Chapter 6 · Application of Derivatives

Exercise 6.1 · Q13: Balloon of diameter $\tfrac32(2x+1)$ ⁠: rate of volume with $x$

Exercise 6.1Rate of change of quantities · Section 6.2

The question

A balloon, which always remains spherical, has a variable diameter $\dfrac{3}{2}(2x + 1)$ ⁠. Find the rate of change of its volume with respect to $x$ ⁠.

Exam answer

Diameter $= \frac32(2x + 1)$ ⁠, so the radius is $r = \frac34(2x + 1)$ ⁠, where $x > -\frac12$ ⁠.

\begin{aligned} V &= \frac43\pi r^3 \\ &= \frac43\pi\left(\frac34\right)^3(2x + 1)^3 \\ &= \frac{9\pi}{16}(2x + 1)^3 \end{aligned}

Differentiating w.r.t. $x$ ⁠,

\begin{aligned} \frac{dV}{dx} &= \frac{9\pi}{16}\cdot 3(2x + 1)^2\cdot 2 \\ &= \frac{27\pi}{8}(2x + 1)^2 \end{aligned}

$\therefore$ the rate of change of the volume is $\dfrac{27}{8}\pi(2x + 1)^2$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The rate asked for is with respect to $x$ ⁠, not time: we need $\frac{dV}{dx}$ ⁠. Write the volume as a function of $x$ alone. The volume formula uses the radius, which is half the diameter, $r = \frac34(2x+1)$ ⁠. Then $V = \frac43\pi r^3 = \frac{9\pi}{16}(2x+1)^3$ ⁠, and the chain rule (inner function $2x+1$ ⁠) gives $\frac{dV}{dx} = \frac{27\pi}{8}(2x+1)^2$ ⁠.

Given, to find, and the plan

Given: A sphere whose diameter is $D = \frac32(2x+1)$ ⁠, where $x$ is a variable.
To find: $\frac{dV}{dx}$ ⁠, the rate of change of the volume $V$ with respect to $x$ ⁠.
Plan: (1) Note where the formula makes sense. (2) Radius from diameter. (3) Volume in terms of $x$ ⁠. (4) Differentiate with the chain rule. (5) Check by a second route, $\frac{dV}{dr}\cdot\frac{dr}{dx}$ ⁠.

Basics you need

1Sphere: radius, diameter, volume

The radius is half the diameter, $r = \frac{D}{2}$ ⁠, and the volume is $V = \frac43\pi r^3$ ⁠. Basics: area and volume formulas

2Rate of change with respect to any variable

"Rate of change of $V$ with respect to $x$ " means $\frac{dV}{dx}$ (Section 6.2). The variable need not be time. Basics: rate of change as a derivative

3Chain rule for a power of a linear expression

$\frac{d}{dx}(ax+b)^n = n(ax+b)^{n-1}\cdot a$ ⁠: the power rule on the outside, times the slope $a$ of the inside. Basics: the chain rule

4Powers of a product

$\left(k\,u\right)^3 = k^3u^3$ ⁠, so $\left(\frac34(2x+1)\right)^3 = \frac{27}{64}(2x+1)^3$ ⁠.

Solution

Where the formula makes sense

A diameter must be positive:
$\frac32(2x+1) > 0 \iff 2x + 1 > 0 \iff x > -\frac12.$
So we work with $x > -\frac12$ ⁠.

The question gives the diameter as a formula in $x$ ⁠, but only values of $x$ that give a real balloon make sense. The algebra below works for every $x$ ⁠, but the answer has a physical meaning only for $x > -\frac12$ ⁠.
The radius in terms of $x$
$r = \frac{D}{2} = \frac12\cdot\frac32(2x+1) = \frac34(2x+1).$
The volume formula is written in terms of the radius, and the question gives the diameter. Halving first avoids the most common error in this question: putting the diameter into $\frac43\pi r^3$ ⁠.

Radius = diameter $\div$ 2. Basics: area and volume formulas
The volume as a function of $x$
$\begin{aligned} V &= \frac43\pi r^3 \\ &= \frac43\pi\left(\frac34\right)^3(2x+1)^3 \\ &= \frac43\cdot\frac{27}{64}\,\pi\,(2x+1)^3 \\ &= \frac{108}{192}\,\pi\,(2x+1)^3 \\ &= \frac{9\pi}{16}(2x+1)^3. \end{aligned}$
Cubing $\frac34(2x+1)$ cubes both factors: $\left(\frac34\right)^3 = \frac{27}{64}$ ⁠. Then $\frac43\cdot\frac{27}{64} = \frac{108}{192}$ ⁠, which simplifies (divide top and bottom by 12) to $\frac{9}{16}$ ⁠. Now $V$ is a function of $x$ alone, ready to differentiate.

