The rate of change of the volume of water in a large industrial storage tank, R(t)R(t)R(t) in m3/day\text{m}^3/\text{day}m3/day, is modeled by the function:
R(t)=54t2+2t−14,t>0 R(t) = \frac{54}{t^2} + 2t - 14, \quad t > 0 R(t)=t254+2t−14,t>0where t t\,t is the time in days since the start of a maintenance cycle. Using calculus:
Find the set of values of t t\,t for which the rate of change R(t)R(t)R(t) is increasing, giving your answer in the form t>ab3t > a\sqrt[3]{b}t>a3b where a a\,a and b b\,b are integers.
Show that ∫39(54t2+2t−14)dt=0\displaystyle \int_{3}^{9} \left( \frac{54}{t^2} + 2t - 14 \right) dt = 0∫39(t254+2t−14)dt=0.
Given that ∫36(54t2+2t−14)dt=−6\displaystyle \int_{3}^{6} \left( \frac{54}{t^2} + 2t - 14 \right) dt = -6∫36(t254+2t−14)dt=−6:
(i) State the value of ∫69(54t2+2t−14)dt\displaystyle \int_{6}^{9} \left( \frac{54}{t^2} + 2t - 14 \right) dt∫69(t254+2t−14)dt.
(ii) Find the value of the constant k k\,k such that ∫36(54t2+2t+k)dt=0\displaystyle \int_{3}^{6} \left( \frac{54}{t^2} + 2t + k \right) dt = 0∫36(t254+2t+k)dt=0.
