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?

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Answer 1 · 2016-10-29T20:32:20

# (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 #