Question

A thin metallic wire in the shape of a circular ring has its enclosed area increasing at a uniform rate when heated. Show that the rate of change of circumference varies inversely as the radius.

Hint:

Whenever a question asks to show that a quantity $Y$ varies inversely as $X$, always look to find an equation of the form $Y = \frac{\text{Constant}}{X}$. Here, since the area rate is constant, the expansion speed of the boundary slows down as the circle grows wider because the same amount of added area must be spread over a larger perimeter.

Answer 1

Concept: This problem deals with the application of derivatives in determining the rates of change of geometric quantities. Consider a circular ring whose radius at any arbitrary instant of time $t$ is denoted by $r$.
• The total area $A$ enclosed by a circle of radius $r$ is given by the standard geometric formula: \[ A = \pi r^2 \]
• The perimeter or circumference $C$ of the circular ring is given by the formula: \[ C = 2\pi r \] When the metallic ring is heated, it undergoes thermal expansion, causing its radius $r$, area $A$, and circumference $C$ to become functions of time $t$. We can use the chain rule of differentiation to relate their rates of change with respect to time $t$, namely $\frac{dA}{dt}$ and $\frac{dC}{dt}$.

Step 1: Expressing the rate of change of area
Let $A$ be the area enclosed by the circular ring at any instant $t$. The area formula is: \[ A = \pi r^2 \] Differentiating both sides of this equation with respect to time $t$ using the chain rule, we obtain: \[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \pi \cdot \frac{d}{dr}(r^2) \cdot \frac{dr}{dt} \] Applying the power rule $\frac{d}{dr}(r^2) = 2r$, we get: \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \quad \cdots (1) \]

Step 2: Incorporating the given uniform rate constraint
According to the problem statement, the enclosed area is increasing at a uniform (constant) rate. Let this constant rate of increase be denoted by $k$, where $k > 0$. Therefore, we can write: \[ \frac{dA}{dt} = k \] Substituting this into Equation (1), we establish a relationship for the rate of change of the radius: \[ k = 2\pi r \frac{dr}{dt} \] Isolating the term $\frac{dr}{dt}$ by dividing both sides by $2\pi r$, we get: \[ \frac{dr}{dt} = \frac{k}{2\pi r} \quad \cdots (2) \]

Step 3: Finding the rate of change of circumference
Let $C$ be the circumference of the circular ring at any instant $t$. The formula for the circumference is: \[ C = 2\pi r \] Now, differentiate both sides of this equation with respect to time $t$: \[ \frac{dC}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt} \quad \cdots (3) \]

Step 4: Substituting $\frac{dr{dt}$ into the circumference rate equation}
We can now substitute the expression for $\frac{dr}{dt}$ from Equation (2) into Equation (3): \[ \frac{dC}{dt} = 2\pi \left( \frac{k}{2\pi r} \right) \] Canceling out the common factor of $2\pi$ from both the numerator and the denominator, the expression simplifies to: \[ \frac{dC}{dt} = \frac{k}{r} \]

Step 5: Concluding the inverse variation property
Since $k$ is a uniform constant value, the equation $\frac{dC}{dt} = \frac{k}{r}$ can be rewritten in terms of a proportionality relationship: \[ \frac{dC}{dt} \propto \frac{1}{r} \] This mathematically proves that the rate of change of the circumference of the circular ring varies inversely as its radius $r$.