Question

A hypothetical square grows so that the length of it's diagonals are increasing at a rate of 8 m/min. How fast is the area of the square increasing when the diagonals are 4m each?

Answer 1

Answer:

32$m^{2}/min$

Step-by-step explanation:

We let the length of the square be l, the diagonal be z and the area be A.

Then by Pythagoras theorem;

$l^{2}+l^{2}=z^{2}\\2l^{2}=z^{2}$

The are of a square of length l is given as;

$A=l^{2}$

Combining these two equations yields;

$z^{2}=2A$

$A=\frac{z^{2} }{2}$

Next, we have been given;

$\frac{dz}{dt}=8$

We are required to find;

$\frac{dA}{dt}$ when z = 4

By chain rule;

$\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt}$

Differentiating $A=\frac{z^{2} }{2}$ with respect to z yields;

$\frac{dA}{dz}=z$

Therefore,

$\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt}$ = 8z

When z is 4m the area of the square will be increasing at a rate of;

8m/min * 4m = 32$m^{2}/min$