Question 1202240
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The term first principles in this context refers to the limit definition of the derivative.


Let's first determine V(t+h).
*[tex \Large V(t) = 5t(2-t)]


*[tex \Large V(t) = 10t-5t^2]


*[tex \Large V(t+h) = 10(t+h)-5(t+h)^2]


*[tex \Large V(t+h) = 10(t+h)-5(t^2+2th+h^2)]


*[tex \Large V(t+h) = 10t+10h-5t^2-10th-5h^2]


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Then we can compute the derivative to get the instantaneous rate of change.
*[tex \Large V'(t) = \frac{dV}{dt}= \text{instantaneous rate of change of the volume}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} \frac{V(t+h)-V(t)}{h}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} \frac{(10t+10h-5t^2-10th-5h^2)-(10t-5t^2)}{h}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} \frac{10t+10h-5t^2-10th-5h^2-10t+5t^2}{h}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} \frac{10h-10th-5h^2}{h}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} \frac{h(10-10t-5h)}{h}]


*[tex \Large \displaystyle V'(t) = \lim_{h \to 0} (10-10t-5h)]


*[tex \Large \displaystyle V'(t) = 10-10t-5*0]


*[tex \Large \displaystyle V'(t) = 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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