APPLICATION OF DERIVATIVES

(SUM 16 TO 20)

ALL EDUCATION BOARD

Question 16 The Area Of Expanding Rectangle Is Increasing At The Rate Of 48 `cm^{2}`/`sec`.The Length Of The Rectangle Is Equle To The Square Of The Breadth. At What Rate Is The Length Increasing At The Instant When The Breadth Is 4.5 `cm`

 Answer Let `x` and `y` the length and breadth of the given rectangle respectively And A be the area of the rectangle at time `t`.

`x`=` y^{2}`     And `A`= `x \times y` = ` y^{2}y` =  `y^{3}`

Rate of change (increase) of area  w.r.t    `t`=48 `cm^{2}`/`sec`

 `\frac{dA}{dt}`  = 48   

so, `\frac{d}{dt} ( y^{3} )`=48                [A= `y^{3}` ]

or  `3y^{2} \frac{dy}{dt}`  = 48

 `\Rightarrow `       `\frac{dy}{dt}`= `\frac{16}{ y^{2}}`................(1)

rate of change of length w.r.t   `t`= ` \frac{dx}{dt} `=  `\frac{d}{dt} ( y^{2}) `

  =`2y  \frac{dy}{dt}` = `2y \frac{16}{ y^{2} }` =` \frac{32}{y} `

                                                                              [Using 1]   

    When breadth  `y`= `4.5cm`, the rate of increase of length

             = `\frac{32}{4.5}`= `\frac{320}{45}`= 7.11 `cm`/`sec`

  

Question 17 A Stone Is Dropped In To A Quiet Lake And Wave Moves In Circle At The Speed Of 4 `cm` /`sec` At The Instant ,When The Radius Of Circular Wave Is 10`cm`,How Fast Is The Enclosed Area Increasing ?

Answer 

             Let `r`be the radius of the circular wave Ana `A` be the area of region enclosed between the circular wave at any time `t`.

now         `A`= `\pi  r^{2}`       `\Rightarrow`  `\frac{dA}{dt}`

= `2 \pi r \frac{dr}{dt}`

Rate of change of radius of wave = 4 `cm`/`sec       [Given]

` \frac{dr}{dt}`=4 

so  `\frac{dA}{dt}`=`2 \pi  \times r \times 4`=`8 \pi r`

  Rate of increase of enclosed area when radius is 10`cm`=`80 \pi  cm^{2}`/`sec`

Question 18 A Man 160 `cm` Tall ,Walks Away From A Source Of Light Situated At the Top Of A Pole 6 `m` High ,At The Rate Of 1.1  `m`/ `sec`.How Fast Is The Length Of His Dhadow Increasing When He Is 1 `m`Away From The Pole.

Answer 

Let the pole and the man be represented by `OB`And `PQ` respectively in  `\Delta` `OAB`,as shown in the figure. `\Delta`'s  `APQ` and `ABO` are similar triangles

fig 1

so  `\frac{PQ}{OB}` =  `\frac{QA}{OA}` 

Let  `OQ`=`l`   and  `QA`=`s`

 `\frac{1.6}{6}`= `\frac{s}{l + s}`

 `\Rightarrow `      `\frac{s}{l + s}`= `\frac{4}{15}` 

 `\Rightarrow`         14`s`= 4`l` +  4`s`   

 `\Rightarrow`         4`l` = 11`s`

Differentiating w.r.t.   `t`  ` 4 \frac{dl}{dt}` = `11  \frac{ds}{dt}` ...........(1)

but         `\frac{dl}{dt}`=1.1 =  `\frac{11}{10}`        ..........[given]

Then from (1)  we have   `11  \frac{ds}{dt}` = `4 \times  \frac{11}{10}` 

 `\Rightarrow`       `\frac{ds }{dt}` = `\frac{4}{10}` =0.4 `m`/ `sec`

hence the length of show is increasing at the rate of 0.4 `m`/`sec`.

   

Question 19 A Man 2 `m` Tall Walk At A Uniform Speed Of 5 `km`/`hr` Away From A Lamp Post 6 `m` high.Find The Rate At Which The Length Of His Shadow Increases.

Answer 

 

  Let`AB` represents the lamp Post ,`PQ`represents the man and `PR` be the shadow of the man .Now it is given that `AB`=6 `m` and  `PQ`=2 `m`.

 `\Delta 's` `ABR` and `PQR` are similar

 `\frac{PQ}{AB}`= `\frac{PR}{AR}`=` \frac{PR}{AP + PR}`

fig 2

Let     `AP` = `l`  ,     `PR`=`s`

 `\frac{PQ}{AB}`=  `\frac{s}{l + s}`= `\frac{2}{6}`=  `\frac{1}{3}`

 `\Rightarrow`     3`s` = `l` + `s`  `\Rightarrow`      2`s` =`l`        

                                                               .......................(1)

 `\frac{dl}{dt}` = `2 \frac{ds}{dt}`      [Differentiating 1] 

 `\Rightarrow `      `\frac{ds}{dt}` = `\frac{1}{2} \frac{dl}{dt}` 

now  

 

`\frac{dl}{dt}`= 5 `km`/`hr` =2.5 `km` / `hr`      [Given]

so  `\frac{ds}{dt}` =  `\frac{1}{2}  \times 5` = `\frac{5}{2}` `km `/ `hr`

Hence ,the length of the shadow is increasing at the rate of 2.5 `km` / `hr`

 

Example 20 The Total Cost `C(x)` Associated With The Production Of `x` Units Of An Item Is Given By `C(x)` = `0.005  x^{3}`  - `0.02  x^{2}`  + `30 x` + `5000` Find The Marginal Cost When 3 units are produced?

Answer 

Total cost function,  `C(x)` = `0.005  x^{3}`  - `0.02  x^{2}`  + `30 x` + `5000`

Marginal cost  `(MC)` is the rate of change of totel cost with respect to output

`MC` =  `\frac{dC}{dx}`

`MC` =   `\frac{d}{dx}``(0.005  x^{3}`  - `0.02  x^{2}`  + `30 x` + `5000)`

 `\Rightarrow`        `MC` =  `0.015 x^{2}`  - `0.04x`  +  30

Marginal cost when 3 units are produced

`MC\mid _{x=3}` = ` 0.015( 3^{2} ) - 0.04(3)` + 30

=0.135 -0.12 + 30 = 30.015 =30.02    (approx)

Hence ,marginal cost when 3 units of the item are produced is `Rs`. 30.02 (approx)