Problem 159 - Calculus 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: Understanding the Problem

We are given the function \( F(x, y, \theta) = x^2 + y^2 - 2xy \cos \theta \) representing the square of the third side of a triangle. We need to find the partial derivatives \( \frac{\partial F}{\partial \theta} \) and \( \frac{\partial F}{\partial x} \) given specific values for \( x, y, \) and \( \theta \).

Step 2: Applying Partial Differentiation with Respect to \( \theta \)

To find \( \frac{\partial F}{\partial \theta} \), treat \( x \) and \( y \) as constants. The derivative of \( F \) with respect to \( \theta \) focuses on the term \( -2xy \cos \theta \). By applying the derivative rules for cosine, we have:\[ \frac{\partial F}{\partial \theta} = 2xy \sin \theta \]

Step 3: Substitute Values into \( \frac{\partial F}{\partial \theta} \)

Using the values \( x = 2 \), \( y = 3 \), and \( \theta = \frac{\pi}{6} \), substitute these into the expression:\[ \frac{\partial F}{\partial \theta} = 2(2)(3) \sin\left(\frac{\pi}{6}\right) \]Since \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), this simplifies to:\[ \frac{\partial F}{\partial \theta} = 12 \times \frac{1}{2} = 6 \]

Step 4: Applying Partial Differentiation with Respect to \( x \)

To find \( \frac{\partial F}{\partial x} \), treat \( y \) and \( \theta \) as constants. Differentiate \( F \) with respect to \( x \):\[ \frac{\partial F}{\partial x} = 2x - 2y \cos \theta \]

Step 5: Substitute Values into \( \frac{\partial F}{\partial x} \)

Using the values \( x = 2 \), \( y = 3 \), and \( \theta = \frac{\pi}{6} \), substitute these into the expression:\[ \frac{\partial F}{\partial x} = 2(2) - 2(3) \cos\left(\frac{\pi}{6}\right) \]Since \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \), this simplifies to:\[ \frac{\partial F}{\partial x} = 4 - 3\sqrt{3} \]

Key Concepts

Law of Cosines

The Law of Cosines is a fundamental principle in geometry that relates the sides of a triangle to the cosine of one of its angles. In any triangle with sides labeled as \(x\), \(y\), and \(z\) with an angle \(\theta\) opposite to \(z\), the law is expressed as:

• \(z^2 = x^2 + y^2 - 2xy \cos \theta\)

This equation helps us understand the relationship between the lengths of the sides of a triangle and its angles.
In the context of partial differentiation, consider this formula as a function of the three variables \(x\), \(y\), and \(\theta\). From here, finding how each part of the triangle changes with respect to these variables becomes a key point of interest.
The Law of Cosines plays a similar role in non-right triangles as the Pythagorean theorem does in right triangles. It is particularly useful in algebraic manipulation scenarios, such as when solving for unknown triangle side lengths or angles.

Partial Differentiation

Partial differentiation is a crucial tool in calculus, especially when dealing with functions of multiple variables. Here, we are working with a function \(F(x, y, \theta) = x^2 + y^2 - 2xy \cos \theta\) and are interested in how this function changes partially with respect to \(x\) and \(\theta\).

• For \(\frac{\partial F}{\partial \theta}\), consider \(x\) and \(y\) as constants. This method focuses on differentiating the \(-2xy \cos \theta\) part with respect to \(\theta\).
• While finding \(\frac{\partial F}{\partial x}\), \(y\) and \(\theta\) are constants, so focus on differentiating \(2x\) and \(-2y \cos \theta\).

Partial differentiation allows us to see how changing one variable while keeping others constant affects the value of the function. This is particularly useful for optimization problems and in understanding dynamic systems within multivariable contexts. By calculating these partial derivatives, you can explore how the shape and dimensions of a triangle change relative to specific variables.

Trigonometric Functions

Trigonometric functions are essential in understanding the relationships between angles and sides in triangles. Both the sine and cosine functions are pivotal in the formula used in the Law of Cosines.
Let's review two important trigonometric functions used here:

• Sine (\(\sin\)): \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). This value is critical when calculating \(\frac{\partial F}{\partial \theta}\), as it directly impacts the rate of change concerning the angle \(\theta\).
• Cosine (\(\cos\)): \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). The cosine function appears in the Law of Cosines formula, affecting calculations for both \(\frac{\partial F}{\partial \theta}\) and \(\frac{\partial F}{\partial x}\).

These values result from evaluating trigonometric functions at specific angles, which is a common practice in mathematical problems involving circles or wave functions. Understanding these basic trigonometric relationships helps in leveraging them for calculus operations and aids in modeling periodic phenomena across various scientific contexts.