A container is in the shape of an inverted co | vedclass

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(A) Let $V$ be the volume,$r$ be the radius,and $h$ be the height of the water in the inverted cone at any time $t$.Given,$\frac{dV}{dt} = 3 \ m^3/min

Vedclass · Accepted Answer

$\frac{3}{4 \pi}$

Vedclass · Answer

(A) Let $V$ be the volume,$r$ be the radius,and $h$ be the height of the water in the inverted cone at any time $t$.Given,$\frac{dV}{dt} = 3 \ m^3/min$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.From the similarity of triangles in the cone,we have $\frac{r}{h} = \frac{4}{6} = \frac{2}{3}$,which implies $r = \frac{2}{3}h$.Substituting $r$ in the volume formula:$V = \frac{1}{3} \pi \left(\frac{2}{3}h\right)^2 h = \frac{1}{3} \pi \left(\frac{4}{9}h^2\right) h = \frac{4}{27} \pi h^3$.Differentiating with respect to $t$:$\frac{dV}{dt} = \frac{4}{27} \pi (3h^2) \frac{dh}{dt} = \frac{4}{9} \pi h^2 \frac{dh}{dt}$.Given $\frac{dV}{dt} = 3$ and we need to find $\frac{dh}{dt}$ when $h = 3 \ m$:$3 = \frac{4}{9} \pi (3)^2 \frac{dh}{dt}$$3 = \frac{4}{9} \pi (9) \frac{dh}{dt}$$3 = 4 \pi \frac{dh}{dt}$$\frac{dh}{dt} = \frac{3}{4 \pi} \ m/min$.

$A$ container is in the shape of an inverted cone. Its height is $6 \ m$ and the radius at the top is $4 \ m$. If it is filled with water at the rate of $3 \ m^3/min$,then the rate of change of the height of the water (in $m/min$) when the water level is $3 \ m$,is

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