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Related rate expanding rectangle

by synergix
Tags: expanding, rate, rectangle
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synergix
#1
Oct25-09, 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?
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Dick
#2
Oct25-09, 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???
synergix
#3
Oct25-09, 02:12 PM
P: 178
Quote Quote by Dick View Post
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???
I have really just not been using my brain lately. They would be one if they were being added which they're not. (x + a)'= 2 does it not? if they are both variables.

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)

Dick
#4
Oct25-09, 02:19 PM
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Related rate expanding rectangle

Quote Quote by synergix View Post
I have really just not been using my brain lately. They would be one if they were being added which they're not. (x + a)'= 2 does it not? if they are both variables.

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)
Ok, you've got the product rule. But (x+a)'=x'+a' if by ' you mean d/dt. Whether that's 2 or not depends what x and a are. dx/dx=1 but dx/dt doesn't necessarily equal 1.
synergix
#5
Oct25-09, 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?
Dick
#6
Oct25-09, 02:24 PM
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d(x(t)^1)/dt=1*x(t)^0*dx/dt, chain rule again.
synergix
#7
Oct25-09, 02:29 PM
P: 178
Could you explain quickly why you need to use the chain rule here?
Dick
#8
Oct25-09, 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.
synergix
#9
Oct25-09, 02:37 PM
P: 178
Quote Quote by Dick View Post
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*dx/dt. dx/dx is always 1. dx/dt is not.
Ok thank you its starting to come together.


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