Question

The diagonal of a square is changing at a rate of $0.5 \mathrm{~cm} / \mathrm{sec}$. Then the rate of change of area, when the area is $400 \mathrm{~cm}^{2}$, is equal to
1 . $20\sqrt{2}\dfrac{c{{m}^{2}}}{\sec }$
2 . $10\sqrt{2}\dfrac{c{{m}^{2}}}{\sec }$
3 . $\dfrac{1}{10\sqrt{2}}\dfrac{c{{m}^{2}}}{\sec }$
4. $[10\sqrt{2}]\dfrac{c{{m}^{2}}}{\sec }$

Answer 1

Hint: In this question, we have given a changing value of the diagonal of a square. So first we use the Pythagoras theorem for determining the value of the side of the square. Further, equating the given value of area with the formula of area of the square to determine the value of diagonal. Finally, differentiate the formula of the area by doing the needful substitution to get the final result.

Formula Used:
$h^{2}=b^{2}+p^{2}$ where b is the base, p is the perpendicular and h is the hypotenuse.
$A=a^{2}$

Complete Step-by-step Solution:
Let the diagonal of the square be ‘D’ and the side of the square be ‘a’
Then the rate of change in the diagonal is
$\mathrm{dD} / \mathrm{dt}=0.5 \mathrm{~cm} / \mathrm{sec}$
A diagonal of a square divides it into two right-angle triangles where the diagonal acts as a hypotenuse of a triangle. By, applying Pythagoras' theorem, we can write
$D^{2}=a^{2}+a^{2}$ (all sides of the square are the same. So both base and perpendicular will be a.)
$\Rightarrow~D=\sqrt{2a^{2}}$
$\Rightarrow~D=a\sqrt{2}$
$\Rightarrow~a=\frac{D}{\sqrt{2}}$
Given that the area is $400 \mathrm{~cm}^{2}$. Also, we know the area of a square is $A=a^{2}$.
Therefore, $a^{2}=400$
$\Rightarrow~\lgroup~\frac{D}{\sqrt{2}}\rgroup^{2}=400$
$\Rightarrow~D^{2}=800$
$\Rightarrow~D=20\sqrt{2}$
The rate of change in area is
$\frac{\text{d}A}{\text{d}t}=\frac{\text{d}\frac{D^{2}}{2}}{\text{d}t}$
$\Rightarrow~\mathrm{d} A / \mathrm{dt}=\mathrm{D}(\mathrm{dD} / \mathrm{dt})$
We have $\mathrm{dD} / \mathrm{dt}=0.5 \mathrm{~cm} / \mathrm{sec}$
Substitute the values
$\mathrm{d} A / \mathrm{dt}=20\sqrt{2}\times0.5$
$\mathrm{d} A / \mathrm{dt}=10\sqrt{2} cm^{2}/sec$
The rate of change in area is $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$
So the correct answer is option B.

Note: Remember that when the question is given in the form of rate of change of area or velocity, we have to differentiate the given equation instead of putting the formula directly. Students made the mistake of directly putting the formula of area and velocity. First, we differentiate it then we put the values to find our answer.