Question

True or False If the radius of a circle is expanding at a constant rate, then its area is increasing at a constant rate. Justify your answer.

Short answer

False. Even though the radius is increasing at a constant rate, the area is not increasing at a constant rate. The rate of increase of the area depends on the radius of the circle at that time, and increases as the radius becomes larger.

Step-by-step solution

Understand the concept

Given that the radius of the circle is expanding at a constant rate, it is assumed that this means the radius \(r\) is increasing linearly with time \(t\). We denote this rate of change as \(dr/dt\), which is a constant. The area \(A\) of a circle is given by the formula \(A = \pi r^2\). To see if the area is also increasing at a constant rate, we need to derive an expression for \(dA/dt\), the rate of change of the area with respect to time.

Obtain the expression for dA/dt

The rate of change of the area with respect to time can be obtained using the chain rule in calculus. Differentiating \(A = \pi r^2\) by \(t\), we get \(dA/dt = 2 \pi r (dr/dt)\). Since \(dr/dt\) is constant, we see that \(dA/dt\) is proportional to the radius \(r\). This means that the rate of increase of the area of the circle depends on the radius of the circle.

Conclusion

From the obtained expression, it can be seen that the rate of increase of the area of the circle, \(dA/dt\), depends on the radius at that instant. Even though the radius is increasing at a constant rate, the area increases more quickly as the radius becomes larger. Therefore, the area is not increasing at a constant rate, which means the statement is False.

Key concepts

Derivative Application

Understanding derivative application is crucial when we are dealing with rates of change in calculus. In the context of the given exercise, we’re exploring how changes in one quantity affect another. To be more precise, we’re looking at how a constant rate of change in the radius of a circle influences the rate of change of the circle’s area.

The application of derivatives is essentially about understanding how a function's output changes as its input changes. For a circle with a changing radius, we express the radius as a function of time, and then we use the derivative to find how fast the area of the circle is changing with respect to time. The derivative of the area with respect to time, denoted by \(dA/dt\), gives us the rate at which the area of the circle is increasing. This helps in understanding dynamic scenarios and is widely used in fields like physics, engineering, and economics to model real-world situations.

Chain Rule

The chain rule is a fundamental tool in calculus used to calculate the derivative of a composite function. In our exercise, it’s used to relate the rates of change of the area of a circle to the rate of change of its radius.

Understanding the Chain Rule

Suppose you have a function \(h(x)\) that is composed of two functions \(f(g(x))\). The chain rule tells us that the derivative of \(h\) with respect to \(x\) is the derivative of \(f\) with respect to \(g\) multiplied by the derivative of \(g\) with respect to \(x\), or \(h'(x) = f'(g(x)) \times g'(x)\). In the context of our problem, \(A = \pi r^2\) is the function \(f\), where \(f\) is a function of \(r\), and \(r\) is a function of time, \(t\), the second function \(g\). Applying the chain rule allows us to find the relationship between the rate of change of the area (\(\frac{{dA}}{{dt}}\)) and the rate of change of the radius (\(dr/dt\)).

Area Differentiation

In our exercise, area differentiation involves finding the rate at which the area of the circle is changing over time. To differentiate the area of a circle, \(A = \pi r^2\), with respect to time, we are essentially looking for \(\frac{dA}{dt}\).

As the radius \(r\) is a function of time \(t\), we need to use implicit differentiation, where we treat \(r\) as \(r(t)\), and differentiate both sides of the area equation with respect to \(t\). This takes into account the chain rule, as we’ve previously discussed.

Implications of Area Differentiation

The result of this differentiation process shows us that the area’s rate of change, \(\frac{dA}{dt}\), isn't constant, but rather it varies depending on the radius \(r\) at any given time. This is because it includes the term \(2\pi r \frac{dr}{dt}\), indicating that as \(r\) increases, so does the value of \(\frac{dA}{dt}\). This key insight tells us that even though \(r\) may increase at a constant rate (\(\frac{dr}{dt}\) being constant), the area does not, revealing a deeper understanding of how different rates of change interact with each other.