Question

A spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3\text{ mm}$ and $1\text{ hour}$ later has been reduced to $2\text{ mm}$, then the expression of radius $r$ of the raindrop at any time $t$ is (where $0 \le t \lt 3$)

• A. $r = t + 5$
• B. $r = t - 5$
• C. $r = 3 - t$
• D. $r = t + 3$

Hint:

Whenever the rate of evaporation of a geometric sphere is proportional to its surface area, the radius decreases at a completely constant linear rate ($\frac{dr}{dt} = \text{constant}$). Since it dropped from $3\text{ mm}$ to $2\text{ mm}$ in 1 hour, it loses exactly $1\text{ mm}$ per hour, giving $r = 3 - 1t$ instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are analyzing a spherical raindrop losing volume due to evaporation. The rate of change of its volume $V$ over time $t$ is directly proportional to its surface area $A$. We are given initial conditions to find a time-dependent linear function expressing the radius $r$.

Step 2: Key Formula or Approach:
For a sphere, the volume and surface area equations in terms of radius are: $$V = \frac{4}{3}\pi r^3 \quad \text{and} \quad A = 4\pi r^2$$ The problem states that $\frac{dV}{dt} \propto -A$ (negative since volume is decreasing). We will differentiate $V$ with respect to $t$ using the chain rule and find a differential equation for $\frac{dr}{dt}$.

Step 3: Detailed Explanation:
Differentiating the volume equation with respect to time $t$: $$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$ According to the problem's criteria: $$\frac{dV}{dt} = -k \cdot A$$ where $k$ is a positive proportionality constant. Substitute our expressions into this equation: $$4\pi r^2 \frac{dr}{dt} = -k(4\pi r^2)$$ Assuming $r \neq 0$, we can cancel out the matching terms $4\pi r^2$ from both sides: $$\frac{dr}{dt} = -k$$ This is a simple separable differential equation. Integrating both sides with respect to $t$: $$\int dr = \int -k \, dt \implies r = -kt + C \quad \text{--- (Equation 1)}$$ Now, let's use our given physical conditions to solve for constants $C$ and $k$: 1. Originally ($t = 0$), the radius is $3\text{ mm}$: $$3 = -k(0) + C \implies C = 3$$ Our equation updates to: $r = -kt + 3$. 2. After $1\text{ hour}$ ($t = 1$), the radius shrinks to $2\text{ mm}$: $$2 = -k(1) + 3 \implies k = 3 - 2 = 1$$ Substituting $k = 1$ and $C = 3$ back into Equation 1 yields our final expression: $$r = 3 - t$$

Step 4: Final Answer:
The geometric expression for the radius at any time $t$ is $r = 3 - t$, matching option (C).