Related rate expanding rectangle

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Homework Help Overview

The problem involves related rates concerning the area of a rectangle, where the length and width are both increasing at specified rates. The original poster seeks to determine how fast the area is increasing given the rates of change of length and width at specific dimensions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the product rule for differentiation to find the rate of change of the area. Some participants question the application of derivatives to the variables involved, particularly regarding the use of the product rule and the nature of the derivatives of L and W.

Discussion Status

Participants are actively discussing the application of differentiation rules, including the product rule and chain rule. There is a recognition of the need to clarify the relationship between the variables and their rates of change. Some guidance has been offered regarding the correct application of these rules, and participants are exploring their understanding of the concepts involved.

Contextual Notes

There is an ongoing discussion about the assumptions made regarding the differentiation of variables that change with time. Participants are also reflecting on their understanding of the mathematical principles involved, indicating a learning process in progress.

synergix
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Homework Statement


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?

Homework Equations



V=LW


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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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?
 
Dick said:
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)
 
synergix said:
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.
 
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?
 
d(x(t)^1)/dt=1*x(t)^0*dx/dt, chain rule again.
 
Could you explain quickly why you need to use the chain rule here?
 
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.
 
Last edited:
Dick said:
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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