SOLUTION: The volume of grains in a silo at a particular time (measured in hours) is given by V (t) = 4t(3-t) m3. Find the rate of change of the volume of grains in the silo from first pri

Algebra ->  Trigonometry-basics -> SOLUTION: The volume of grains in a silo at a particular time (measured in hours) is given by V (t) = 4t(3-t) m3. Find the rate of change of the volume of grains in the silo from first pri      Log On


   

Question 868034: The volume of grains in a silo at a particular time (measured in hours) is given
by V (t) = 4t(3-t) m3. Find the rate of change of the volume of grains in the
silo from first principles (using the definition of the rate of change).
Found 3 solutions by Fombitz, stanbon, Ra:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

V%28t%29=12t-4t%5E2
V%28t%2Bdt%29=12%28t%2Bdt%29-4%28t%2Bdt%29%5E2
You have a mistake here,
V%28t%2Bdt%29=12%28t%2Bdt%29-4%28t%5E2%2B2tdt%2Bdt%5E2%29
V%28t%2Bdt%29=12t%2B12dt-4t%5E2-8tdt-4dt%5E2
V%28t%2Bdt%29-V%28t%29=12dt-8tdt-4dt%5E2
%28V%28t%2Bdt%29-V%28t%29%29%2Fdt=12-8t-4dt
dV%2Fdt=lim%28t-%3E0%2C%28V%28t%2Bdt%29-V%28t%29%29%2Fdt%29=12-8t

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
My answer is:
V(t)=4t(3-t)
Rate of change of volume of V(t) is V'(t)
V(t)=4t(3-t)
-------------------------
V(t) = 12t-4t^2
-------------------------
V(t+D) = 12(t+D) - 4(t+D)^2
= 12t+12D -4(t^2+ 2tD + D^2)
= 12t +12D -4t^2 -8tD -4D^2
-------------------------------------------
[V(t+∆t)- V(t)] / ∆t
= [ 12D - 8(tD) -4(D)^2]/D
= [12 - 8t - 4D]
lim of V(t)/Dt as V goes to 0 = 12-8t
================
Cheers,
Stan H.
===============

Answer by Ra(1) About Me  (Show Source):
You can put this solution on YOUR website!
So you can use the power rule which is easier, if ur tutor doesnt want u to show all work. using the formula like the other guys.
x%5En+=+nx%5E%28n-1%29
so
12t-4t%5E2
12%28d%2Fdx%29+-%284x2t%5E%282-1%29%29+=+12-8t
Heaps easier!!!