Question

If by dropping a stone in a quiet lake a wave moves in circle at a speed of \(3.5 \mathrm{~cm} / \mathrm{sec}\), then the rate of increase of the enclosed circular region when the radius of the circular wave is \(10 \mathrm{~cm}\), is \(\left(\pi=\frac{22}{7}\right)\) (1) 220 sq. \(\mathrm{cm} / \mathrm{sec}\) (2) 110 sq. \(\mathrm{cm} / \mathrm{sec}\) (3) 35 sq. \(\mathrm{cm} / \mathrm{sec}\) (4) 350 sq. \(\mathrm{cm} / \mathrm{sec}\)

Short answer

The rate of increase of the enclosed circular region when the radius of the circular wave is 10 cm is 220 sq. cm/sec.

Step-by-step solution

Identify Given Values

Given that the speed of the wave is 3.5 cm/sec, and the current radius of the circular wave is 10 cm.

Express Area of Circle in Terms of Radius

The area of a circle is given by the formula: \[ A = \pi r^2 \].

Differentiate Area with Respect to Time

Since we are looking for the rate of increase of the enclosed circular region, differentiate the area function with respect to time (t): \[ \frac{dA}{dt} = 2 \pi r \frac{dr}{dt} \].

Substitute Given Values into Derived Formula

We know that \( \frac{dr}{dt} = 3.5 \mathrm{~cm/sec} \) and \( r = 10 \mathrm{~cm} \). Substitute these values into the formula: \[ \frac{dA}{dt} = 2 \pi (10 \text{ cm} \times 3.5 \text{ cm/sec}) \].

Calculate the Rate of Increase

Plug in the value of \( \pi \) as given \( \pi = \frac{22}{7} \): \[ \frac{dA}{dt} = 2 \times \frac{22}{7} \times 10 \times 3.5 \]. Simplify this to get: \[ \frac{dA}{dt} = 220 \text{ sq. cm/sec} \].

Key concepts

circular wave propagation

Circular wave propagation refers to the movement of waves in a circular pattern, commonly observed when a stone is dropped into water, creating ripples. These waves spread out from the point of impact, forming an expanding circle. As the radius of the wave increases, so does the area of the circle that the wave covers.

For example, if a stone is dropped in a quiet lake and waves start to spread, those waves are considered to be moving in a circular propagation. This concept is crucial in understanding how the area impacted by the waves changes over time.

differentiation with respect to time

Differentiation with respect to time is a key concept in calculus. It helps to find the rate at which quantities change over time. In this case, we are interested in how fast the area of the expanding wave is growing.

To find the rate of change of the area of a circle with respect to time, we start with the formula for the area of a circle: \[A = \pi r^2\]. Differentiating this with respect to time (t) gives us the formula: \[ \frac{dA}{dt} = 2 \pi r \frac{dr}{dt} \]. Here, \frac{dA}{dt}\ is the rate of change of the area, and \frac{dr}{dt}\ is the rate of change of the radius. By knowing these rates, we can determine how quickly the area of the wave is increasing.

application of calculus in physics

Calculus is a powerful tool used in physics to solve problems involving changing quantities. In the context of wave motion, calculus helps us understand how different physical properties, like the area covered by waves, change over time.

In our example, we applied calculus to find the rate of increase in the area enclosed by a circular wave. By differentiating the area formula with respect to time and substituting given values, we calculated that the rate of increase of the enclosed region when the radius is 10 cm and the wave speed is 3.5 cm/sec is 220 sq. cm/sec.

This demonstrates how differentiating physical equations allows physicists to predict and describe dynamic systems accurately, which is fundamental in many fields of science and engineering.