A cone of radius 𝑟 centimeters and height ℎ centimeters is lowered point first at
a rate of 1 cm/s into a tall cylinder of radius 𝑅 centimeters that is partially filled with
water. How fast is the water level rising at the instant the cone is completely
submerged?
V−volume of cone in watery−heigth of cone in waterV=(yh)3⋅13πr2h=πr23h2y3z−heigth of water in cylinder above the initial levelz=VπR2=πr23h2y3πR2=r23h2R2y3dzdt=r23h2R2⋅3y2dydt=[y=h]=r2R2dydt=r2R2V-volume\,\,of\,\,cone\,\,in\,\,water\\y-heigth\,\,of\,\,cone\,\,in\,\,water\\V=\left( \frac{y}{h} \right) ^3\cdot \frac{1}{3}\pi r^2h=\frac{\pi r^2}{3h^2}y^3\\z-heigth\,\,of\,\,water\,\,in\,\,cylinder\,\,above\,\,the\,\,initial\,\,level\\z=\frac{V}{\pi R^2}=\frac{\frac{\pi r^2}{3h^2}y^3}{\pi R^2}=\frac{r^2}{3h^2R^2}y^3\\\frac{dz}{dt}=\frac{r^2}{3h^2R^2}\cdot 3y^2\frac{dy}{dt}=\left[ y=h \right] =\frac{r^2}{R^2}\frac{dy}{dt}=\frac{r^2}{R^2}V−volumeofconeinwatery−heigthofconeinwaterV=(hy)3⋅31πr2h=3h2πr2y3z−heigthofwaterincylinderabovetheinitiallevelz=πR2V=πR23h2πr2y3=3h2R2r2y3dtdz=3h2R2r2⋅3y2dtdy=[y=h]=R2r2dtdy=R2r2
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