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} $

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