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Related rate expanding rectangle 
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#1
Oct2509, 01:47 PM

P: 178

1. The problem statement, all variables and given/known data
The length of a rectangle is increasing at a rate of 8cm/s and its width is increasing at a rate of 3cm/s when the length is 20 cm and the width is 10cm, how fast is the area of the rectangle increasing? 2. Relevant equations V=LW 3. The attempt at a solution dL/dt=8 dW/dt=3 l=20 w= 10 dA/dt=? (V=LW)' dA/dt= dL/dt * dW/dt before I go any further is this correct both the L and the W become one when derived right? 


#2
Oct2509, 02:03 PM

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A(t)=L(t)W(t), right? Take a look at the 'product rule'. And why would the derivative of L and W be 1???



#3
Oct2509, 02:12 PM

P: 178

And you have t in parenthesis to make clear that L and W and A change with time and are therefore not constants right? so according to the product rule> dA/dt=L(dW/dt)+W(dW/dt) 


#4
Oct2509, 02:19 PM

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Related rate expanding rectangle



#5
Oct2509, 02:22 PM

P: 178

Ok I was just using the power rule (x^1 + a^1)'= 1*x^0 + 1*a^0= 1+1
am I assuming something here I shouldn't be? 


#6
Oct2509, 02:24 PM

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d(x(t)^1)/dt=1*x(t)^0*dx/dt, chain rule again.



#7
Oct2509, 02:29 PM

P: 178

Could you explain quickly why you need to use the chain rule here?



#8
Oct2509, 02:34 PM

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Because x(t) is a function of t. d(x^1)/dx is 1*x^0 (though you could still throw the chain rule in and write it as 1*x^0*dx/dx, but dx/dx=1). d(x^1)/dt=1*x^0*dx/dt. dx/dx is always 1. dx/dt is not.



#9
Oct2509, 02:37 PM

P: 178




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