Problem 9 - Thomas Calculus

Question

Area The area \(A\) of a triangle with sides of lengths \(a\) and \(b\) en- closing 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) \( \frac{dA}{dt} = \frac{1}{2} ab \cos\theta \frac{d\theta}{dt} \); b) \( \frac{dA}{dt} = \frac{1}{2} (b\sin\theta \frac{da}{dt} + ab\cos\theta \frac{d\theta}{dt}) \); c) \( \frac{dA}{dt} = \frac{1}{2} (b\sin\theta \frac{da}{dt} + a\sin\theta \frac{db}{dt} + ab\cos\theta \frac{d\theta}{dt}) \)."}

Step 1: Understanding the given expression

The expression for the area of a triangle is given by \( A = \frac{1}{2}ab\sin\theta \). This formula relates the area \( A \) of a triangle to the lengths of two sides \( a \) and \( b \), and the sine of the angle \( \theta \) between them.

Step 2: Derivative of A with respect to time when a and b are constants

To find \( \frac{dA}{dt} \) when \( a \) and \( b \) are constants, apply the chain rule to differentiate with respect to \( t \):\[\frac{dA}{dt} = \frac{1}{2} ab \frac{d}{dt} (\sin \theta) = \frac{1}{2} ab \cos \theta \frac{d\theta}{dt}\]Here, only \( \theta \) changes with time, so its derivative involves \( \frac{d\theta}{dt} \).

Step 3: Derivative of A with respect to time when only b is constant

In this scenario, both \( a \) and \( \theta \) might change with time. To differentiate \( A \):\[\frac{dA}{dt} = \frac{1}{2} \left( b \sin \theta \frac{da}{dt} + ab \cos \theta \frac{d\theta}{dt} \right)\]This expression reflects the relationship of \( \frac{dA}{dt} \) to both \( \frac{da}{dt} \) and \( \frac{d\theta}{dt} \).

Step 4: Derivative of A with respect to time when none are constant

If \( a \), \( b \), and \( \theta \) all change with time, we must differentiate with respect to all variables:\[\frac{dA}{dt} = \frac{1}{2} \left( b \sin \theta \frac{da}{dt} + a \sin \theta \frac{db}{dt} + ab \cos \theta \frac{d\theta}{dt} \right)\]Here \( \frac{dA}{dt} \) includes contributions from \( \frac{da}{dt} \), \( \frac{db}{dt} \), and \( \frac{d\theta}{dt} \), each affected by corresponding sides and angle.

Key Concepts

Triangle Area

The area of a triangle can be calculated using different formulas, depending on what information you have about the triangle. For a triangle with sides of lengths \(a\) and \(b\) with an angle \(\theta\) between them, the area \(A\) is given by: \[ A = \frac{1}{2}ab\sin\theta \] This formula uses the basic principle that the area of a triangle is equivalent to half the product of two sides multiplied by the sine of the included angle.

• The term \(\sin\theta\) relates to the height of the triangle when one side is considered as the base.
• The division by 2 arises from the need to find the actual area of the triangular shape derived from a rectangle with the same base and height.

Understanding the geometric relationship this equation represents can be very useful in solving problems related to triangle areas in trigonometry.

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. When we have a function that depends on another function, we apply the chain rule. For the area \(A\) of the triangle where \(a\) and \(b\) are constants, the chain rule helps us determine how \(\theta\), the angle included between sides, affects the area over time: Given: \[\frac{dA}{dt} = \frac{1}{2} ab \cos\theta \frac{d\theta}{dt}\] Here, the chain rule allows us to express the rate of change of area \(dA/dt\) in terms of the rate of change of the angle \(d\theta/dt\).

• Since \(a\) and \(b\) don't change, the derivative focuses only on \(\theta\).
• The use of cosine \(\cos\theta\) comes from differentiating the sine function.\(\sin\theta\).

This makes the chain rule incredibly powerful when dealing with related rates.

Trigonometry

Trigonometry, the study of triangles and the relationships between their sides and angles, plays a crucial role in solving problems involving triangles. In the given problem, trigonometry is employed through the use of the sine function. The sine function helps in determining the area when given two sides and an included angle.

• The sine function, \(\sin\theta\), measures the opposite side's length over the hypotenuse in a right triangle.
• When applied to the area formula, the sine function effectively incorporates the height relative to the base formed by sides \(a\) and \(b\).

Trigonometry links algebra, geometry, and calculus providing the tools necessary to manage angles and dimensions in physical space.

Derivatives

Derivatives in calculus represent the rate at which a function is changing at any given point. In the context of the triangle area, derivatives help find how the area changes over time as the sides and angle of the triangle change. When the problem asks for \(dA/dt\), the derivative of the area with respect to time, it wants to know how quickly the area changes as \(a\), \(b\), or \(\theta\) change.

• If only \(\theta\) changes, we get: \[\frac{dA}{dt} = \frac{1}{2} ab \cos\theta \frac{d\theta}{dt}\]
• If only \(b\) is constant, the change depends on both \(a\) and \(\theta\): \[\frac{dA}{dt} = \frac{1}{2} \left( b \sin\theta \frac{da}{dt} + ab \cos\theta \frac{d\theta}{dt} \right)\]
• When all variables change, each derivative adds to the rate of area change.

Understanding derivatives allows us to analyze problems where variables are dynamic and interconnected, as is common in real-world applications.