How fast is the area of a square increasing when the side is 6cm in length and growing at a rate of 0.3cm/s? Help!?

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Answer 1 · 2017-08-17T08:03:40

#3.6cm^2s^(-1)#

For this question let #A=#the area of the square & #" "x=#the length of the sides at time #t#

We are asked for how fast the area of the square is increasing,

i.e. the rate of increase of the area#=(dA)/(dt)#

we want #(dA)/(dt)# and we know #(dx)/(dt)=0.3cm s^(-1)#

we can connect these by the chain rule

#color(blue)((dA)/(dt)=(dx)/(dt)xx(dA)/(dx))#

we are missing #(dA)/(dx)#

for a square #A=x^2=>(dx)/(dt)=2x#

#:.(dA)/(dt)=0.3xx2x=0.6x#

so at #x=6cm#

#:.[(dA)/(dt)]_(x=6)=0.6 xx 6=3.6cm^2s^(-1)#