Question Number 83524 by liki last updated on 03/Mar/20
Commented by liki last updated on 03/Mar/20
$$…{help}\:{me}\:{plz}\:{qns}\:{no}.\:\mathrm{5},\mathrm{7}\:{and}\:\mathrm{9} \\ $$
Commented by liki last updated on 03/Mar/20
$$…{plz}\:{help}\:{me}\:\:….. \\ $$
Answered by Kunal12588 last updated on 03/Mar/20
Commented by Kunal12588 last updated on 03/Mar/20
$$\frac{{dr}_{\mathrm{2}} }{{dt}}=\mathrm{1}\:{cm}/{s}\:\:,\:\:\frac{{dr}_{\mathrm{1}} }{{dt}}=\mathrm{2}\:{cm}/{s} \\ $$$${A}_{{shaded}} =\pi{r}_{\mathrm{2}} ^{\mathrm{2}} −\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{A}_{{shaded}} =\pi\left({r}_{\mathrm{2}} ^{\mathrm{2}} −{r}_{\mathrm{1}} ^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{dA}}{{dt}}=\mathrm{2}\pi\left({r}_{\mathrm{2}} ×\frac{{dr}_{\mathrm{1}} }{{dt}}−{r}_{\mathrm{1}} ×\frac{{dr}_{\mathrm{2}} }{{dt}}\right) \\ $$$$\Rightarrow\frac{{dA}}{{dt}}=\mathrm{2}\pi\left({r}_{\mathrm{2}} \left(\mathrm{1}\right)−{r}_{\mathrm{1}} \left(\mathrm{2}\right)\right) \\ $$$$\Rightarrow\frac{{dA}}{{dt}}=\mathrm{2}\pi\left({r}_{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} \right) \\ $$$${when}\:{r}_{\mathrm{1}} =\mathrm{8}\:{cm},\:{r}_{\mathrm{2}} =\mathrm{12}\:{cm} \\ $$$$\frac{{dA}}{{dt}}=\mathrm{2}\pi\left(\mathrm{12}−\mathrm{16}\right) \\ $$$$\frac{{dA}}{{dt}}=−\mathrm{8}\pi\:{cm}/{s} \\ $$
Answered by Kunal12588 last updated on 03/Mar/20
Commented by Kunal12588 last updated on 03/Mar/20
$$\frac{{r}}{{h}}=\frac{\mathrm{12}}{\mathrm{18}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{h}}{\mathrm{3}} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{r}^{\mathrm{2}} {h} \\ $$$$\Rightarrow{V}=\frac{\mathrm{1}}{\mathrm{3}}\pi×\frac{\mathrm{4}}{\mathrm{9}}{h}^{\mathrm{3}} \\ $$$$\Rightarrow{V}=\frac{\mathrm{4}}{\mathrm{27}}\pi{h}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{dV}}{{dh}}=\frac{\mathrm{4}}{\mathrm{9}}\pi{h}^{\mathrm{2}} ×\frac{{dh}}{{dt}} \\ $$$$\Rightarrow\frac{{dV}}{{dh}}=\frac{\mathrm{4}}{\mathrm{9}}\pi×\mathrm{36}×\frac{{dh}}{{dt}} \\ $$$$\Rightarrow\mathrm{2}=\mathrm{16}\pi\frac{{dh}}{{dt}} \\ $$$$\Rightarrow\frac{{dh}}{{dt}}=\frac{\mathrm{1}}{\mathrm{8}\pi}{cm}/{s} \\ $$