Question 745144
{{{h}}}= height of the cone = diameter of the base
The volume of a cone, {{{V}}}, is calculated as
{{{V=(1/3)*B*h=(1/3)*pi*r^2*h}}} based on the height and
{{{B}}}= area of the base, or {{{r}}}= radius of the base
In the case of the pile of sand, {{{r=h/2}}}, so
{{{V=(1/3)*pi*(h/2)^2*h=(1/3)*pi*(h^2/4)*h=(1/12)*pi*h^3}}}
When {{{V=144pi}}}, {{{(1/12)*pi*h^3=144pi}}} --> {{{h^3=144*12=12^3}}} --> {{{h=12}}}
 
For all other times, solving for {{{h}}} we get
{{{V=(1/12)*pi*h^3}}} --> {{{12V/pi=h^3}}} --> {{{h=root(3,12V/pi)}}} or {{{h=(12V/pi)^(1/3)}}}
{{{V}}} is a linear function of time
If {{{t}}} time after {{{V=144pi}}} (with {{{V}}} in {{{m^3}}} and {{{t}}} in minutes,
{{{V=144pi+10t}}} and {{{h=(12*(144pi+10t)/pi)^(1/3)=((1728pi+120t)/pi)^(1/3)=(1728+120t/pi)^(1/3)}}}
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.
Without calculus, we can get approximate values.
 
WITH CALCULUS:
{{{dh/dt=(1/3)(1728+120t/pi)^(-2/3)*(120/pi)}}},
and at {{{t=0}}}, {{{dh/dt=(1/3)*(1728+0/pi)^(-2/3)*(120/pi)=(1/3)*(1/12^2)*(120/pi)=40/144pi=highlight(5/18pi)=highlight(about0.088419)}}}
 
WITHOUT CALCULUS:
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}}}.
For example, between {{{t=0}}} and {{{t=0.1}}} minutes, {{{V}}} changes from
{{{V(0)=144pi}}} to {{{V(0.1)=144pi+1}}} and {{{h}}} changes from
{{{h(0)=12}}} to {{{h(0.1)=(1728+12/pi)^(1/3)=about12.00883543}}}
The average rate of change is {{{0.00883543/0.1=0.0883543}}}
If we use {{{t=0}}} and {{{t=0.01}}} with
{{{h(0)=12}}} and {{{h(0.01)=(1728+1.2/pi)^(1/3)=about12.00088413}}},
we get an average rate of change of
{{{0.00088413/0.01=0.88413}}}
Either way, we see that {{{highlight(0.884)}}} meters per minute is a good estimate of the rate of increase for the height of the pile when {{{V=144pi}}}.
 
NOTE:
We did not really need to keep using {{{V=(1/12)*pi*h^3}}}, or {{{h=(12V/pi)^(1/3)}}}, or {{{h=(1728+120t/pi)^(1/3)}}} to calculate an approximate rate of change without calculus. All we needed to know is that for {{{V=144pi}}} {{{h=12}}}.
We know that the ratio of volumes of similar solids is the cube of the ratio of lengths for any dimension measured, so
{{{V(t)/V(0)=(h(t)/h(0))^3}}} <--> {{{h(t)/h(0)=(V(t)/V(0))^(1/3)=root(3,V(t)/V(0))}}}
That would let us calculate the height for any other volume
Since {{{V(0.1)=144pi+1}}}, {{{h(0.1)/h(0)=root(3,(144pi+1)/144)=root(3,1.002210485)=0.000736286}}},
{{{V(0.1)=V(0)*0.000736286=12*0.000736286=12.00883543}}}
and the average rate of change
{{{(h(0.1)-h(0))/(0.1-0)=(12.00883543-12)/0.1=0.0883543}}}