Exercise 9.3 · Q19: Balloon with volume changing at a constant rate: $r = (63t + 27)^{1/3}$

Basics you need

Solution

1. Name the variables and write the model

   Let $t$ be the time in seconds, $r$ the radius in units and $V$ the volume in cubic units. The volume changes at a constant rate:

   $$\frac{dV}{dt} = k,\qquad k \text{ a constant.}$$
   "Rate of change" of the volume with time is $\frac{dV}{dt}$⁠; "constant" means it does not depend on $t$⁠. The balloon is being inflated, so $k > 0$⁠, but its value is not given: we find it from the data. This is a first order equation of degree 1.
   A rate of change is a derivative (Chapter 6). Basics: y′ is the slope

2. Rewrite the rate in terms of the radius

   $V = \frac{4}{3}\pi r^3$ and $r$ depends on $t$⁠, so by the chain rule

   $$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = k.$$
   The question asks for the radius, so we want an equation in $r$ and $t$⁠. $\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2$⁠, times $\frac{dr}{dt}$⁠.
   Chain rule (Chapter 5). Basics: y′ is the slope

3. Separate the variables and integrate

   $$\begin{aligned} 4\pi r^2\,dr &= k\,dt \\ \int 4\pi r^2\,dr &= \int k\,dt \\ \frac{4}{3}\pi r^3 &= kt + C. \end{aligned}$$
   The $r$ terms are already with $dr$⁠, so no division is needed. $\int r^2\,dr = \frac{r^3}{3}$ and $\int k\,dt = kt$⁠. The constants of both sides combine into one arbitrary constant $C$⁠. (Integrating $\frac{dV}{dt} = k$ directly gives the same line: $V = kt + C$⁠.)
   Section 9.4.1; Section 7.2, formula (i). Lesson: separate, then integrate

4. Use the initial radius

   At $t = 0$⁠, $r = 3$⁠: $\frac{4}{3}\pi\cdot 27 = 0 + C$⁠, so

   $$C = 36\pi.$$
   $3^3 = 27$ and $\frac{4}{3}\cdot 27 = 36$⁠. $C$ is the starting volume, $36\pi$ cubic units.
   Section 9.3: a condition gives the constant a value. Basics: one constant, one condition

5. Use the radius after 3 seconds to find $k$

   At $t = 3$⁠, $r = 6$⁠: $\frac{4}{3}\pi\cdot 216 = 3k + 36\pi$⁠, that is,

   $$\begin{aligned} 288\pi &= 3k + 36\pi \\ \Rightarrow\ 3k &= 252\pi \;\Rightarrow\; k = 84\pi. \end{aligned}$$
   $6^3 = 216$ and $\frac{4}{3}\cdot 216 = 288$⁠. The rate is $84\pi \approx 263.89$ cubic units per second. Two data, two unknowns: $C$ from the start, $k$ from the second reading.
   Solving one linear equation for $k$⁠. Lesson: general and particular solutions

6. Solve for $r$⁠, and check

   $$\begin{aligned} \frac{4}{3}\pi r^3 &= 84\pi t + 36\pi \\ r^3 &= 63t + 27 \\ r &= (63t + 27)^{\frac{1}{3}}. \end{aligned}$$

   Check: $r(0) = 27^{1/3} = 3$ and $r(3) = 216^{1/3} = 6$⁠. Also $V = \frac{4}{3}\pi(63t + 27) = 84\pi t + 36\pi$⁠, whose rate is $\frac{dV}{dt} = 84\pi$⁠, a constant.

   Multiplying by $\frac{3}{4\pi}$ gives $84\cdot\frac{3}{4} = 63$ and $36\cdot\frac{3}{4} = 27$⁠. The real cube root is taken, and $63t + 27 > 0$ for $t \ge 0$⁠. Both given radii and the "constant rate" condition are satisfied.
   A solution satisfies the equation and the conditions (Section 9.3). Lesson: general and particular solutions

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