Problem 67 - Calculus Early Transcendentals

Question

All triangles satisfy the Law of Cosines \(c^{2}=a^{2}+b^{2}-2 a b \cos \theta\) (see figure). Notice that when \(\theta=\pi / 2,\) the Law of cosines becomes the Pythagorean Theorem. Consider all triangles with a fixed angle \(\theta=\pi / 3,\) in which case, \(c\) is a function of \(a\) and \(b,\) where \(a>0\) and \(b>0\) a. Compute \(\frac{\partial c}{\partial a}\) and \(\frac{\partial c}{\partial b}\) by solving for \(c\) and differentiating. b. Compute \(\frac{\partial c}{\partial a}\) and \(\frac{\partial c}{\partial b}\) by implicit differentiation. Check for agreement with part (a). c. What relationship between \(a\) and \(b\) makes \(c\) an increasing function of \(a\) (for constant \(b\) )?

Step-by-Step Solution

Answer

Short answer: Given a triangle with sides a, b, and c, and an angle θ = π/3 between sides a and b, we found that the partial derivatives of side c with respect to sides a and b are: - \(\frac{\partial c}{\partial a} = \frac{1}{2\sqrt{a^2+b^2-ab}}(2a - b)\) - \(\frac{\partial c}{\partial b} = \frac{1}{2\sqrt{a^2+b^2-ab}}(-a + 2b)\) Side c is an increasing function of side a when the inequality \(2a > b\) holds.

Step 1: Part a.1) Solve for c #

To compute the partial derivatives of \(c\) with respect to \(a\) and \(b\), we first need to solve for \(c\) from the given Law of Cosines equation, considering \(\theta = \pi/3\). $$ c^2 = a^2 + b^2 - 2ab\cos(\frac{\pi}{3}) $$ Substitute the cosine value: $$ c^2 = a^2 + b^2 - ab $$ Take the square root of both sides: $$ c = \sqrt{a^2 + b^2 - ab} $$ Now, we have c expressed in terms of \(a\) and \(b\).

Step 2: Part a.2) Compute the partial derivatives of c with respect to a and b #

Now that we have found the function \(c(a, b)\), we can compute the partial derivatives of \(c\) with respect to \(a\) and \(b\) by differentiating the above expression: $$ \frac{\partial c}{\partial a} = \frac{\partial}{\partial a} (\sqrt{a^2 + b^2 - ab}) \\ \frac{\partial c}{\partial b} = \frac{\partial}{\partial b} (\sqrt{a^2 + b^2 - ab}) $$ Let's find each partial derivative. For \(\frac{\partial c}{\partial a}\): $$ \frac{\partial c}{\partial a} = \frac{1}{2\sqrt{a^2+b^2-ab}}(2a - b) $$ For \(\frac{\partial c}{\partial b}\): $$ \frac{\partial c}{\partial b} = \frac{1}{2\sqrt{a^2+b^2-ab}}(-a + 2b) $$ Thus, we have the two partial derivatives.

Step 3: Part b) Compute the partial derivatives using implicit differentiation #

To compute the partial derivatives of \(c\) with respect to \(a\) and \(b\) using implicit differentiation, we will differentiate the original Law of Cosines equation with \(\theta = \pi/3\): $$ c^2 = a^2 + b^2 - ab $$ Let's differentiate with respect to \(a\): $$ \frac{\partial c^2}{\partial a} = \frac{\partial}{\partial a} (a^2 + b^2 - ab) \\ 2c\frac{\partial c}{\partial a} = 2a - b \\ \frac{\partial c}{\partial a} = \frac{1}{2\sqrt{a^2+b^2-ab}}(2a - b) $$ Now let's differentiate with respect to \(b\): $$ \frac{\partial c^2}{\partial b} = \frac{\partial}{\partial b} (a^2 + b^2 - ab) \\ 2c\frac{\partial c}{\partial b} = -a + 2b \\ \frac{\partial c}{\partial b} = \frac{1}{2\sqrt{a^2+b^2-ab}}(-a + 2b) $$ The partial derivatives found using implicit differentiation agree with the ones found in part (a).

Step 4: Part c) Determine when c is an increasing function of a #

To determine when \(c\) is an increasing function of \(a\), we need to find when \(\frac{\partial c}{\partial a} > 0\). Using the partial derivative found in part (a), we have: $$ \frac{1}{2\sqrt{a^2+b^2-ab}}(2a - b) > 0 $$ Since \(2\sqrt{a^2+b^2-ab}>0\) \(a>0\) and \(b>0\), we can ignore the denominator and focus on the numerator $$ 2a - b > 0 $$ This inequality tells us that \(c\) is an increasing function of \(a\) when \(2a > b\).

Key Concepts

Partial Differentiation

Partial differentiation is a technique used to compute the derivative of a multivariable function with respect to one variable while holding the others constant. When dealing with a function like \( c(a, b) = \sqrt{a^2 + b^2 - ab} \), you want to determine how slight changes in one variable affect the function \( c \), assuming the other variable is constant.

To find the partial derivative \( \frac{\partial c}{\partial a} \), you treat \( b \) as if it's a constant, and differentiate the function with respect to \( a \):
\[ \frac{\partial}{\partial a} (\sqrt{a^2 + b^2 - ab}) = \frac{1}{2\sqrt{a^2+b^2-ab}}(2a - b) \]

This expression tells us the rate at which \( c \) changes as \( a \) changes. Similarly, to find \( \frac{\partial c}{\partial b} \), treat \( a \) as constant:

\[ \frac{\partial}{\partial b} (\sqrt{a^2 + b^2 - ab}) = \frac{1}{2\sqrt{a^2+b^2-ab}}(-a + 2b) \]

These derivatives help in analyzing how \( c \), which represents one triangle side, depends on the other two sides.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry stating that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In mathematical terms, for a triangle with sides \( a \), \( b \), and hypotenuse \( c \), it can be expressed as:

\[ c^2 = a^2 + b^2 \]

When the angle \( \theta = \pi/2 \), the Law of Cosines simplifies to become the Pythagorean Theorem. For non-right triangles, the Law of Cosines is used instead, which adjusts the relationship between the sides based on the angle. In our context, the angle \( \pi/3 \) adjusts the relationship to:

\[ c^2 = a^2 + b^2 - ab \]

The Pythagorean Theorem is a special case of the Law of Cosines, highlighting its importance and wide applicability. It greatly aids in understanding triangle properties, especially in trigonometry and geometry.

Implicit Differentiation

Implicit differentiation is a method used to find derivatives of functions that are not easily expressed in the standard \( y = f(x) \) format. It is particularly useful when dealing with equations that define \( y \) implicitly in terms of \( x \).

In our exercise, the function \( c \) is expressed implicitly by the Law of Cosines with the angle \( \pi/3 \):

\[ c^2 = a^2 + b^2 - ab \]

To find \( \frac{\partial c}{\partial a} \) using implicit differentiation, you differentiate both sides of the equation with respect to \( a \), remembering that \( c \) is also a function of \( a \) and thus its derivative applies:

\[ 2c \frac{\partial c}{\partial a} = 2a - b \]

Solve for \( \frac{\partial c}{\partial a} \) to get the same result obtained from explicit partial differentiation.

This technique can simplify finding derivatives without needing to explicitly solve for \( c \), illustrating its effectiveness for handling complex relationships in calculus.