APPLICATION OF DERIVATIVES
(SUM 11 TO 15)
ALL EDUCATION BOARD
Question 11.The Volume of spherical balloon is increasing at the rate of 20` cm^{3}` /sec20.Find the rate of change of its surface area at the instant when its radius is 8 cm.``
Answer
Let .`r` be the radius of the balloon, .`V`
Be the volume and .`S`be the surface area at any time t.
So. `V`=`\frac{4}{3} \pi r^{3}` and `S`=`4 \pi r^{2}`
Rate of change( increase) of volume w.r.t.t
=20 `cm^{3}` /sec. (Given)
`\frac{dV}{dt}` =20
now, `V`= `\frac{4}{3} \pi r^{3}`
`\Rightarrow` `\frac{dV}{dt}`= `4 \pi r^{2}. \frac{dr}{dt}`
`\Rightarrow` 20=`4 \pi r^{2}. \frac{dr}{dt}`
`\Rightarrow` `\frac{dr}{dt}`
= `\frac{20}{4 \pi r^{2} }`= `\frac{5}{ \pi r^{2} }`
Also , `S`= `4 \pi r^{2}`
`\Rightarrow` `\frac{dS}{dt}` = `8 \pi r. \frac{dr}{dt}`
=`8 \pi r. \frac{5}{ \pi r^{2} }`= `\frac{40}{r}`
[ `\frac{dr}{dt}`= `\frac{5}{ \pi r^{2} }` ]
Question 12 The Radius Of A Sphere Is Increasing At The Rate Of 0.02 Cm Per Minute.At What Rate Is The Weight Varying When The Radius Is 15Cm And The Material Weighs 0.3 Gm Per c.c.
Answer
Let `r` be the radius of the sphere and `V` be the volume of the sphere at time t.
`V`= `\frac{4}{3} \pi r^{3}`
Rate Of change (increasing) of radius w.r.t.t=
`\frac{dr}{dt}`=0.02 cm/min [Given]
If `m` be the density,then weight of sphere =`w`
=`\frac{4}{3} \pi r^{3}.m`
Rate of variation (change) of weight of sphere
= `\frac{dW}{dt}`
= ` \frac{d}{dr} ( \frac{4}{3} \pi r^{3}.m ) \frac{dr}{dt}`
= `\frac{4}{3}.3 \pi r^{2}m \frac{dr}{dt}`
=`4 \pi r^{2}m. \frac{dr}{dt}`
Now when `r`=15 `cm`, `m`=0.03 `gm`/cc,
`\frac{dr}{dt}` =0.02 `cm`/min
`\frac{dW}{dt}` = `4 \pi (15)^{2} \times (0.3) \times (0.02)`
=`4 \pi \times 225 \times \frac{3}{10} \times \frac{2}{100}`
=` \frac{27 \pi }{5}``gm`/`min`
Question 13 An Edge Of A Variable Cube Is Increasing At The Rate Of 3 Cm Per Second.How Fast Is The Volume Of Cube Increasing When The Edge Is 10 Cm Long.
Answer
Let `a` be the edge and `V` be the volume of the cube at time `t`.
`V`= `a^{3}`
Rate of change (increase) of edge of cube =3 `cm`/`sec`
[Given]
` \frac{da}{dt}` =3
Rate of change of volume
= `\frac{dV}{dt}`=` \frac{d}{dt}( a^{3} )`
= `3 a^{2} \frac{da}{dt}` =`9 a^{2} cm^{3}` /`sec`
[ `\frac{da}{dt}`=3 ]
Rate of change of volume when the edge is 10 `cm` long
=`9 \times (10)^{2}` =900 `cm^{3}` /`sec`
Question 14 The Volume Of Cube Is Increasing At The Rate Of 9 Cubic Centimetres Per Second.How Fast Is The Surface Area Increasing When The Length Of An Edge Is 10 `cms`.?
Answer
Let `a`be the edge ,`V` be the volume And `S`Be the surface area of the cube at any time t.
`S`=6 `a^{2}` And `V`= `a^{3}`
Rate of change (increase) Of volume w.r.t. `t` =9 `cm^{3}`/`sec `
`\frac{dV}{dt}`=9
now `V`= `a^{3}` `\Rightarrow` `\frac{dV}{dt}` =3 `a^{2} \frac{da}{dt}`
`\Rightarrow` 9= `3 a^{2} \frac{da}{dt}`
`\Rightarrow` `\frac{da}{dt}` = `\frac{9}{3 a^{2} }`= `\frac{3}{ a^{2} }`
Also `S`= 6 `a^{2}`
`\frac{dS}{dt}`= 12`a` `\frac{da}{dt}`
=12`a` `\times \frac{3}{ a^{2} }`
= `\frac{36}{a}`
when the length of edge is 10 `cm` , the rate of change of the surface area
= `\frac{36}{10}`=`3.6 cm^{2}` /`sec`
Question 15 The Length `x` of a Rectangle is Decreasing At The Rate of 3 `cm`/Minute and the Width `y` is Increasing at the Rate Of 2 `cm` /Minute .When `x`=10 `cm` and `y`= 6 `cm`,Find The Rate Of Change Of
(`1`) The Perimeter And (`2`) The area of rectangle .
Answer Here `x` is the length and `y` is the width of rectangle which are decreasing at the rate of 3 `cm` /`min` and increasing at the rate of 2 `cm` /`min` .respectively.
`\frac{dx}{dt}` = - 3 `cm`/ `min` And `\frac{dy}{dt}` =2 `cm`/ `min`
Let `P` be the perimeter and `A` be the area of rectangle at any time `t`
(`1`) Perimeter of rectangle , `P` = `2(x + y)`
`\frac{dP}{dt}`=`2 ( \frac{dx}{dt} + \frac{dy}{dt} )` =`2( - 3 +2 )` `cm`/`min`
= - 2 `cm` / `min`
(`2`) Area of rectangle , `A`=`xy`
`\frac{dA}{dt}` =`x \frac{dy}{dt} + y \frac{dx}{dt}`
When `x`=10 `cm` and `y`= 6 `cm`, the rate of change of area
=`10 \times 2` + ` 6 \times (-3)` = `20 - 18` = `2 cm^{2}` /`min`.
Hence ,the rate of decrease of perimeter of the rectangle is 2 `cm` / `min`
0 Comments