1.7.4 Rates of Change - GCEALevelMaths9709

In this topic we will learn how to:

apply differentiation to rates of change

Rate of change refers to the speed at which a variable changes. We can use differentiation to determine the rate at which a variable changes. If we were to investigate the rate at which a variable $x$ was changing we would denote this using the notation,
$\frac{dx}{dt}$ To find the rate at which a variable is changing we will use the chain rule.

Let’s walk through some past paper questions.

1. A curve is such that $\frac{dy}{dx} = \frac{6}{(3x - 2)^{3}}$ and $A(1, -3)$ lies on the curve. A point is moving along the curve and at $A$ the $y$ -coordinate of the point is increasing at $3$ units per second. Find the rate of increase at $A$ of the $x$ -coordinate of the point. (9709/12/F/M/21 number 6)

Let’s write out, mathematically, all the information we have been given,
$\frac{dy}{dx} = \frac{6}{(3x - 2)^{3}}\ \ \ \ \ \ \frac{dy}{dt} = 3\ \ \ \ \ \ \frac{dx}{dt} = ?\ \ \ \ \ \ A(1, -3)$ The question requires us to find $\frac{dx}{dt}$ . Let’s define a chain rule in terms of $\frac{dx}{dt}$ ,
$\frac{dx}{dt} = \frac{dx}{dy} \times \frac{dy}{dt}$ We already have $\frac{dy}{dt}$ . Let’s find $\frac{dx}{dy}$ , using $\frac{dy}{dx}$ ,
$\frac{\textcolor{#2192ff}{dy}}{\textcolor{#0f0}{dx}} = \frac{\textcolor{#2192ff}{6}}{\textcolor{#0f0}{(3x - 2)^{3}}}$ $\frac{\textcolor{#0f0}{dx}}{\textcolor{#2192ff}{dy}} = \frac{\textcolor{#0f0}{(3x - 2)^{3}}}{\textcolor{#2192ff}{6}}$ Note: Rates of change behave like fractions. So we can flip them to find their reciprocal.

Let’s substitute the $x$ -value at $A(1, -3)$ into $\frac{dx}{dy}$ ,
$\frac{dx}{dy} = \frac{(3x - 2)^{3}}{6}$ $\frac{dx}{dy} = \frac{(3(1) - 2)^{3}}{6}$ $\frac{dx}{dy} = \frac{1}{6}$ Now let’s go back to our chain rule,
$\frac{dx}{dt} = \textcolor{#2192ff}{\frac{dx}{dy}} \times \textcolor{#0f0}{\frac{dy}{dt}}$ Evaluate $\frac{dx}{dt}$ ,
$\frac{dx}{dt} = \textcolor{#2192ff}{\frac{1}{6}} \times \textcolor{#0f0}{3}$ $\frac{dx}{dt} = \frac{1}{2}$ Therefore, the final answer is,
$\frac{dx}{dt} = \frac{1}{2} \textmd{ units per second}$
Note: Though you may not be marked down for it, it is important to always give the units in the final answer.

2. A weather balloon in the shape of a sphere is being inflated by a pump. The volume of the balloon is increasing at a constant rate of $600$ cm $^{3}$ per second. The balloon was empty at the start of pumping. (9709/12/M/J/20 number 3)

(a) Find the radius of the balloon after $30$ seconds.

To find the radius we need the formula for volume of a sphere,
$V = \frac{4}{3}\pi r^{3}$ Note: You do not need to know it, it is given the list of formulae booklet MF19.

We are told that the volume is increasing at a constant rate of $600$ cm $^{3}$ per second. So we can multiply by that number by $30$ to determine the volume after $30$ seconds,
$V = 600 \times 30$ $V = 18\ 000$ Now let’s equate $18\ 000$ to the formula for volume of a sphere and evaluate $r$ ,
$V = 18\ 000\ \ \ \ \ \ V = \frac{4}{3}\pi r^{3}$ $18\ 000 = \frac{4}{3}\pi r^{3}$ Make $r^{3}$ the subject of the formula,
$r^{3} = \frac{3 \times 18\ 000}{4\pi}$ $r^{3} = \frac{13\ 500}{\pi}$ Solve for $r$ ,
$r = \left(\frac{13\ 500}{\pi}\right)^{\frac{1}{3}}$ $r = 16.2577821$ $r = 16.3$ Therefore, the final answer is,
$r = 16.3 \textmd{ cm}$ (b) Find the rate of increase of the radius after $30$ seconds.

The question requires us to find,
$\frac{dr}{dt}$ Let’s write the information from the stem of the question that can help us to define a chain rule,
$V = \frac{4}{3}\pi r^{3}\ \ \ \ \ \ \frac{dV}{dt} = 600 \ \ \ \ \ \ \textmd{\ \ after 30 seconds }r = 16.2578$ Let’s construct a chain for $\frac{dr}{dt}$ ,
$\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt}$ We already have $\frac{dV}{dt}$ . To find $\frac{dr}{dV}$ let’s first find $\frac{dV}{dr}$ ,
$V = \frac{4}{3}\pi r^{3}$ $\frac{dV}{dr} = (3)\left(\frac{4}{3}\right)\pi r^{2}$ $\frac{dV}{dr} = 4\pi r^{2}$ Flip $\frac{dV}{dr}$ to get $\frac{dr}{dV}$ ,
$\frac{dr}{dV} = \frac{1}{4\pi r^{2}}$ Substitute $r = 16.2578$ ,
$\frac{dr}{dV} = \frac{1}{4\pi (16.2578)^{2}}$ $\frac{dr}{dV} = \frac{1}{4\pi (16.2578)^{2}}$ $\frac{dr}{dV} = 0.00030106937$ Note: Use the either the exact value of $r$ or the value of $r$ correct to four decimal places or more to get the correct answer.

