# Varying rate changes

1. Nov 20, 2005

### skivail

the length of a rectangle is decreasing at 5 cm/sec and the width is increasing at 3cm/sec. naive thinking would lead one to conclude that the area of the rectangle is decreasing becasue the length is decreasing faster than the width increases. is this conlcusion true of false at a time when the rectangle is 100 cm long and 50 cm wide? justify your answer.

2. Nov 20, 2005

### sanitykey

If you take x to be the time in seconds then you can work out two equations which tell you the height and width of the rectangle at any time. If you then multiply both of these equations together you can find the area of the rectangle at any time. You should end up with a quadratic that you can differentiate to find the rate of change of the area of the rectangle. You should then be able to see whether the area of the rectangle is decreasing or increasing. There would be a point where the length would start to increase again but your earlier equation for calculating the length at any time would show a negative value, which would lead to a negative area. You can't have negative areas...i think...so the way around that would be to say that you're only going to take the modulus of your area function, this means that you'd also take the modulus of your differentiated function i think.

Anyway hope that helps...
(please correct me if i'm wrong)

3. Nov 20, 2005

### Fermat

You could do it that way, yes, but it's a bit fiddly.

The usual way of doing these types of problems, involving rates of change, is like this ...

A = L*W

where both L and W are functions of time. L = L(t) and W = W(t).

Using the chain rule,

dA/dt = L.dW/dt + W.dL/dt
dA/dt = 3L - 5H (using dW/dt = 3 cm/s and dL/dt = - 5 cm/s)

at L = 100 and W = 50,

dA/dt = 3*100 - 5*50 = 300 - 250
dA/dt = 50 cm²/s
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which is positive, so the area is actually increasing!

Last edited: Nov 20, 2005