Problem 19 - University Calculus: Early Transcendentals

Question

Area The area \(A\) of a triangle with sides of lengths \(a\) and \(b\) enclosing an angle of measure \(\theta\) is $$A=\frac{1}{2} a b \sin \theta$$ a. How is \(d A / d t\) related to \(d \theta / d t\) if \(a\) and \(b\) are constant? b. How is \(d A / d t\) related to \(d \theta / d t\) and \(d a / d t\) if only \(b\) is constant? c. How is \(d A / d t\) related to \(d \theta / d t, d a / d t,\) and \(d b / d t\) if none of \(a, b,\) and \(\theta\) are constant?

Step-by-Step Solution

Answer

(a) It relates through \( \frac{dA}{dt} = \frac{1}{2} a b \cos \theta \frac{d\theta}{dt} \). (b) \( \frac{dA}{dt} = \frac{1}{2} (b \sin \theta \frac{da}{dt} + a b \cos \theta \frac{d\theta}{dt}) \). (c) \( \frac{dA}{dt} = \frac{1}{2} (b \sin \theta \frac{da}{dt} + a \sin \theta \frac{db}{dt} + a b \cos \theta \frac{d\theta}{dt}) \).

Step 1: Understanding the Relationship for Part (a)

Given the formula for the area of the triangle, we have \[ A = \frac{1}{2} a b \sin \theta \]For part (a), the problem states that \(a\) and \(b\) are constants. Therefore, the only variable affecting the area is \(\theta\). Differentiating \(A\) with respect to time \(t\) using the chain rule gives:\[ \frac{dA}{dt} = \frac{1}{2} a b \cos \theta \frac{d\theta}{dt} \]This expression shows how the rate of change of the angle \(\theta\) relates to the rate of change of the area \(A\).

Step 2: Differentiating for Part (b)

Here, \(a\) can change, but \(b\) is constant. The area is given by:\[ A = \frac{1}{2} a b \sin \theta \]We need to differentiate this expression with respect to \(t\), where both \(a\) and \(\theta\) could change:\[ \frac{dA}{dt} = \frac{1}{2} \left( b \sin \theta \frac{da}{dt} + a b \cos \theta \frac{d\theta}{dt} \right) \]This result accounts for changes in both \(a\) and \(\theta\) affecting the area.

Step 3: Finding the Relationship for Part (c)

In this case, all three variables \(a\), \(b\), and \(\theta\) can change. The formula for the area remains:\[ A = \frac{1}{2} a b \sin \theta \]Differentiating this expression with respect to \(t\) gives:\[ \frac{dA}{dt} = \frac{1}{2} \left( b \sin \theta \frac{da}{dt} + a \sin \theta \frac{db}{dt} + a b \cos \theta \frac{d\theta}{dt} \right) \]This expression captures the influence of changes in \(a\), \(b\), and \(\theta\) on the change in \(A\).

Key Concepts

Differentiation

Differentiation is a core concept in calculus used to find the rate of change of a function. It provides valuable insights into how one quantity changes in relation to another. In the context of related rates, we're interested in determining how changes in one variable affect another. This is particularly important when differentiating equations where multiple variables are involved.
To differentiate a function, we apply certain rules. Specifically, when multiple variables are involved, we often use the chain rule. The chain rule helps to differentiate composite functions, allowing us to account for each variable's contribution to the overall change.

• If a variable is constant, its derivative is zero, simplifying the differentiation process.
• When differentiating, always consider the rate of change with respect to time (denoted as \( \frac{d}{dt} \)).

Understanding differentiation is vital because it lays the foundation for analyzing how and why changes occur in problems involving related rates.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in many mathematical applications, including related rates problems. Specifically, they come into play when dealing with angles in geometric shapes like triangles.
In this context, the sine function is used to determine the area of a triangle when given two sides and the included angle. The formula \( A = \frac{1}{2} a b \sin \theta \) leverages the sine of the angle \( \theta \) to calculate the area.
When differentiating with respect to time, the derivative of the sine function, \( \cos \theta \), becomes crucial:

• \( \sin \theta \) describes how the angle affects the triangle's area.
• \( \cos \theta \) is needed when finding the rate of change of the area concerning the angle \( \theta \).

These trigonometric identities are integral for solving problems where angles and their rates of change play a role.

Triangle Area

The area of a triangle can be found using various formulas, but one particularly useful expression is when you have two sides and the included angle. This formula is:

• \( A = \frac{1}{2} a b \sin \theta \)

This formula works because it incorporates both the lengths of the sides and the angle. It's a classic example of how geometry and trigonometry can intersect to provide solutions in related rates problems.
When determining the rate of change of the area over time (\( \frac{dA}{dt} \)), we account for how any change in the sides or the angle alters the area. Depending on which variables are constant or changing:

• If both sides \(a\) and \(b\) are constant, only changes in \(\theta\) affect \(\frac{dA}{dt}\).
• If one side is constant and the other changes, both the changing side and \(\theta\) contribute to \(\frac{dA}{dt}\).
• If neither side nor \(\theta\) is constant, every variable plays a part in the rate of change of the area.

Understanding these relationships is crucial for accurately calculating how the triangle’s area evolves.