SOLUTION: The volume of water in a tank at a particular time (measured in
seconds) is given by V (t) = 5t(2 â t) m3
. Find the rate of change of the
volume of water in the tank from fir
Algebra.Com
Question 1202240: The volume of water in a tank at a particular time (measured in
seconds) is given by V (t) = 5t(2 â t) m3
. Find the rate of change of the
volume of water in the tank from first principles (using the definition of the rate
of change).
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52866) (Show Source): You can put this solution on YOUR website!
.
It is about finding the derivative of the given function,
and usually people solve such problems using standard rules of Calculus,
without using "first principles".
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
The term first principles in this context refers to the limit definition of the derivative.
Let's first determine V(t+h).
 = 5t(2-t))
 = 10t-5t^2)
 = 10(t+h)-5(t+h)^2)
 = 10(t+h)-5(t^2+2th+h^2))
 = 10t+10h-5t^2-10th-5h^2)
-------------------
Then we can compute the derivative to get the instantaneous rate of change.
 = \frac{dV}{dt}= \text{instantaneous rate of change of the volume})
 = \lim_{h \to 0} \frac{V(t+h)-V(t)}{h})
 = \lim_{h \to 0} \frac{(10t+10h-5t^2-10th-5h^2)-(10t-5t^2)}{h})
 = \lim_{h \to 0} \frac{10t+10h-5t^2-10th-5h^2-10t+5t^2}{h})
 = \lim_{h \to 0} \frac{10h-10th-5h^2}{h})
 = \lim_{h \to 0} \frac{h(10-10t-5h)}{h})
 = \lim_{h \to 0} (10-10t-5h))
 = 10-10t-5*0)
 = 10-10t)
The instantaneous rate of change of the volume is V'(t) = 10-10t
The units for the rate of change is "cubic meters per second".
Extra info:
1 cubic meter = 1000 liters = 1 kiloliter
1 meter = 3.2808 feet approximately
1 cubic meter = (3.2808)^3 = 35.313 cubic feet approximately
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