Question

A rectangular wire frame has one of its dimensions moving at the rate of $0.5cm$/second. Its width is constant and equal to $4cm$. If at $t=0$ the length of the rectangle is $10cm$,
(a) What is the length at time $t$
(b) Write a formula for the area $A$ of the rectangle in terms of $t$
(c) Write a formula for the perimeter $P$ of the rectangle in terms of $t$.
(d) As $x$ increases, which one the perimeter or the area increases faster?

Answer 1

$\frac{dx}{dt}=0.5㎝/sec$

Now, length of rectangle,$x=10+\frac{dx}{dt}\times$(time)

a)Length at time $t=10+0.5t$

b)Area of rectangle=lxb

$=(10+0.5t)4$

Area$=40+2t$

c)Perimeter P of rectangle$=2(l+b)$

$=2(10+0.5t+4)$

$P=28+t$

d) Area$=l\times b=xb\Rightarrow 40+2t$

Perimeter$=2(l+b)=2(x+b)\Rightarrow 28+t$

Now, as x increases, it is increased due to increase in time

$\frac{dA}{dt}=2$ and $\frac{dP}{dt}=1$

Therefore, Area increases twice faster than Perimeter.