$(ku)^3 = k^3u^3$ (laws of exponents).
Differentiate with respect to $x$ (chain rule)

Inner function $u = 2x + 1$ ⁠, with $\frac{du}{dx} = 2$ ⁠. Outer function $V = \frac{9\pi}{16}u^3$ ⁠, with $\frac{dV}{du} = \frac{9\pi}{16}\cdot 3u^2 = \frac{27\pi}{16}u^2$ ⁠. By the chain rule,
$\begin{aligned} \frac{dV}{dx} &= \frac{dV}{du}\cdot\frac{du}{dx} \\ &= \frac{27\pi}{16}(2x+1)^2\cdot 2 \\ &= \frac{27\pi}{8}(2x+1)^2. \end{aligned}$
$(2x+1)^3$ is a cube of an expression, not of $x$ ⁠, so after the power rule we must multiply by the derivative of the inside, which is 2. That factor 2 turns $\frac{27}{16}$ into $\frac{27}{8}$ ⁠.

Chain rule (Theorem 4, Section 5.3.1) and $\frac{d}{du}\left(u^3\right) = 3u^2$ (Table 5.3). Basics: the chain rule
Check by a second route: through the radius

Treat $V$ as a function of $r$ and $r$ as a function of $x$ ⁠:
$\frac{dV}{dr} = 4\pi r^2, \qquad \frac{dr}{dx} = \frac34\cdot 2 = \frac32.$
So
$\begin{aligned} \frac{dV}{dx} &= 4\pi r^2\cdot\frac32 = 6\pi r^2 \\ &= 6\pi\cdot\frac{9}{16}(2x+1)^2 \\ &= \frac{27\pi}{8}(2x+1)^2. \end{aligned}$
The same answer.

Two independent routes agreeing is strong evidence that no factor was lost. This route uses the same $\frac{dV}{dr} = 4\pi r^2$ that appears in Q8, Q9 and Q12, multiplied here by the rate of $r$ with respect to $x$ instead of time.

Chain rule: $\frac{dV}{dx} = \frac{dV}{dr}\cdot\frac{dr}{dx}$ ⁠. Lesson: rate of change of quantities

The diameter is

D = \frac32(2x+1)

⁠, but the volume formula needs the radius

r = \frac{D}{2} = \frac34(2x+1)

⁠.

Result

\frac{dV}{dx} = \frac{27}{8}\,\pi\,(2x+1)^2 \qquad \left(x > -\tfrac12\right).

(Textbook answer: $\frac{27}{8}\pi(2x+1)^2$ ⁠.) No units are given for $x$ ⁠, so the rate is in units of volume per unit of $x$ ⁠. It is positive for every $x > -\frac12$ (a non-zero square times $\frac{27\pi}{8}$ ⁠), so the volume increases as $x$ increases.

Check it with numbers

Take $x = 1$ ⁠: $D = 4.5$ ⁠, $r = 2.25$ ⁠, and the formula gives $\frac{dV}{dx} = \frac{27\pi}{8}\cdot 9$ ⁠, which is $\frac{243\pi}{8} \approx 95.43$ ⁠. Numerically, with $V(x) = \frac{9\pi}{16}(2x+1)^3$ ⁠, the difference quotient $\frac{V(1.0001) - V(0.9999)}{0.0002}$ is about 95.4259, in agreement.

Common mistakes

Using the diameter as the radius. $\frac43\pi D^3$ leads to $27\pi(2x+1)^2$ ⁠, which is 8 times too big (at $x = 1$ ⁠: about 763.41 instead of 95.43).
Dropping the inner derivative. Forgetting the factor 2 from $2x+1$ gives $\frac{27\pi}{16}(2x+1)^2$ ⁠, half the answer (about 47.71 at $x = 1$ ⁠).
Cubing only part of the radius. Both $\frac34$ and $(2x+1)$ must be cubed.
Differentiating with respect to time. The question asks for the rate with respect to $x$ ⁠; no time rate is involved.

What to take away

"Rate of change of $V$ with respect to $x$ " is simply $\frac{dV}{dx}$ ⁠; write $V$ in terms of $x$ and differentiate.
Convert diameter to radius before using sphere formulas.
A power of a linear expression needs the chain rule: multiply by the slope of the inside.

Every fact used, and where it comes from

Fact	Used in	Source
Rate of change of $V$ with respect to $x$ is $\frac{dV}{dx}$	Whole solution	Section 6.2 (Lesson: rate of change of quantities)
Sphere: $r = \frac{D}{2}$ ⁠, $V = \frac43\pi r^3$	Steps 2, 3, 5	Mensuration (Classes IX and X, Surface Areas and Volumes)
Chain rule	Steps 4, 5	Theorem 4, Section 5.3.1
$\frac{d}{du}\left(u^n\right) = nu^{n-1}$	Steps 4, 5	Table 5.3, Section 5.3
$(ku)^3 = k^3u^3$	Step 3	Laws of exponents

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 6.1 · Q13: Balloon of diameter 32(2x+1)\tfrac32(2x+1)23​(2x+1)⁠: rate of volume with xxx

Basics you need

Solution

Where the formula makes sense

The radius in terms of xxx

The volume as a function of xxx

Differentiate with respect to xxx (chain rule)

Check by a second route: through the radius

Every fact used, and where it comes from

Exercise 6.1 · Q13: Balloon of diameter $\tfrac32(2x+1)$ ⁠: rate of volume with $x$

The radius in terms of $x$

The volume as a function of $x$

Differentiate with respect to $x$ (chain rule)