Problem 27 - Calculus Single Variable

Question

Let \(P\) denote the position of particle that, from its initial point \(Q=(1,0)\), moves counterclockwise along the unit circle. If \(\theta\) denotes the angle subtended at the origin \(O\) by the circular arc \(Q P,\) then the radial velocity \(d \theta / d t\) of the particle is constantly equal to \(6 .\) How fast is the area of the sector \(Q P O\) swept out by the particle changing?

Step-by-Step Solution

Answer

The area of the sector is changing at a rate of 3 square units per unit time.

Step 1: Understand the Problem

We have a particle moving counterclockwise along a unit circle from an initial point \(Q = (1, 0)\). The radial velocity \(\frac{d\theta}{dt} = 6\) is given as constant. We need to calculate how quickly the area of the sector \(QPO\) changes as the particle moves.

Step 2: Formula for Area of Sector

The formula for the area \(A\) of a sector of a circle with radius \(r\) and central angle \(\theta\) in radians is given by \(A = \frac{1}{2} r^2 \theta\). Since the circle is a unit circle, \(r = 1\), so \(A = \frac{1}{2} \theta\).

Step 3: Differentiate the Area Formula

Differentiate the area formula with respect to time \(t\): \( \frac{dA}{dt} = \frac{d}{dt} \left( \frac{1}{2} \theta \right) = \frac{1}{2} \cdot \frac{d\theta}{dt}.\)

Step 4: Substitute Given Radial Velocity

Substitute the given radial velocity \(\frac{d\theta}{dt} = 6\) into the differentiated formula: \( \frac{dA}{dt} = \frac{1}{2} \cdot 6 = 3.\)

Step 5: Final Step: Solve and Conclude

The rate at which the area of the sector \(QPO\) is changing as the particle moves is \(3\) square units per unit time.

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that focuses on the concept of change. It's all about understanding how functions change over time or in space. This field is crucial for solving problems related to motion, growth, and other dynamic processes. When we talk about rates of change, like velocity or acceleration, we rely heavily on differential calculus. By taking derivatives, we can find how one quantity affects another as they change. In this exercise, differential calculus is used to find out how quickly the area of a sector changes as the angle subtended changes. This involves differentiating the formula for the area of a sector with respect to time. By doing this, we get the rate of change of the area, given radial velocity.

Radial Velocity

Radial velocity is a measure of how fast an angle, \( \theta \), changes over time. It's often used when discussing objects moving in circular paths because it describes rotational speed. In this scenario, radial velocity is given as \( \frac{d\theta}{dt} = 6\). This means that the angle is changing at a constant rate. This constant rate is key to solving many problems involving rotation because it gives us a stable parameter to work from.Understanding radial velocity helps us link angular movement to other dynamic changes, like the area of a sector in this example. We used this concept to determine how fast the area is increasing by integrating it into the differentiation of the area formula, leading to a clear solution.

Sector Area

The area of a sector is a portion of a circle, defined by a central angle. It's like a slice of pie, with both straight and curved edges. To find the area of a sector, we use the formula: \( A = \frac{1}{2} r^2 \theta \). For a unit circle, where \( r = 1 \), this simplifies to \( A = \frac{1}{2} \theta \). In the context of this problem, the area changes as the particle moves along the circle, increasing the central angle \( \theta \). By taking the derivative of the sector area formula, we find how quickly the area changes over time. This derivative incorporates the radial velocity to root out the final rate of change, effectively linking all these beautiful concepts together, showing calculus in action.