Imagine you have a cylindrical container which is 5 metres in diameter and 10 metres tall. If liquid is pouring in to the container at a rate of 1000 litres per second what is the rate of change of depth?
The volume of liquid is given by $V = \pi r^2 y$ where $y$ is the depth of liquid.
| $V$ | $=$ | $\pi r^2 y$ |
| Differentiating both sides with respect to time $t$ we get | ||
| $\dfrac{dV}{dt}$ | $=$ | $\pi r^2 \dfrac{dy}{dt}$ |
| So the rate of change of depth is given by | ||
| $\dfrac{dy}{dt}$ | $=$ | $\dfrac{1}{\pi r^2} \dfrac{dV}{dt}$ |
| 1000 litres = 1 cubic metre so | ||
| $\dfrac{dy}{dt}$ | $=$ | $\dfrac{1}{\pi r^2}$ |
| $=$ | $0.051$ m/s or $51$ mm/s | |
The radius of a cylinder is constant, it is not dependent on the depth of liquid. What would happen if we had a conical funnel instead of a cylinder?
Using the dimensions from above, imagine we have a conical vessel with a maximum diameter of 5 metres and a height of 10 metres. If the small end is downwards and liquid is pouring into the vessel at 1000 litres per second what is the rate of change of depth when the depth is 2 metres?
Unlike the case of the cylinder the radius of the top of the liquid varies with the depth of the liquid. This means we have to replace $r$ with $x$ where $x=f(y)$.
| $V$ | $=$ | $\dfrac{\pi x^2 y}{3}$ |
| From similar triangles we can see $x/y = r/h$ | ||
| so $V$ | $=$ | $\dfrac{\pi r^2 y^3}{3h^2}$ |
| Differentiating both sides with respect to time $t$ we get | ||
| $\dfrac{dV}{dt}$ | $=$ | $\dfrac{\pi r^2y^2}{h^2} \dfrac{dy}{dt}$ |
| So the rate of change of depth is given by | ||
| $\dfrac{dy}{dt}$ | $=$ | $\dfrac{h^2}{\pi r^2 y^2} \dfrac{dV}{dt}$ |
| 1000 litres = 1 cubic metre so | ||
| $\dfrac{dy}{dt}$ | $=$ | $\dfrac{10^2}{\pi 2.5^2 2^2}$ |
| $=$ | $1.27$ m/s | |
The volume of liquid in a spherical container is given by $\pi(r y^2 - y^3/3)$ where $r$ is the radius and $y$ is the depth of liquid.
| $V$ | $=$ | $\pi(r y^2 - y^3/3)$ |
| Differentiating both sides with respect to time $t$ we get | ||
| $\dfrac{dV}{dt}$ | $=$ | $\pi(2ry-y^2) \dfrac{dy}{dt}$ |
| so just like the case of the cone the rate of change of depth is dependent on the depth of liquid. The rate of change of depth is given by: | ||
| $\dfrac{dy}{dt}$ | $=$ | $\dfrac{1}{\pi(2ry-y^2)} \dfrac{dV}{dt}$ |
In the following table the left hand column contains functions of $x$ ($y=f(x)$). The right hand column contains the differentials of the functions with respect to $x$ ($dy/dx=f'(x)$).
| $y=f(x)$ | $dy/dx = f'(x)$ |
|---|---|
| $x^n$ | $nx^{n-1}$ |
| $sin(ax)$ | $acos(ax)$ |
| $cos(ax)$ | $-asin(ax)$ |
| $tan(ax)$ | $asec^2(ax)$ |
| $e^{ax}$ | $ae^{ax}$ |
| $ln(ax)$ | $1/x$ |