This is (b) part of the multi-part question q7

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- If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
- $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1:

The rate of change in length is $\large\frac{dx}{dt}$$=-5cm/min$.

Negative sign indicates that the length is decreasing.

The rate of change in width is $\large\frac{dy}{dt}=$$4cm/min$

Also it is given that the length and width are $x=8cm$ and $y=6cm$ respectively.

Area of the triangle is $xy$

$A=xy$

Step 2:

Differentiate on both sides w.r.t $t$

(Apply product rule)

$\large\frac{d}{dt}$$(uv)=v\large\frac{du}{dt}+$$u.\large\frac{dv}{dt}$

$\large\frac{dA}{dt}=$$x.\large\frac{dy}{dt}+$$y.\large\frac{dx}{dt}$

Step 3:

Substituting the values for $x,y,\large\frac{dy}{dt}$ and $\large\frac{dx}{dt}$ is

$\large\frac{dA}{dt}=$$8\times (4)+4\times (-5)$

$\quad\;\;\;=32-30$

$\quad\;\;\;=2cm^2/min$

Hence the area of the rectangle increases at the rate of $2cm^2/min$

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