The isosceles triangle shown has a height of 7 meters and base
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The rate at which the water from the container is being drained is 24 inches per second.
Given radius of right circular cone 4 inches .height being 5 inches, height of water is 2 inches and rate at which surface area is falling is 2 inches per second.
Looking at the image we can use similar triangle propert to derive the relationship:
r/R=h/H
where dh/dt=2.
Thus r/5=2/5
r=2 inches
Now from r/R=h/H
we have to write with initial values of cone and differentiate:
r/5=h/5
5r=5h
differentiating with respect to t
5 dr/dt=5 dh/dt
dh/dt is given as 2
5 dr/dt=5*-2
dr/dt=-2
Volume of cone is 1/3 π
We can find the rate at which the water is to be drained by using partial differentiation on the volume equation.
Thus
dv/dt=1/3 π(2rh*dr/dt)+(*dh/dt)
Putting the values which are given and calculated we get
dv/dt=1/3π(2*2*2*2)+(4*2)
=1/3*3.14*(16+8)
=3.14*24/3.14
=24 inches per second
Hence the rate at which the water is drained from the container is 24 inches per second.
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