Question

The area of an expanding rectangle is increasing at the rate of \[48c{{m}^{2}}/\sec \]. The length of the rectangle is always equal to the square of its breadth. At what rate is the length increasing at the instant when the breadth is 4.5cm?

Answer 1

Hint: Apply the formula: - Area = l \[\times \] b, where l and b are the length and breadth of the rectangle respectively, to find the area of the rectangle. Differentiate both sides with respect to time and substitute \[\dfrac{dA}{dt}=48\], where A is the area. Now, form a relation using the information given in the question between l and b and substitute its value in the differential equation to get the answer.

Complete step-by-step solution
Here, we have been given that the area of a rectangle is increasing at the rate of \[48c{{m}^{2}}/\sec \]. We have to find the rate of increase in length.
Now, let us assume the length and breadth of the rectangle as ‘l’ and ‘b’ respectively. We know that area of a rectangle is given as: - A = l \[\times \] b, where ‘A’ is the area, therefore we have,
\[\Rightarrow A=l~\times b\]
Differentiating both sides with respect to time (t), we get,
\[\Rightarrow \dfrac{dA}{dt}=\dfrac{d\left[ l\times b \right]}{dt}\]
Using product rule of differentiation given as: - \[\dfrac{d\left( u.v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow \dfrac{dA}{dt}=l\dfrac{db}{dt}+b\dfrac{dl}{dt}\]
Substituting \[\dfrac{dA}{dt}=48\], we get,
\[\Rightarrow b\dfrac{dl}{dt}+l\dfrac{db}{dt}=48\] - (1)
Now, it is given that the length of the rectangle is always equal to the square of its breadth. Therefore, we have,
\[\Rightarrow l={{b}^{2}}\]
Differentiating both sides with respect to time (t), we get,
\[\Rightarrow \dfrac{dl}{dt}=2b\dfrac{db}{dt}\]
\[\Rightarrow \dfrac{db}{dt}=\dfrac{1}{2b}\dfrac{dl}{dt}\]
So, substituting the value of l and \[\dfrac{db}{dt}\] in equation (1), we get,
\[\begin{align}
  & \Rightarrow b\dfrac{dl}{dt}+{{b}^{2}}\times \dfrac{1}{2b}\dfrac{dl}{dt}=48 \\
 & \Rightarrow \dfrac{3b}{2}\left( \dfrac{dl}{dt} \right)=48 \\
 & \Rightarrow \dfrac{dl}{dt}=\dfrac{32}{b} \\
\end{align}\]
Here, we have to find the rate of increase in length when the breadth of the rectangle is 4.5cm. So, substituting b = 4.5 in the above relation, we get,
\[\begin{align}
  & \Rightarrow \dfrac{dl}{dt}=\dfrac{32}{4.5} \\
 & \Rightarrow \dfrac{dl}{dt}=\dfrac{32\times 10}{45} \\
\end{align}\]
\[\Rightarrow \dfrac{dl}{dt}=\dfrac{64}{9}\]cm/sec
Hence, the length of the rectangle is increasing at the rate of \[\dfrac{64}{9}\] cm/sec.

Note: One may note that we have substituted the value of the relation \[l={{b}^{2}}\] after forming the differential equation (1). You can also substitute it at the initial step before forming the differential equation and then differentiate. It will give the same answer. Note that we do not have to substitute b = 4.5cm in the relation \[A=l\times b\] because we have to find \[\dfrac{dl}{dt}\] at b = 4.5cm. So first, we have to differentiate the area relation. Remember the product rule of differentiation to solve the question.