Answer:
See solutions for detail.
Step-by-step explanation:
a. [tex]\frac{dV}{dt}[/tex] is the instantaneous rate of change of volume given with respect to time, t.
The volume's rate of change is written as a function of time.
-[tex]\frac{dh}{dt}[/tex] is the rate of change in the height of water in the tank with respect to time, t.
b. [tex]\frac{dV}{dt}-[/tex] is the only constant. Water flows into the constant at a constant rate, say [tex]6cm^3[/tex] per minute.
c. [tex]\frac{dV}{dt}[/tex] is positive. Volume water in the take is increasing from time to time.
-The volume at time t=1 is greater than the volume at t=0, hence, it's a positive rate of change.
d. [tex]\frac{dh}{dt}[/tex] is a positive rate. The initial height of water in the tank is zero.
-The final height at time t is 0.25h. The height is increasing with time.
Hence, it is positive.
