\) becomes \(2xy \sin \theta\). This arises because the derivative of cosine is negative sine, illustrating how these functions swap roles during calculus operations.

Understanding these trigonometric components is critical in applications involving oscillation, waves, and any scenario where angles determine system behavior. Their properties, like periodicity and the relationships between sine and cosine, are essential for thorough analysis in both pure and applied mathematics.