Question

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

Miss.Sue (mathematics teacher)

(Please ignore the language beneath English)

Answer 1

`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`

Therefore, `20/40=r/h`, ie , `r=h/2`........`[1]`

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

`V=[pih^3]/[12]`......`[2]`. Differentiating wrt `t`,

`dV/dt=[pih^2]/[4]dh/dt`, but it is given that `dV/dt=5`

So, `5=[pih^2]/[4]dh/dt`, solving for `dh/dt`'

`dh/dt=[[20]/[pih^2]]`. When `h=20`, `dh/dt=[1/[20pi]]`.

Hope this was helpful.