Let’s go back to the chain rule we defined,
$\frac{dr}{dt} = \textcolor{#2192ff}{\frac{dr}{dV}} \times \textcolor{#0f0}{\frac{dV}{dt}}$ Substitute and evaluate $\frac{dr}{dt}$ ,
$\frac{dr}{dt} = \textcolor{#2192ff}{0.00030106937} \times \textcolor{#0f0}{600}$ $\frac{dr}{dt} = 0.181$ Therefore, the final answer is,
$\frac{dr}{dt} = 0.181 \textmd{ cm per second}$ 3. The Volume $V\ m^{3}$ of a large circular mound of iron ore of radius $r$ m is modelled by the equation $V = \frac{3}{2}\left(r - \frac{1}{2}\right)^{3} - 1$ for $r \ge 2$ . Iron ore is added to the mound at a constant rate of $1.5\ m^{3}$ per second. (9709/12/O/N/21 number 9)

(a) Find the rate at which the radius of the mound is increasing at the instant when the radius is $5.5$ m.

Let’s write out, mathematically, all the information we have been given,
$V = \frac{3}{2}\left(r - \frac{1}{2}\right)^{3} - 1\ \ \ \ \ \ \frac{dV}{dt} = 1.5\ \ \ \ \ \ \frac{dr}{dt} = ?\ \ \ \ \ \ r = 5.5$ Using the information above, let’s define a chain rule for $\frac{dr}{dt}$ ,
$\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt}$ We already have $\frac{dV}{dt}$ . To find $\frac{dr}{dV}$ let’s first find $\frac{dV}{dr}$ ,
$V = \frac{3}{2}\left(r - \frac{1}{2}\right)^{3} - 1$ $\frac{dV}{dr} = (3)\left(\frac{3}{2}\right)\left(r - \frac{1}{2}\right)^{2}$ $\frac{dV}{dr} = \frac{9}{2}\left(r - \frac{1}{2}\right)^{2}$ Flip $\frac{dV}{dr}$ to get $\frac{dr}{dV}$ ,
$\frac{dr}{dV} = \frac{2}{9\left(r - \frac{1}{2}\right)^{2}}$ Substitute $r = 5.5$ ,
$\frac{dr}{dV} = \frac{2}{9\left(5.5 - \frac{1}{2}\right)^{2}}$ $\frac{dr}{dV} = \frac{2}{225}$ Let’s go back to the chain rule we defined,
$\frac{dr}{dt} = \textcolor{#2192ff}{\frac{dr}{dV}} \times \textcolor{#0f0}{\frac{dV}{dt}}$ Substitute and evaluate $\frac{dr}{dt}$ ,
$\frac{dr}{dt} = \textcolor{#2192ff}{\frac{2}{225}} \times \textcolor{#0f0}{1.5}$ $\frac{dr}{dt} = \frac{1}{75}$ Therefore, the final answer is,
$\frac{dr}{dt} = \frac{1}{75} \textmd{ m per second}$ (b) Find the volume of the mound at the instant when the radius is increasing at $0.1$ m per second.

Let’s write out, mathematically, all the information we have been given,
$V = \frac{3}{2}\left(r - \frac{1}{2}\right)^{3} - 1\ \ \ \ \ \ \frac{dV}{dt} = 1.5\ \ \ \ \ \ \frac{dr}{dt} = 0.1\ \ \ \ \ \ r = 5.5$ Using the information above, let’s define a chain rule for $\frac{dr}{dt}$ ,
$\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt}$ Substitute and evaluate $\frac{dr}{dV}$ ,
$0.1 = \frac{dr}{dV} \times 1.5$ $\frac{dr}{dV} = \frac{0.1}{1.5}$ $\frac{dr}{dV} = \frac{1}{15}$ Remember that in part $(a)$ we found that,
$\frac{dr}{dV} = \frac{2}{9\left(r - \frac{1}{2}\right)^{2}}$ Let’s equate $\frac{dr}{dV}$ to $\frac{1}{15}$ ,
$\frac{1}{15} = \frac{2}{9\left(r - \frac{1}{2}\right)^{2}}$ Cross multiply and simplify,
$\frac{2(15)}{9} = \left(r - \frac{1}{2}\right)$ $\frac{10}{3} = \left(r - \frac{1}{2}\right)^{2}$ Take the square roots of both sides,
$\pm\sqrt{\frac{10}{3}} = \sqrt{\left(r - \frac{1}{2}\right)^{2}}$ $\pm\sqrt{\frac{10}{3}} = r - \frac{1}{2}$ Make $r$ the subject of the formula,
$r = \frac{1}{2} \pm \sqrt{\frac{10}{3}}$ $r = \frac{1}{2} + \sqrt{\frac{10}{3}}$ Note: We disregard $r = \frac{1}{2} - \sqrt{\frac{10}{3}}$ because it gives a negative radius. Radius is a measurement, therefore, it is a positive quantity.

We can now substitute the radius into the formula for volume, to find the volume of the mound,
$V = \frac{3}{2}\left(r - \frac{1}{2}\right)^{3} - 1$ $V = \frac{3}{2}\left(\frac{1}{2} + \sqrt{\frac{10}{3}} - \frac{1}{2}\right)^{3} - 1$ $V = \frac{-3 + 5\sqrt{30}}{3}$ $V = 8.13$ Therefore, the final answer is,
$V = 8.13 \textmd{ m}^{3}$

GCE_AS_Level_Differentiation__Rates_of_Change Download