Modelling with differentiation

​​In a nutshell

Differentiating an equation finds the gradient of the equation's curve. This represents the rate of change of one variable with respect to the other. Using differentiation in real-life scenarios can show how variables change, for example, over time.

A look at notation

The "$$\text d$$" in $$\frac{\text dy}{\text dx}$$ can be thought of as meaning "a small change in". This means that, $$\dfrac{\text dy}{\text dx}=\text{a small change in }y \div\text{a small change in }x$$ ​which is the rate of change of $$y$$ with respect to $$x$$. 

Now suppose $$y$$ and $$x$$ were replaced with variables that represented some real-life quantities. The derivative would now represent the rate of change of the $$y$$-replacement with respect to the $$x$$-replacement.

Example 1

The equation for the area of a circle, $$A$$, is given in terms of its radius, $$r$$:

$$A=\pi r^2$$​​

Find an equation for the rate of change of the area with respect to the radius.

This is given by the first derivative of $$A$$ with respect to $$r$$:

$$\underline{\dfrac{\text dA}{\text dr}=2\pi r}$$​​

​​Example 2

A cuboid is being filled with water from a tap. The volume in $$\text{cm}^3$$​ of water in the cuboid is given by $$V=5t^2$$ where $$t$$ is the time in seconds since the tap was turned on. It takes ten seconds to fill the cuboid. Find the rate of change of the volume with respect to time after three seconds.

The rate of change is found by differentiating $$V$$ with respect to $$t$$:

$$\dfrac{\text dV}{\text dt}=10t$$​​

Find the rate of change when $$t=3$$:

  $$\dfrac{\text dV}{\text dt}=10(3)=\underline{30 \text{ cm}^3\text s^{-1}}$$. 

Note: The unit has come from the fact that you have a volume unit ($$\text{cm}^3$$) per time unit ($$\text s$$).

Optimisation

Models of real-life scenarios can help you maximise or minimise different quantities. Recall that maxima and minima have zero gradients. They are classified using the second derivative. 

Example 3

A cuboid has two perpendicular sides of length $$x$$ metres and the third perpendicular side of length $$y$$ metres. You have that the total surface area of the cuboid is $$54\text{ m}^2$$. Given that the value $$x$$ can vary, find the maximum volume of the cuboid. Justify why it is the maximum. 

Start with the equation for the volume of a cuboid: 

​$$V_{\text{cuboid}}=\text{height}\times\text{base}\times\text{length}$$​​

Hence for this example you have:

$$V=x^2y$$​​

Now the surface area:

$$A_{\text{cuboid}}=2(\text{base}\times \text{height})+2(\text{length}\times \text{height})+2(\text{base}\times \text{length})$$​​

Hence for this example you have:

$$54=2x^2+4xy\\\space\\27=x^2+2xy$$​​

Given that $$x$$ can vary, and you are looking for how volume changes, use these equations to find an equation for $$V$$​ in terms of $$x$$:

$$27=x^2+2xy\\\space\\y=\dfrac{27-x^2}{x}=\frac{27}x-x$$​​

Insert this into the equation for volume:

$$V=x^2y\\\space\\V=x^2\bigg(\dfrac{27}x-x\bigg)=27x-x^3$$​​

Differentiate $$V$$ with respect to $$x$$:

$$\dfrac{\text dV}{\text dx}=27-3x^2$$​​

Set this equal to zero and solve for $$x$$:

$$0=27-3x^2\\\space\\3x^2=27\\\space\\x^2=9\\\space\\x=\pm3$$​​

Since $$x$$ is a measurement of distance, it follows that it must be positive, so $$x=3$$.

To confirm that this represents a maximum $$V$$, find the second derivative and insert $$x=3$$ to get a maximum:

$$\dfrac{\text d^2V}{\text dx^2}=-6x\\\space\\\dfrac{\text d^2V}{\text dx^2}=-6(3)=-18<0$$​​

Thus when $$x=3$$, $$V$$ is at a maximum. Insert this $$x$$ into the equation for $$V$$:

$$V=30x-x^3=30(3)-(3)^3=63$$​​

The maximum volume is $$\underline{63 \text{ m}^3}$$.