APPLICATION OF DERIVATIVES
(SUM 5 TO 10)
ALL EDUCATION BOARD
Question 5. The Side Of A Square Is Increasing At a Rate Of 0.2 cm/Sec .Find The Rate Of Increase Of the Perimeter Of The Square.
Answer.
Let the side of square be `x` and the Perimeter of squre be R at time t .
so P= `4x`
Rate of change (increase) of side x w.r.t. t =0.2 cm/sec (given)
`\frac{dx}{dt} `=0.2
so Rate of change of perimeter =`\frac{dP}{dt} `
= `\frac{d}{dt}(4x)=4. \frac{dx}{dt} `
= 4 `times` 0.2
=0.8 cm/sec.
Question 6. The Radius Of The Circle is Increasing Uniformly At The Rate Of 3 Cm Per Second .At What Rate is The Area Increasing Whe The Radius Is 10 Cm.
Answer
Let `r` be the radius of the circle And `A ` be the area of the circle at time `t` .
so A= `/pi r^{2}`
so Rate of change (increase) of radius w.r.t. `t=
`\frac{dr}{dt} `=3 cm/sec. {given}
so Rate of change of area =`\frac{dA}{dt} `
=`\frac{d}{dr} ( \pi r^{2} ). \frac{dr}{dt}`
=`2 \pi r. \frac{dr}{dt}`
=`2 \pi r \times 3`
=`6 \pi r cm^{2}` /sec.
When r =10,the rate of change of area =`6 \pi \times 10` =`60 \pi cm^{2}` /sec
Question 7. A Spherical Soap Buble Is Expanding So That Its Radius Is Increasing At The Rate Of 0.02 Centimeters Per Sec.At What Rate Is The Surface Area Increasing When Its Radius Is 4 cm? (Take `\pi `=3.14).
Answer
Let r be the radius And S be the surface area of spherical soap bubble at time t.
S=`4\pi r^{2}`
Rate of change (increase) of radius w.r.t. time t
=`\frac{dr}{dt}`=0.02 cm/sec ( Which is given)
Rate of change of surface area =`\frac{dS}{dt}`
=`\frac{d}{dr} (4 \pi r^{2}) \frac{dr}{dt}`
=`8 \pi r \frac{dr}{dt}`
=`8 \pi r(0.02)`
=`0.16 \pi r` [ `\frac{dr}{dt}`=0.02 cm/sec ]
when r =4 ,the rate of change of surface area
=0.16`\times`3.14`\times`4 =2.0096 `cm^{2}`/sec.
Question 8 The Radius Of A Circular Soap Bubble Is Increasing At The Rate Of 0.02 cm /sec .Find The Rate Of Increase Of Its Volume When The Radius Is 4 cm.
Answer
Let r be the radius and V be the Volume of The Circular Soap Bubble At Time t.
V=`\frac{4}{3} \pi r^{3}`
` \frac{dV}{dr}` =`\frac{4}{3} \times 3\pi r^{2}`
= `4 \pi r^{2} `
Rate of change (increase) of radius w.r.t t = `\frac{dr}{dt}`
=0.2 cm/sec {Given}
So rate of change of volume = `\frac{dV}{dt}`
= `\frac{d}{dr}(\frac{4}{3} \pi r^{3}) .\frac{dr}{dt}`
=` \frac{4}{3}` ` \times 3 \pi r^{2}. \frac{dr}{dt}`
=`4 \pi r^{2}. \frac{dr}{dt}`
=`4 \pi r^{2} \times 0.2` = `0.8 \pi r^{2} cm^{3}` /sec
{ `\frac{dr}{dt}`=0.2 cm/sec}
When r=4, rate of change of volume =
`0.8 \times (4)^{2} cm^{3}` /sec
=`12.8 \pi cm^{3}` /sec
`Question 9 The Surface Area Of A Spherical Bubble Is Increasing At The Rate Of 2` cm^{3}`/sec.Find The Rate At Which The Volume Of The Bubble Is Increasing At The Instant Its Radius Is 6 cm.
Answer
Let r be the radius ,S be the surface area And V be the volume of the spherical bubble at time t.
S=`4 \pi r^{2}` And V= `\frac{4}{3} \pi r^{3}`
Rate of change(increase) of Surface area w.r.t. t =2`cm^{2}`/sec [Given]
`\frac{dS}{dt}`=2
`\Rightarrow` `\frac{d}{dt}(4 \pi r^{2} )`=2
`\Rightarrow` `8 \pi r. \frac{dr}{dt}`=2
`\Rightarrow` `\frac{dr}{dt}` = `\frac{1}{4 \pi r}`
...........................1
Also V= `\frac{4}{3} \pi r^{3}`
So Rate of change of Volume w.r.t. t= `\frac{dV}{dt}`
=` \frac{d}{dt}(\frac{4}{3} \pi r^{3}) `
=`4 \pi r^{2} . \frac{dr}{dt}`
=`4 \pi r^{2} \times \frac{1}{4 \pi r}`=r [Using 1]
Answer
Let `r` be the radius and `R` be the circumfrence of the circle at time t.
`R`= `2 \pi r`
Rate of change (increase) of Radius w.r.t. t =0.7 cm/sec
[Given]
`\frac{dr}{dt}` =0.7
Rate of change (increase) of circumfrence
= `\frac{dR}{dt}`
= `\frac{d}{dr}(2 \pi r). \frac{dr}{dt}`
=`2 \pi \times 0.7`
= `2 \times \frac{22}{7} \times \frac{7}{10}`
= 4.4 cm/sec.
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