A research probe is submerged in a fluid where the vertical force, F F\,F kilonewtons, exerted on its hull is modeled by the function
F(d)=54d2+4d−26,d>0 F(d) = \frac{54}{d^2} + 4d - 26, \quad d > 0 F(d)=d254+4d−26,d>0where d d\,d is the depth in metres below the surface.
Using calculus,
determine the range of depths for which the vertical force F(d)F(d)F(d) is increasing.
show that ∫39(54d2+4d−26)dd=0\displaystyle \int_{3}^{9} \left( \frac{54}{d^2} + 4d - 26 \right) dd = 0∫39(d254+4d−26)dd=0.
The points A(3,−6)A(3, -6)A(3,−6) and B(6,−0.5)B(6, -0.5)B(6,−0.5) lie on the curve F(d)F(d)F(d).
Given that ∫36(54d2+4d−26)dd=−15\displaystyle \int_{3}^{6} \left( \frac{54}{d^2} + 4d - 26 \right) dd = -15∫36(d254+4d−26)dd=−15.
(i) state the value of ∫69(54d2+4d−26)dd\displaystyle \int_{6}^{9} \left( \frac{54}{d^2} + 4d - 26 \right) dd∫69(d254+4d−26)dd.
(ii) find the value of the constant k k\,k such that ∫36(54d2+4d+k)dd=0\displaystyle \int_{3}^{6} \left( \frac{54}{d^2} + 4d + k \right) dd = 0∫36(d254+4d+k)dd=0.
