Question 366108
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Let *[tex \Large r'] represent the rate of change of the fill rate; in this case *[tex \Large 1\ ml/sec^2].  Let *[tex \Large r] be the instantaneous fill rate, and let *[tex \Large V] be the instantaneous volume filled.  And finally, let *[tex \Large t] represent the elapsed time in seconds since I started to fill the bottle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \int\,1\ \text{ml/sec^2}\ dt\ =\ t\ \text{ml/sec}\ +\ C]


But the constant of integration in this case is the initial flow rate, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ t\ +\ r_o\ \text{ml/sec}]


Then the instantaneous volume is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \int\ r\ dt\ =\ \int\ t\ +\ r_o\ \text{ml/sec}\ dt\ =\ \frac{t^2}{2}\ +\ r_ot\ +\ V_o\ \text{ml}].


Since both the initial volume, *[tex \Large V_o], and the initial fill rate, *[tex \Large r_o], are both zero, if *[tex \Large V_p] is the volume of the pyramid bottle, then we only need solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{t^2}{2}\ =\ V_p]


for *[tex \Large t], namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \sqrt{\frac{V_p}{2}}\ \text{sec}]


since we can assume for the precision required for this problem that cubic centimeters are equivalent to milliliters.


The volume of any pyramid is given by *[tex \Large \frac{1}{3}Bh] where *[tex \Large h] is the height, and *[tex \Large B] is the area of the base.


To find the area of a regular polygon knowing the measure of a side, use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{1}{4}ns^2\cot\left(\frac{180^\circ}{n}\right)]


Which, for your regular pentagon reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{pentagon}\ =\ \frac{5}{4}s^2\cot\left(36^\circ\right)]


You should be able to handle it from here.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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