### Applications of DIfferentiation Part IV

Connected rates of change

Example 1
Liquid is poured into a bucket at a rate of $20cm^{-3}s^{-1}$ The volume of the liquid in the bucket $V cm^3$ when the depth of the liquid is $x$ is given by $V = 0.1x^3 + x^2 + 100x$

a) Find the rate of increase in the depth of liquid when $x = 10$

b) Find the depth of liquid when the rate of increase in depth is $0. 08 cm s^{-1}$

$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$

a)
Given $\frac{dV}{dt} = 20$

To find $\frac{dV}{dx}$ differentiate V with respect to x

$\frac{dV}{dx} = 0.3x^2 + 2x + 100$

When x = 10, $\frac{dV}{dx} = 30 + 20 + 100 = 150$

$\frac{dx}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dx}} = \frac{20}{150} = \frac{2}{15}$

The rate of increase in the depth of liquid is $\frac{2}{15} \:cm\:s^{-1}$

b)

\begin{aligned} \frac{dV}{dt} &= \frac{dV}{dx} \times \frac{dx}{dt} \\ 10 &= \frac{dV}{dx} \times 0.08 \\ \frac{dV}{dx} &= 125 \\ 0.3x^2 + 2x + 100 &= 125 \\ 0.3x^2 + 2x - 25 &=0 \\ x &= 6.38 \\ \end{aligned}

The depth of liquid is 6.38 cm when the rate of increase in depth is $0. 08 cm s^{-1}$