Rate of decrease: always negative?

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Rana1997
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Consider the following problem: Water starts running out of a small hole in the bottom of a bowl so that the depth of the water in the bowl at time t is x. The rate at which the volume of water is decreasing is proportional to x.

Why did the book represent the relationship as: -dV/dt ∝ x. knowing that as x decreases the volume decreases as well. Thus putting a negative sign will make the decrease in x cause an increase in V.
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ghostwalker
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(Original post by Rana1997)
Consider the following problem: Water starts running out of a small hole in the bottom of a bowl so that the depth of the water in the bowl at time t is x. The rate at which the volume of water is decreasing is proportional to x.

Why did the book represent the relationship as: -dV/dt ∝ x. knowing that as x decreases the volume decreases as well.

dV/dt is the rate of increase in V with respect to t - it could be positive or negative, and -dV/dt is the rate of decrease, which again could be positive or negative. In this example dV/dt will be negative and -dV/dt positive.

But, it doesn't overly matter here, since this is "proportional to", rather than "equal to". To turn it into an equation you'd need to add the coefficient of proportionality (-dV/dt = kx). With the minus sign that constant will be positive; without it, it would be negative.

Thus putting a negative sign will make the decrease in x cause an increase in V.
That doesn't follow.

As x gets smaller towards zero, -dV/dt will tend to zero, i.e. the rate at which the volume changes will tend to zero.

If you work through and solve the equation, you will find that there is no problem either way, except without the -ve sign the coefficient of proportionality will be negative, and although that's not wrong per se., it's usual, if not an actual requirment, that it's +ve.
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