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$$(5/1000)P(t+h)h < P(t) - P(t+h) < (5/1000)P(t)h$$

Where

  • $P(t)$ stands for the # of gallons of pollutants in a cistern at time $t$ minutes after a spill.

  • 5 is the rate the cistern fills with gallons of clean water per minute

  • 1000 is the size in gallons of the cistern

I understand that since $P(t)$ is the initial amount of pollutants in the cistern, $P(t+h)$ is the amount of pollutants at a later time and therefore would be less. $P(t)$ - $P(t+h)$ is the actual amount of pollutants at a given time after the spill.

One thing I don't understand is why $h$ is being multiplied at the left and right equations. Also, why it is beneficial to set up an equality like this.

How does one get the derivative $P'(t) = -(5/1000)P(t)$?

P.S.: (This is my first exposure to differential equations/integrals) I may have trouble understanding terminology. Please explain gently so I can follow THANKS!

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3 Answers

up vote 1 down vote accepted

Divide both inequalities by $h$. Then you get $$(5/1000)P(t+h) < \frac{P(t) - P(t+h)}{h} < (5/1000)P(t)$$ This tells you the rate of variation of $P(t)$.

About the derivative, what is the definition of a derivative? Apply it to the function $P(t)$. The above inequalities become very useful! Hint: consider what happens when $h$ is really small, and then when $h$ tends to 0.

Is this a homework question? If so, please tag it 'homework'.

Edit:

Using http://en.wikipedia.org/wiki/Derivative: formally, the derivative of the function $f$ at $a$ is the limit

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$ So replace $f$ by $P$ and $a$ by $t$, and you'll get what you want.

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Hi, Derivative from my understanding is the slope of a function, instant growth rate, that tells whether or not the function is increasing or decreasing. Ok, so if h tends to 0...than P(t+h) is near equal to P(t)...Hence why (5/1000)P(t) is the derivative? –  koloa Feb 21 '13 at 22:35
    
See my above edit: take the definition of the derivative & apply. This is basic differential calculus... or even introduction to calculus. Yes, the derivative is the slope and there are a few ways of finding a slope. The above is directly from the definition of a derivative: the slope (at a point) is the change in the y-axis divided by the change in the x-axis... when the latter tends to zero. So it's the slope of the tangent line to the curve $f$ at point $a$. –  Gene Arboit Feb 21 '13 at 22:44
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Try to divide by $h$. Then you'll find a difference quotient. Assuming the functions are continuous you can conclude differentiability and the equation for the derivative. Maybe this is what helps to understand the estimates. No the differential equation describes the growth depending on $P(t)$

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I don't think the inequality is strict, i.e. you really have

$$\frac{5h P(t+h)}{1000} \leq P(t) - P(t+h) \leq \frac{5h P(t)}{1000}$$

Rearrange the inequality, dividing by $h$: $$\frac{5 P(t+h)}{1000} \leq \frac{P(t) - P(t+h)}{h} \leq \frac{5 P(t)}{1000}$$

and take limits as $h \to 0$, getting $P'(t)$ in the middle squeezed to $ \frac{5P(t)}{1000}$

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