Related Rates with a Rectangle? (Calculus)
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Related Rates with a Rectangle? (Calculus)

[From: ] [author: ] [Date: 13-01-23] [Hit: ]
Thank you so much!dw/dt = -0.The question establishes that the area is constant, so dA/dt = 0.0 = 2*w- 0.0.......
Hi! I have a sort of conceptual/hypothetical question about related rates. I don't have exact numbers for a problem but if someone could explain how to do this that'd be AMAZING :D

Let's say you have a rectangle that has a constant area (we can say 100 square units). Its width is decreasing at a rate of let's say .25 and the length is increasing by 2.
How/would you be able to find the measurement of the width?


Thank you so much! :)

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lw = 100

dw/dt = -0.25
dl/dt = 2

Start by finding the derivative of the area:

A = lw
dA/dt = l*dw/dt + w*dl/dt (using the product rule)

The question establishes that the area is constant, so dA/dt = 0.
0 = l*dw/dt + w*dl/dt
0 = 2*w - 0.25*l
0.25*l = 2*w

Now eliminate one of the variables, lets say l, because we're trying to solve for w:
l = A/w
l = 100/w

0.25*100/w = 2*w
2*w^2 = 25
w^2 = 25/2
w = 5/sqrt(2) = (5/2)*sqrt(2)

w = 3.536 units
l = 28.28 units

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Differential equations.
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