Help pls...Question 22)b) ??

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Help pls...Question 22)b) ??

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Answer 1 · 2018-05-28T21:14:15

$\frac{1}{20} π$ $1/20pi$ cms s^-1

Explanation:

Assuming this is a right circular cone, then by proportion, we can say that the radius and height of the large cone is in proportion to the radius and height of the smaller cone of height $h$ $h$

Therefore, $\frac{20}{40} = \frac{r}{h}$ $20/40=r/h$ , ie , $r = \frac{h}{2}$ $r=h/2$ ........ $[1]$ $[1]$

Volume of the cone is given by $V = \frac{1}{3} π r^{2} h$ $V=1/3pir^2h$ , and substituting for $r$ $r$ from ..... $[1]$ $[1]$ , $V = \frac{1}{3} π {[\frac{h}{2}]}^{2} h$ $V=1/3pi[h/2]^2h$ = $\frac{π h^{3}}{12}$ $[pih^3]/[12]$ .

$V = \frac{π h^{3}}{12}$ $V=[pih^3]/[12]$ ...... $[2]$ $[2]$ . Differentiating wrt $t$ $t$ ,

$d \frac{V}{d t} = \frac{π h^{2}}{4} d \frac{h}{d t}$ $dV/dt=[pih^2]/[4]dh/dt$ , but it is given that $d \frac{V}{d t} = 5$ $dV/dt=5$

So, $5 = \frac{π h^{2}}{4} d \frac{h}{d t}$ $5=[pih^2]/[4]dh/dt$ , solving for $d \frac{h}{d t}$ $dh/dt$ '

$d \frac{h}{d t} = [\frac{20}{π h^{2}}]$ $dh/dt=[[20]/[pih^2]]$ . When $h = 20$ $h=20$ , $d \frac{h}{d t} = [\frac{1}{20 π}]$ $dh/dt=[1/[20pi]]$ .

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Help pls...Question 22)b) ??

(Please ignore the language beneath English)

1 Answer

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