### 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} $

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} $

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