Question

A gardener is digging a plot of land. As he gets tired, he works slowly. After ‘t’ minutes he is digging at a rate of $\dfrac{2}{\sqrt{t}}$ square meters per minute. How long will it take for him to dig an area of 40 square meters?
(a) 10 minutes
(b) 20 minutes
(c) 100 minutes
(d) 30 minutes

Answer 1

Hint: Let the area which had been dug by the gardener till time t be A sq meters. So, according to the question the rate of digging is given, i.e., $\dfrac{dA}{dt}$ is given. Form the differential equation and solve it to reach the answer. The limits of integration should be from 0 to t minutes on time and 0 to A for area dug, where A is the area dug and t is the time corresponding to A.

Complete step by step solution:
Let us start the solution to the above question by letting the area which had been dug by the gardener till time t be A sq meters.
As it is given that after ‘t’ minutes he is digging at a rate of $\dfrac{2}{\sqrt{t}}$ square meters per minute. So, if we form the differential equation of it, we get
$\dfrac{dA}{dt}=\dfrac{2}{\sqrt{t}}$
$\Rightarrow dA=\dfrac{2}{\sqrt{t}}dt$
Now we will integrate both sides of the equation. The limits of the integration on time should be 0 to t while the limits of area dug is 0 to A, where A is the area dug and t is the time corresponding to A.
$\int\limits_{0}^{A}{dA}=\int\limits_{0}^{t}{\dfrac{2}{\sqrt{t}}dt}$
Now we know that $\int{dA=A}$ and $\int{\dfrac{1}{\sqrt{t}}}dt=2\sqrt{t}$ . So, if we use this in our expression, we get
$\left( A \right)_{0}^{A}=\left( 2\times 2\sqrt{t} \right)_{0}^{t}$
$\Rightarrow A=4\sqrt{t}$
Now to get the time by which 40 sq meters of area is dug, we will put A to be 40. On doing so, we get
$40=4\sqrt{t}$
$\Rightarrow \sqrt{t}=10$
We will square both the sides of the equation. On doing so, we get
$t=100\text{ minutes}$
Hence, the answer to the above question is option (c).

Note: Always remember that when you differentiate the equation with respect to time, the variables are taken at any time t, i.e., the variables are also changing continuously with. It is also important that you choose the limits of the integration for solving the differential equation correctly, for differentials with 2 variables you are free to choose the limits for one of the variables, but once you choose the limits of one of the variables the limits of the other must be chosen accordingly.