Calculating the Rate of Change of a Square's Area

[Cbray]

Homework Statement

Find the rate of change of the area of a square whose side is 8 cm long, if the side length is increasing at 2cm/min

Homework Equations

A(t) = xy
x(0)=8
y(0)=8
dx/dt=2
dy/dt=2

The Attempt at a Solution

dA/dt = dx/dt * x * dy/dt * y
Using the product rule...
dA/dt = dx/dt * x * y + dy/dt * y * x
What happens now? Does the dx/dt * x = 1?

or is it..
dA/dt = dx/dt * d/dx * x * dy/dt * d/dy * y
Using the product rule..
dA/dt = dx/dt * d/dx * x * y + dy/dt * d/dy * y * x
Therefore the d/dx = 1, therefore..
dA/dt = dx/dt * y + dy/dt * x
dA/dt = 2*8 + 2*8
dA/dt = 32 cm^2/min ?? If it is correct is what I'm thinking correct (ie my working?)Help, thanks!

 

[NewtonianAlch]

I got 32t + 4t^2 - I did the d(xy)/dt = dx/dy*y + dy/dt*x thing like you did, which gave me 32t.

I wrote out the values for the next few minutes and tacked on the 4t^2. I wouldn't know how to do it the proper way, been awhile.

So for the first minute the increase is 32 + 4 = 36, which is for a square of 10 by 10 (36 + initial 64)

Then for the next minute it would be 64 + 16 = 80, (80 + initial 64 = 144; 12 by 12)

Edit: Also, it's area so it's cm^2/min - so yea it would be 32cm^2/min

 

[Rokas_P]

Huh? Wouldn't it work a simpler way:

$A(t)=\left(8+2t\right)^2$
$A'(t)=2\left(8+2t\right)\cdot2$
$A'(t)=32+8t$?

here t is in minutes.