APPLICATION OF DERIVATIVES
(SUM 25 TO 28)
ALL EDUCATION BOARD
uestion 26
A conical vessel whose height is `10` `meter` and the radius of whose base is `5` meter is being filled with water at the uniform rate of `1.5` cubic `meter`/`min`Find the rate at which the the level of water in the vessel is rising when the depth is `4` meteres.
Answer
`ABC` is the cone with radius `OC`=`5`` m` and height `OA`=`10` `m`
At any time `t` ,Let `V `be the volume of water .Now the vessel is being filled with water at the uniform rate of `1.5` `m^{3}`/`min` .
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| fig4 |
Rate of change of volume is
` \frac{3}{2} m^{3}`/`min` , `\frac{dV}{dt}`= `\frac{3}{2}`.At that instant suppose the water forms a cone whose height `D`=`h` and radius `DE`=`r` `meters`.
Now `\Delta`'s `OCA and DEA are similar
so `\frac{DE}{OC}`= `\frac{DA}{OA}`
` \Rightarrow ` `\frac{r}{5}`= `\frac{h}{10}`
` \Rightarrow ` `r`=` \frac{h}{2}`
Now `V`= ` \frac{1}{3} \pi r^{2} h`
= `\frac{1}{3} \pi ( \frac{h}{2} )^{2} .h`
= `\frac{ \pi h^{3} }{12} `
`[r= \frac{h}{2} ]`
so `\frac{dV}{dt}`= `\frac{3 \pi h^{2} }{12}. \frac{dh}{dt}`
` \Rightarrow ` `\frac{3}{2}`= `\frac{ \pi h^{2} }{4}. \frac{dh}{dt}`
` \Rightarrow ` `\frac{dh}{dt}`= `\frac{6}{ \pi h^{2} }`
` [ \frac{dV}{dt}=\frac{3}{2} ] `
when `h`=`4`,Rate of change (increase) in water level
= `\frac{6}{ \pi \times 16}` = ` \frac{3}{8 \pi }meters`/ `minute`
Question 27
From a cyclinderical drum containing oil and kept vertical ,the oil is leaking at the rate of `10` `cm^{3}`/`sec` .If the radius of the drum is `10` `cm` and height is `50``cm`,find the rate at which level of oil is changing when oil level is `20` `cm`.
Answer
Let `r`be the radius ,`h` be the height and `V` be the volume of the cylindrical drum at any time `t`
`v`= `\pi r^{2} h`
Now it is given that `r`=`10``cm`
`V`= `\pi \times (10)^{2} \times h`=`100` \pi h
` \Rightarrow ` `\frac{dV}{dt}`=`100 \pi \frac{dh}{dt}`
..................(1)
The oil is leaking from the drum at the rate of `10` ` cm^{3}`/`sec`
Rate of change (decrease) of volume = `10` ` cm^{3}`/`sec`
`\frac{dV}{dt}`=`-10`
[ `\frac{dV}{dt}` is `- \nue` ` V` decrease as t increse ]
`-10`=`100 \pi \frac{dh}{dt}` {using 1}
` \Rightarrow ` `\frac{dh}{dt}`=` \frac{-10}{100 \pi }`
=`- \frac{1}{10 \pi } ``cm`/`sec`
Hence rate of change of oil level is `\frac{1}{10 \pi }``cm`/`sec`
Quetion 28
A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost Its semi vertical angle is `tan^{-1} (0.5)`.Water is poured into it at a constant rate of `5` cubic metre per hour .Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is `4m`.
Answer
Let `ABC` represent the water tank which is in shap of a right circular cone.
`r`be the radius ,`h` be the height and `\alpha `
be the semi vertical angle of the cone `ABC`.then
`tan \alpha` = `\frac{r}{h}`
` \Rightarrow ` ` \alpha `=` tan^{-1}( \frac{r}{h})`
now, ` \alpha `=` tan^{-1}(0.5)`
`[Given]`
` \frac{r}{h}`=`0.5`
` \Rightarrow ` `r`=` \frac{1}{2}h`
Let `V` be the volume of the cone
`V`=`\frac{1}{3} \pi r^{2}h`
` \Rightarrow ` `V`= `\frac{1}{3} \pi ( \frac{1}{2}h)^{2}.h`
= `\frac{ \pi }{2} h^{3}`
now water is poured in to cone at a constant rate of `5 m^{3}`/`h`
` \frac{dV}{dt}` =`5 m^{3}`/`h`
` \Rightarrow ` `\frac{d}{dt}( \frac{ \pi }{12} h^{3})=`5`
` \Rightarrow ` ` \frac{ \pi }{12}3 h^{2}. \frac{dh}{dt}`=`5`
` \Rightarrow ` `\frac{ \pi h^{2} }{4}. \frac{dh}{dt} =`5`
` \Rightarrow ` `\frac{dh}{dt}`=` \frac{20}{ \pi h^{2}}`
when `h`=`4m`,the rate of change of volume
=` \frac{20}{ \pi (4)x^{2} }`
= `\frac{20}{16} \times \frac{7}{22}`
=` \frac{35}{88} `m`/`hr`
Hence the rate of change of water level is `\frac{35}{88} `m`/ `hr`
