A rectangular plate is expanding. Its length x is increasing at the rate 1 cm/s and its width y is decreasing at the rate 0.5 cm/s. At the moment when $x=4$ and $y=3$, find the rate of change of its area, perimeter and diagonal.
Length of rectangular plate $=x \; cm$
Width of rectangular plate $=y \; cm$
Given: $x=4 \; cm; \;\;\; y=3 \; cm$ $\;\; \cdots$ (1)
Rate of increase of length $= \dfrac{dx}{dt} = 1 \; cm/s$ $\;\; \cdots$ (2)
Rate of decrease of width $=\dfrac{dy}{dt} = -0.5 \; cm/s$ $\;\; \cdots$ (3)
Area of the rectangular plate $=A=xy$ $\;\; \cdots$ (4)
Differentiating equation (4) w.r.t time t gives
$\begin{aligned}
\dfrac{dA}{dt} & = x \dfrac{dy}{dt} + y \dfrac{dx}{dt} \\
& = 4 \times \left(-0.5\right) + 3 \times 1 \;\; \left[\text{from equations (1), (2) and (3)}\right] \\
& = 1 \; cm^2 / s
\end{aligned}$
Perimeter of rectangular plate $=P=2\left(x+y\right)$ $\;\; \cdots$ (5)
Differentiating equation (5) w.r.t time t gives
$\begin{aligned}
\dfrac{dP}{dt} & = 2 \left(\dfrac{dx}{dt} + \dfrac{dy}{dt}\right) \\
& = 2 \times \left(1-0.5\right) \;\; \left[\text{from equtions (2) and (3)}\right] \\
& = 1 \; cm/s
\end{aligned}$
Diagonal of rectangular plate $= D = \sqrt{x^2 +y^2}$ $\;\; \cdots$ (6)
Differentiating equation (6) w.r.t time t gives
$\begin{aligned}
\dfrac{dD}{dt} & = \dfrac{1}{2\sqrt{x^2+y^2}} \times \left(2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}\right) \\
& = \dfrac{1}{\sqrt{4^2 + 3^2}} \times \left[4 \times 1 + 3 \times \left(-0.5\right)\right] \;\; \left[\text{from equations (1), (2) and (3)}\right] \\
& = 0.5 \; cm/s
\end{aligned}$
$\therefore$ $\dfrac{dA}{dt}=1 \; cm^2/s; \; \dfrac{dP}{dt}=1 cm/s; \; \dfrac{dD}{dt}=0.5 \; cm/s$