dV/dt = surface area * dh/dt (geometry , no calculus really required, draw picture)
dV/dt = pi r^2 dh/dt
6. A parabaloid with height h and radius r has volume 1/2 �pieR^2H. Consider a parabaloid-shaped water
tank with height of 7m and radius of 4m. If V is the volme of water in the tank and h is the depth
of water,
find the relationship between dV/dt and dhdt.
How is the volume changing when h = 3m and
dh/dt = 0.2m/s?
2 answers
consider a cross-section of the paraboloid placed with vertex at the origin an opening upwards
then its equation would be
y = ax^2
when x=4, y = 7
y = (7/16)x^2
or
h = (7/16)r^2 , using r and h
r^2 = 16h/7
V = (1/2)πr^2h
= (1/2)π(16h/7)h
= (8π/7) h^2
dV/dt = 16π/7 h dh/dt ----> the relationship between dV/dt and dh/dt
Whe h=3 and dh/dt = .2
dV/dt = (16π/7)(3)(.2) = ......
I will let you do the button pushing.
then its equation would be
y = ax^2
when x=4, y = 7
y = (7/16)x^2
or
h = (7/16)r^2 , using r and h
r^2 = 16h/7
V = (1/2)πr^2h
= (1/2)π(16h/7)h
= (8π/7) h^2
dV/dt = 16π/7 h dh/dt ----> the relationship between dV/dt and dh/dt
Whe h=3 and dh/dt = .2
dV/dt = (16π/7)(3)(.2) = ......
I will let you do the button pushing.
2 answers