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)
0 Comments