APPLICATION OF DERIVATIVES

(SUM 21TO 25)

ALL EDUCATION BOARD


Question 21 The Totel Revenue Recived From The Sale Of `x` Units Of The Product Is Given By `R(`x`)` = `3 x^{2} + 36`x` + 5 `. Find The Marginal Revenue When `x`=5.


Answer 
Totel revenue function `R(`x`)` = `3 x^{2} + 36`x` + 5 `
Marginal revenue (MR) is the rate of change of total revenue with respect to the numbers of units sold.

`MR`= `\frac{dR}{dx}`

`MR`= ` \frac{d}{dx}`( 3 x^{2} + 36`x` + 5)`

` \Rightarrow `       `MR` = `6x  + 36`

Marginal revenue ,when `5` Units are sold 

`MR` ` \mid_{x=5} =`6(5) +  36` = `30 + 36`=`66`

Hence , Marginal revenue When  `x`= `5` is Rs.66




Question 22 A particular moves along the curve find the points on the curve at which the y coordinate is changing twice as fast as the x coordinate.

Answer  Here `y`= `\frac{2}{3}  x^{3}`  + `1`
                                            ...................(2)

 `\frac{dy}{dx}`= `\frac{2}{3}.3 x^{2} . \frac{dx}{dt} `
              
                     = `2 x^{2}`  = `2  x^{2}   \frac{dx}{dt} `
                                                   ...........................(2)

Now y co-ordinate changes 2 times as fast `x` co ordinate 

     `\frac{dy}{dt}` = `2 \frac{dx}{dt} `

Putting the value of  `\frac{dy}{dt}`in (2)  we have

     ` 2 \frac{dx}{dt}`= `2  x^{2}  \frac{dx}{dt} `  

` \Rightarrow `     ` x^{2}  =1`

` \Rightarrow `         `x`=  `\pm ` 1

When `x` =`1`,then from  (1),     

                 `y`=  `\frac{2}{3} ` + `1`=`\frac{5}{5}`

    And When  `x`=`-1 ` then from (1), =` -  \frac{2}{3}  + 1`

     =` \frac{1}{3} 


Hence the required points Are `(1, \frac{5}{3} )` and `(-1, \frac{1}{3} )`




Question 23 A Hemisphere is constructed on a circlar base .If the radius of the base is increasing at the rate of  `0.5` `cm`/`sec`,find the rate at which the volume of the hmisphere is increasing when the radius is `10 ` `cm`.


Answer Let `r` be the radius And `V ` be the volumn of  the hemisphere at any time `t`.

so  `V` =  `\frac{2}{3}` `\pi  x^{3}` 

radius of base is increasing at the rate of 0.5 `cm`/`sec`

` \frac{dr}{dt}`=`0.5 `


rate of change of volume= `\frac{dV}{dt}` = `\frac{2}{3} .\pi.3 r^{2}. \frac{dr}{dt}`


                             =`2 \pi  r^{2} \times (0.5)`= `\pir^{2}` 

rate of change of volume when radius is 
`10cm`=` \pi  \times  (10)^{2}`

                                    =`100 \pi `

     =`100 \times 3.1428`= `314.28` `cm^{3} `/`sec



Question 24  The radius of a cylinder is increasing at the rate `2` `cm` /`sec` and its altitude is decreasing at the rate of `3` `cm` /`sec` .Find the rate of change of volume when radius is `3` `cm` and altitude `5` `cm`.
Answer 
    Let `r` be the the radius  and `h` be the altitude of the cyclinder at the time `t`.

rate of change (increase) in radius  = `\frac{dr}{dt}`=`2` 
                                                        
                                                                ............(1)  given

rate of change (decrease) in altitude= `\frac{dh}{dt}`=`-3` 

                                                                   ................(2)

now let `V` be the volume of the cyclinder at the time `t`.


                                          `V`= `\pi  r^{2}h` 


Rate of change of volume = `\frac{dV}{dt}`
 = `\frac{d}{dt}`` ( \pi  r^{2}h )`

` \Rightarrow `     `\frac{dV}{dt}`
=` \pi (2r .\frac{dr}{dt} .h ` + `r^{2} . \frac{dh}{dt})`

` \Rightarrow ``\frac{dV}{dt}`=` \pi (2r . 2 . h  +  r^{2} . (-3))`

                                                          [Using (1) and (2)] 

                                                 = ` \pi (4rh - 3 r^{2} ) `  



  when `r`= `3`  and  `h` = `5``cm` , then the rate of change in volume


=` \pi (4 \times 3 \times 5 -3 (3)^{2} )`

=`\pi (60-27)` = `33\pi  cm^{3}`/`sec `


Quetion 25 Water is running out of a conical funnel at the rate of `5` `cm`/`sec`. If the radius of the base of the funnel is  `10``cm` and The altitude is `20` `cm` ,Find the rate at which the water level is dropping when it is `5` `cm` from the top.

Answer Let `r` be the radius and `h`be the height of the surface of the water at time `t`.

Let `V` bethe volumn of the water in funnel.


fig 3



`ABC` Is the cone of height `20``cm` and radius `10cm`.

`OC` =`10` `cm`

`OA` =`20``cm`

In  `\Delta``ADE`, `AD`=`h` and `DE`=`r`

so  `\frac{OC}{OA} `=` \frac{DE}{DA} `

  ` \Rightarrow`   `\frac{10}{20}`=` \frac{r}{h}`

` \Rightarrow `     `r`= ` \frac{h}{2}`

now volume of the cone `V`= `\frac{1}{3} \pi  r^{2}h`

` \Rightarrow `   `V`
=` \frac{1}{3} \pi ( \frac{h}{2} )^{2} h`    

=` \frac{1}{12} \pi  h^{3}`

so  `\frac{dv}{dt}`
= `\frac{1}{12} \pi .3 h^{2} \frac{dh}{dt}`

 `\frac{ \pi }{4}  h^{2}  \frac{h}{dt}`
                                            .............(1)


now water is running out of funnel at the rate of `5`` cm^{3}`/`sec`     


 `\frac{dV}{dt}`=`5`   
`[\frac{dV}{dt} is   -  \nu e since V decreases as t increases ] ` 

Also when water level is `5` `cm` from top then `h`=`20`-`5`=`15`

putting these value in (1) we have


`-5`= ` \frac{ \pi }{4} \times  (15)^{2}  \frac{dh}{dt}` 

` \Rightarrow `    `-5`=` \frac{225 \pi }{4} . \frac{dh}{dt}`

` \Rightarrow ``\frac{dh}{dt}`

                              =`- \frac{20}{225 \pi }`
                

                              =`- \frac{4}{45 \pi }  `cm ``sec`

Hence rate of drop of water level w.r.t   `t`   when `h` is `15`

                     =  `\frac{4}{45 \pi } `  `cm`/`sec`