A hypothetical square grows so that the length of its diagonals are increasing at a rate of 4 m/min. How fast is the area of the square increasing when the diagonals are 14 m each?

Question

Paisley Johnson · Accepted Answer

# (dA)/dt = (7sqrt(6))/6 # #m^2#/min

Let the length of one side be #l#, the diagonal be #z# and the area be #A# Then By Pythagoras, # z^2=l^2 + l^2 # # :. z^2=2l^2 # .......... [1] And, the Area is, #A=l^2#, Substituting this into [1] gives us; # z^2=2A^2 # .......... [2] We are told that #dz/dt=4# (constant), and we want to find #(dA)/dt# when #z=14#. By the chain rule, #(dA)/dt = (dA)/(dz).(dz)/(dt) # .......... [3] Differentiating [2] wrt z (implicitly) gives: # 2A^2 = z^2 # # :. 2z = 2(2A)(dA)/(dz) # # :. 2A(dA)/(dz) = z # # :. (dA)/(dz) = z/(2A) # Substituting this into [3] gives us: # (dA)/dt = (z/(2A)).(4) # # (dA)/dt = (2z)/A # .......... [4] So, When #z=14#, using [2] we have # 2A^2 = 14^2 # # 2A^2 = 196 # # A^2 = 98 # # A = sqrt(98 ) # Substituting #z=14# and # A = sqrt(98) # into [4] we have: # (dA)/dt = (2(14))/sqrt(98) # # (dA)/dt = (7sqrt(6))/6 #

Christopher Agresta · Answer

undefined To find the rate at which the area of the square is increasing when the diagonals are 14 meters each, we can use the fact that the diagonal of a square is related to its side length by the formula:

$$ d = s\sqrt{2} $$

Differentiating both sides with respect to time, we get:

$$ \frac{{dd}}{{dt}} = \frac{{ds}}{{dt}}\sqrt{2} $$

Given that $ \frac{{dd}}{{dt}} = 4 $ m/min, we can rearrange the formula to solve for $ \frac{{ds}}{{dt}} $:

$$ \frac{{ds}}{{dt}} = \frac{{4}}{{\sqrt{2}}} $$

Now, when the diagonals are 14 meters each, we can find the side length of the square using the formula:

$$ s = \frac{{d}}{{\sqrt{2}}} $$

$$ s = \frac{{14}}{{\sqrt{2}}} $$

$$ s = 7\sqrt{2} $$

Then, substituting $ s = 7\sqrt{2} $ and $ \frac{{ds}}{{dt}} = \frac{{4}}{{\sqrt{2}}} $ into the formula for the rate of change of the area of a square with respect to time, we get:

$$ \frac{{dA}}{{dt}} = 2s \frac{{ds}}{{dt}} $$

$$ \frac{{dA}}{{dt}} = 2(7\sqrt{2})\left(\frac{{4}}{{\sqrt{2}}}\right) $$

$$ \frac{{dA}}{{dt}} = 56 \text{ m}^2/\text{min} $$

So, when the diagonals are 14 meters each, the area of the square is increasing at a rate of 56 square meters per minute.