document.write( "Question 745144: Dry sand is being poured into a conical pile at a rate of 10 cubic meters per minute; the diameter of the pile is equal to the height. At what rate is the height of the cone changing when there are 144π cubic meters of sand in the pile? \n" ); document.write( "
Algebra.Com's Answer #453763 by KMST(5328)\"\" \"About 
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\"h\"= height of the cone = diameter of the base
\n" ); document.write( "The volume of a cone, \"V\", is calculated as
\n" ); document.write( "\"V=%281%2F3%29%2AB%2Ah=%281%2F3%29%2Api%2Ar%5E2%2Ah\" based on the height and
\n" ); document.write( "\"B\"= area of the base, or \"r\"= radius of the base
\n" ); document.write( "In the case of the pile of sand, \"r=h%2F2\", so
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\n" ); document.write( "When \"V=144pi\", \"%281%2F12%29%2Api%2Ah%5E3=144pi\" --> \"h%5E3=144%2A12=12%5E3\" --> \"h=12\"
\n" ); document.write( "
\n" ); document.write( "For all other times, solving for \"h\" we get
\n" ); document.write( "\"V=%281%2F12%29%2Api%2Ah%5E3\" --> \"12V%2Fpi=h%5E3\" --> \"h=root%283%2C12V%2Fpi%29\" or \"h=%2812V%2Fpi%29%5E%281%2F3%29\"
\n" ); document.write( "\"V\" is a linear function of time
\n" ); document.write( "If \"t\" time after \"V=144pi\" (with \"V\" in \"m%5E3\" and \"t\" in minutes,
\n" ); document.write( "\"V=144pi%2B10t\" and
\n" ); document.write( "As \"h\" is not a linear function of time, the rate of change for \"h\" changes with time, and for an exact value we have to calculate it using calculus.
\n" ); document.write( "Without calculus, we can get approximate values.
\n" ); document.write( "
\n" ); document.write( "WITH CALCULUS:
\n" ); document.write( "\"dh%2Fdt=%281%2F3%29%281728%2B120t%2Fpi%29%5E%28-2%2F3%29%2A%28120%2Fpi%29\",
\n" ); document.write( "and at \"t=0\",
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\n" ); document.write( "WITHOUT CALCULUS:
\n" ); document.write( "We can get estimates of the instantaneous rate of change in \"h\" when \"V=144pi\" by calculating the average rate of change over short periods of time, when \"V\" is about \"144pi\".
\n" ); document.write( "For example, between \"t=0\" and \"t=0.1\" minutes, \"V\" changes from
\n" ); document.write( "\"V%280%29=144pi\" to \"V%280.1%29=144pi%2B1\" and \"h\" changes from
\n" ); document.write( "\"h%280%29=12\" to \"h%280.1%29=%281728%2B12%2Fpi%29%5E%281%2F3%29=about12.00883543\"
\n" ); document.write( "The average rate of change is \"0.00883543%2F0.1=0.0883543\"
\n" ); document.write( "If we use \"t=0\" and \"t=0.01\" with
\n" ); document.write( "\"h%280%29=12\" and \"h%280.01%29=%281728%2B1.2%2Fpi%29%5E%281%2F3%29=about12.00088413\",
\n" ); document.write( "we get an average rate of change of
\n" ); document.write( "\"0.00088413%2F0.01=0.88413\"
\n" ); document.write( "Either way, we see that \"highlight%280.884%29\" meters per minute is a good estimate of the rate of increase for the height of the pile when \"V=144pi\".
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\n" ); document.write( "NOTE:
\n" ); document.write( "We did not really need to keep using \"V=%281%2F12%29%2Api%2Ah%5E3\", or \"h=%2812V%2Fpi%29%5E%281%2F3%29\", or \"h=%281728%2B120t%2Fpi%29%5E%281%2F3%29\" to calculate an approximate rate of change without calculus. All we needed to know is that for \"V=144pi\" \"h=12\".
\n" ); document.write( "We know that the ratio of volumes of similar solids is the cube of the ratio of lengths for any dimension measured, so
\n" ); document.write( "\"V%28t%29%2FV%280%29=%28h%28t%29%2Fh%280%29%29%5E3\" <-->
\n" ); document.write( "That would let us calculate the height for any other volume
\n" ); document.write( "Since \"V%280.1%29=144pi%2B1\", ,
\n" ); document.write( "\"V%280.1%29=V%280%29%2A0.000736286=12%2A0.000736286=12.00883543\"
\n" ); document.write( "and the average rate of change
\n" ); document.write( "\"%28h%280.1%29-h%280%29%29%2F%280.1-0%29=%2812.00883543-12%29%2F0.1=0.0883543\"
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