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Q1-A-balloon-in-the-shape-of-cone-surrmo-surmounted-by-hemispherical-top-the-daimeter-of-balloon-is-equal-to-hieght-of-cone-find-the-rate-of-change-of-volume-of-the




Question Number 7265 by 2402@gmail.com last updated on 20/Aug/16
Q1. A balloon in the shape of cone surrmo            surmounted by hemispherical           top . the daimeter of balloon is         equal to hieght of cone . find the       rate of change of volume of the     balloone with respect to its total      height  h .
$${Q}\mathrm{1}.\:{A}\:{balloon}\:{in}\:{the}\:{shape}\:{of}\:{cone}\:{surrmo} \\ $$$$\:\:\:\:\:\:\:\:\:\:{surmounted}\:{by}\:{hemispherical} \\ $$$$\:\:\:\:\:\:\:\:\:{top}\:.\:{the}\:{daimeter}\:{of}\:{balloon}\:{is} \\ $$$$\:\:\:\:\:\:\:{equal}\:{to}\:{hieght}\:{of}\:{cone}\:.\:{find}\:{the} \\ $$$$\:\:\:\:\:{rate}\:{of}\:{change}\:{of}\:{volume}\:{of}\:{the} \\ $$$$\:\:\:{balloone}\:{with}\:{respect}\:{to}\:{its}\:{total} \\ $$$$\:\:\:\:{height}\:\:{h}\:. \\ $$
Commented by Yozzia last updated on 20/Aug/16
d=h′=height of cone  r=radius of hemisphere section  V=(1/3)πr^2 h′+((2πr^3 )/3)=(π/3)(r^2 h′+2r^3 )  h=h′+(d/2)=(3/2)h′⇒h′=(2/3)h  r=(d/2)=((h′)/2)=(1/2)×(2/3)h=(1/3)h  V=(π/3)((h^2 /9)×(2/3)h+2×(1/(27))h^3 )  V=(π/3)×(4/(27))h^3   ⇒(dV/dh)=((4π)/(27))h^2
$${d}={h}'={height}\:{of}\:{cone} \\ $$$${r}={radius}\:{of}\:{hemisphere}\:{section} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{r}^{\mathrm{2}} {h}'+\frac{\mathrm{2}\pi{r}^{\mathrm{3}} }{\mathrm{3}}=\frac{\pi}{\mathrm{3}}\left({r}^{\mathrm{2}} {h}'+\mathrm{2}{r}^{\mathrm{3}} \right) \\ $$$${h}={h}'+\frac{{d}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}{h}'\Rightarrow{h}'=\frac{\mathrm{2}}{\mathrm{3}}{h} \\ $$$${r}=\frac{{d}}{\mathrm{2}}=\frac{{h}'}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}{h}=\frac{\mathrm{1}}{\mathrm{3}}{h} \\ $$$${V}=\frac{\pi}{\mathrm{3}}\left(\frac{{h}^{\mathrm{2}} }{\mathrm{9}}×\frac{\mathrm{2}}{\mathrm{3}}{h}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{27}}{h}^{\mathrm{3}} \right) \\ $$$${V}=\frac{\pi}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{27}}{h}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{dV}}{{dh}}=\frac{\mathrm{4}\pi}{\mathrm{27}}{h}^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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