If there is a relationship between two or more variables, then there will also be a relationship between how these variables change. You may need to find how fast one variable changes in relation to another variable that is changing. This is called the rate of change.
We often study the relationship between two or more variables, like:
the area and radius of a circle (\(A=\pi r^{2}\))
the length of a side and volume of a cube (\(V=l^{3}\))
the days since first case and number of people with an infectious disease.
Where there is a relationship between variables, there is also be a relationship between the rates at which the variables change.
Finding the rate of change
We can use differentiation to find the function that defines the rate of change between variables. This involves differentiating with respect to time \(t\).
For the example of area and radius of a circle:
\[\begin{align*} A & = \pi r^{2}\\
\frac{dA}{dr} & = 2\pi r
\end{align*}\]
To find the rate the area changes with respect to time \(t\), we would use:
\[\begin{align*} \frac{dA}{dt} & = \frac{dA}{dr}\times\frac{dr}{dt}\\
& = 2\pi r\times\frac{dr}{dt}
\end{align*}\]
For the example of length of a side and volume of a cube:
\[\begin{align*} V & = l^{3}\\
\frac{dV}{dl} & = 3l^{2}
\end{align*}\]
To find the rate the volume changes with respect to time \(t\), we would use
\[\begin{align*} \frac{dV}{dt} & = \frac{dV}{dl}\times\frac{dl}{dt}\\
& = 3l^{2}\times\frac{dl}{dt}
\end{align*}\]
Example 1 – finding the rate of change
A balloon has a small hole and its volume \(V\) in cubic centimeters after \(t\) seconds is \(V=66-10t-0.01t^{2}\) for \(t>0\). Find the rate of change of volume after \(10\) seconds.
First, we find the derivative.
\[\begin{align*} V & = 66-10t-0.01t^{2}\\
\frac{dV}{dt} & = -10-0.02t
\end{align*}\]
Now, we can substitute \(t=10\) into the derivative to find the rate of change of volume.
\[\begin{align*} \frac{dV}{dt} & = -10-0.02(10)\\
& = -10.2\textrm{ cm}^{3}/{\textrm{s}}
\end{align*}\]
The volume of the balloon is decreasing at a rate of \(10.2\textrm{ cm}^{3}/{\textrm{s}}\).
The pressure \(P\) of a given mass of gas kept at constant temperature, and its volume \(V\) are connected by the equation \(PV=500\). Find \(\frac{dP}{dV}\) when \(V=20\).
Here, we need to rearrange the equation to make \(P\) the subject.
\[\begin{align*} PV & = 500\\
P & = \frac{500}{V}
\end{align*}\]
We will also need this in index form.
\[\frac{500}{V} = 500V^{-1}\]
Again, find the derivative.
\[\frac{dP}{dV} = -500V^{-2}\]
Now, we can substitute \(V=20\) into the derivative to find the rate of change of pressure.
\[\begin{align*} \frac{dV}{dt} & = -500(20)^{-2}\\
& = -1.25
\end{align*}\]
The pressure is decreasing at a rate of \(1.25\).
A ladder \(4.5\textrm{ m}\) long is sliding down a vertical wall with the top of the ladder descending at a rate of \(2.3\textrm{ m/s}\). How fast is the bottom of the ladder moving along the ground when the bottom is \(3\textrm{ m}\) from the wall?
Here, it helps to draw a diagram. Let \(y\) be the height the ladder reaches up the wall and \(x\) be the distance of the bottom of the ladder from the wall.
The rate at which the ladder slides down the wall is defined by \(\dfrac{dy}{dt}=-2.3\textrm{m/s}\). We need to find the rate at which the ladder moves along the ground, defined by \(\dfrac{dx}{dt}\).
We can use Pythagoras' theorem to define the relationship between \(x\) and \(y\), then use Implicit differentiation to find the derivative.
\[\begin{align*} x^{2}+y^{2} & = 4.5^{2}\\
\left(2x\cdot\frac{dx}{dt}\right)+\left(2y\cdot\frac{dy}{dt}\right) & = 0
\end{align*}\]
Let's find the \(y\) when \(x=3\). This is the distance the ladder is up the wall when the bottom is \(3\textrm{ m}\) from the wall.
\[\begin{align*} x^{2}+y^{2} & = 4.5^{2}\\
(3)^{2}+y^{2} & = 20.25\\
y^{2} & = 11.25\\
y & = \pm3.35
\end{align*}\]
\(y\) can only be \(3.35\) since the distance cannot be a negative value.
We know \(\dfrac{dy}{dt}=-2.3\) and now, we have all we need to find \(\dfrac{dx}{dt}\).
\[\begin{align*} \left(2\times3\times\frac{dx}{dt}\right)+\left(2\times3.35\times(-2.3)\right) & = 0\\
6\times\frac{dx}{dt}-15.41 & = 0\\
\frac{dx}{dt} & = 2.57
\end{align*}\]
The ladder is moving along the ground at a speed of \(2.57\textrm{ m/s}\).
Exercise – finding the rate of change
The radius of a spherical balloon is increasing at a rate of \(3\textrm{ cm/min}\). At what rate is the volume increasing when the radius is \(5\textrm{ cm}\)?
If the displacement of an object from a starting point is given by \(s(t)=\sin(t)-2\cos(t)\), find the velocity when \(t=1\). Hint: \(v(t)=s'(t)=\dfrac{ds}{dt}\).
The function \(n(t)=200t-100\sqrt{t}\) describes the spread of a virus where \(t\) is the number of days since the initial infection and \(n\) is the number of people infected. Find the rate at which \(n\) is increasing at the instant when \(t=4\).
If \(y=\left(x-\dfrac{1}{x}\right)^{2}\) find \(\dfrac{dx}{dt}\) when \(x=2\), given \(\dfrac{dy}{dt}=1\).
A hollow right circular cone is held point downwards under a tap leaking at the rate of \(2\textrm{ cm}^{3}/\textrm{s}\). Find the rate of rise of water level when the depth is \(6\textrm{ cm}\) given that the height of the cone is \(18\textrm{ cm}\) and its radius is \(12\textrm{ cm}\). Hint: Use the properties of similar triangles to find a relationship between radius and height.
