Let
l= length of the cylinder
<span>V=πr²l r=11 Vt=(380.133)l </span>
<span>x=√(121-y²) </span>
<span>dv=2x dy </span>
<span>10 </span>
<span>V=∫2l√(121-y²) dy=(373.965)l ft³ </span>
<span>-11 </span>
<span>% of the total capacity being used=(373.965/380.133)100 </span>
<span>=98.38%</span>
280 square feet is my answe
Let
h: height of the water
r: radius of the circular top of the water
V: the volume of water in the cup.
We have:
r/h = 3/10
So,
r = (3/10)*h
the volume of a cone is:
V = (1/3)*π*r^2*h
Rewriting:
V (t) = (1/3)*π*((3/10)*h(t))^2*h(t)
V (t) =(3π/100)*h(t)^3
Using implicit differentiation:
V'(t) = (9π/100)*h(t)^2*h'(t)
Clearing h'(t)
h'(t)=V'(t)/((9π/100)*h(t)^2)
the rate of change of volume is V'(t) = 2 cm3/s when h(t) = 5 cm.
substituting:
h'(t) = 8/(9π) cm/s
Answer:
the water level is rising at a rate of:
h'(t) = 8/(9π) cm/s
Answer:
25
Step-by-step explanation: