### Applications of DIfferentiation Part V

Connected Rates of Change

Example 1

A curve has the equation $y = (5x-2)(\sqrt{2x + 2})$

a) Find $\frac{dy}{dx}$ in the form of $\frac{ ax }{ \sqrt{2x + 2} }$

Using Product rule
\begin{aligned} \frac{dy}{dx} &= (5x-2)\frac{1}{2}( 2x + 2)^{-\frac{1}{2}}( 2 ) + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 }{ ( 2x + 2)^{\frac{1}{2} } } + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 + 2x + 2 }{ \sqrt{2x + 2} } \\ &= \frac{ 7x }{ \sqrt{2x + 2} } \\ \end{aligned}

b) Hence find the rate of change of x when x = 7 when y is change at a rate of 4 units per second

Given $x = 7 \qquad \frac{dy}{dt} = 4$

\begin{aligned} \frac{dy}{dt} &= \frac{dy}{dx} \times \frac{dx}{dt} \\ 4 &= \frac{ 7x }{ \sqrt{2x + 2} } \times \frac{dx}{dt} \\ 4 &= \frac{ 49 }{ \sqrt{14 + 2} } \times \frac{dx}{dt} \\ \frac{dx}{dt} &= \frac{16}{49} \\ \end{aligned}