Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expans

Q: Why do we take the log instead of differentiating directly?

Differentiating directly using the product rule works perfectly fine, but taking the log turns products into sums and powers into coefficients. This makes it much easier to isolate fractional changes like $\frac{1}{\rho}\frac{d\rho}{dt}$ without messy algebra.

Q: Why is the derivative of mass zero?

The problem explicitly states that the 'total mass remains constant' during the expansion. In calculus, the derivative of any constant value with respect to time is always zero.

Q: What does $\frac{dR}{dt}$ represent physically?

$\frac{dR}{dt}$ is the rate at which the radius is increasing. For a point on the surface of the sphere, its distance from the center is $R$, so its outward velocity $v$ is exactly the rate of change of that distance, which is $\frac{dR}{dt}$.

Q: Why is there a negative sign in the velocity equation?

The negative sign indicates an inverse relationship between the rates. As the sphere expands (velocity $v$ is positive), the density must decrease (so $\frac{d\rho}{dt}$ is negative) to keep the total mass constant.

Q: Can this method be used for other shapes?

Absolutely! Whether it's a cylinder, a cube, or any other shape, if the mass is constant, you can use logarithmic differentiation to relate the fractional changes in its dimensions and its density.

Question

Consider an expanding sphere of instantaneous radius $R$ whose total mass remains constant. The expansion is such that the instantaneous density $\rho$ remains uniform throughout the volume. The rate of fractional change in density $\left(\frac{1}{\rho}\frac{d\rho}{dt}\right)$ is constant. The velocity $v$ of any point of the surface of the expanding sphere is proportional to

Options:

(A) $R$

(B) $\frac{1}{R}$

(C) $R^3$

(D) $R^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a balloon expanding uniformly in all directions. As it expands, its volume increases, but we are told that its total mass remains strictly constant. For the mass to remain constant while the volume increases, the density of the material must decrease.

We can express the total mass $m$ of the sphere in terms of its instantaneous radius $R$ and its uniform density $\rho$ using the standard formula for the mass of a sphere:

$$m = \frac{4}{3}\pi R^3 \rho$$

### The Master Equation and Logarithmic Differentiation
We are given information about the *fractional rate of change* of density, which is mathematically written as $\frac{1}{\rho}\frac{d\rho}{dt}$. Whenever a physics problem involves products of variables and asks for fractional changes or percentage errors, **logarithmic differentiation** is your best friend.

Let's take the natural logarithm ($\ln$) of both sides of our mass equation. This brilliant mathematical move transforms the multiplication into simple addition:

$$\ln(m) = \ln\left(\frac{4\pi}{3}\right) + 3\ln(R) + \ln(\rho)$$

### Calculus in Action
Now, we differentiate the entire equation with respect to time $t$. We must remember two critical things: first, the mass $m$ is constant, so its derivative is zero. Second, the term $\frac{4\pi}{3}$ is also a constant, so its derivative is zero as well.

$$0 = 0 + \frac{3}{R}\frac{dR}{dt} + \frac{1}{\rho}\frac{d\rho}{dt}$$

What is $\frac{dR}{dt}$? It is the rate at which the radius is increasing. For any point on the surface of the sphere, its outward velocity $v$ is exactly equal to the rate of change of the radius. Therefore, we can substitute $\frac{dR}{dt} = v$:

$$\frac{3}{R}v + \frac{1}{\rho}\frac{d\rho}{dt} = 0$$

### Final Calculation
Let's rearrange this equation to isolate the velocity $v$:

$$\frac{3}{R}v = -\frac{1}{\rho}\frac{d\rho}{dt}$$

$$v = -\frac{R}{3}\left(\frac{1}{\rho}\frac{d\rho}{dt}\right)$$

The problem explicitly states that the fractional change in density, $\left(\frac{1}{\rho}\frac{d\rho}{dt}\right)$, is a constant. Let's call this constant $k$.

$$v = -\frac{k}{3} R$$

Since $-\frac{k}{3}$ is just another constant, we can clearly see the direct relationship:

**$$v \propto R$$**

The velocity of any point on the surface of the expanding sphere is directly proportional to its instantaneous radius $R$.

Select Answer:

Visualized Solution

$m = \frac{4}{3} π R^{3} ρ$

$ln (m) = ln (\frac{4 π}{3}) + 3 ln (R) + ln (ρ)$

$\frac{d}{d t} (ln m) = 0$

$\frac{d R}{d t} = v$

$v = - \frac{R}{3} (\frac{1}{ρ} \frac{d ρ}{d t})$

$v \propto R$

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup Imagine a balloon expanding uniformly in all directions

The Master Equation and Logarithmic Differentiation We are given information about the fractional rate of change of density, which is mathematically written as $\frac{1}{ρ} \frac{d ρ}{d t}$

Calculus in Action Now, we differentiate the entire equation with respect to time $t$

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=34​πR3ρ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(m)=ln(34π​)+3ln(R)+ln(ρ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd​(lnm)=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdR​=v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=−3R​(ρ1​dtdρ​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v∝R

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup Imagine a balloon expanding uniformly in all directions

The Master Equation and Logarithmic Differentiation We are given information about the fractional rate of change of density, which is mathematically written as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρ1​dtdρ​

Calculus in Action Now, we differentiate the entire equation with respect to time .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t