Application of Derivatives NCERT Solutions
NCERT Exercise 6.1
Question: (1)Find the rate of change of the area of a circle with respect to its radius `r` when (i)`r= 3 cm `and (ii)`r=4cm`.
Solution:
Let`A` be area of the circle.
Now `A= pir^2`
Differentiating both sides w.r.t. `r` ,we have
`(dA)/(dr) = pi xx2r = 2pi r`
(i)Here `r=3 cm.`
`:.(dA)/(dr) = 2pi xx 3 = 6pi cm^2//cm`
(ii)Here `r= 4cm`
`(dA)/(dr) =2pi xx4 = 8pi cm^2//cm`
Question: (2) The volume of a cube is increasing at the rate of `cm^2//` sec. How fast is the surface area increasing when the length of an edge is `2 cm`?
Solution:
Let `x` be the length of an edge of cube, `v` be the volume of cube and a be the surface area of cube at any time `t`.
Then `(dv)/(dt)=8cm^3//sec `
Now `v=x^3 and s=6x^2`
Differentiating both sides w.r.t. `t` ,we have
`(dv)/(dt = 3x^2 (dx)/(dt) `
`=> 8= 3x^2 (dx)/(dt)`
`=>(dx)/(dt)=8/(3x^2)`
Also `(ds)/(dt)=12 xx(dx)/(dt)`
`=>(ds)/(dt)=12x xx 8/(3x^2)= 32/ x`
` :. (ds)/(dt)=(32)/12 = 8/3 cm^2//sec`.
Thus rate of change of surface area is `8/3 cm^2 //sec`.
Question: (3)The radius of a circle is increasing uniformly at the rate of `3 cm// s`. find the rate at which the area of the circle is increasing when the radius is` 10` cm.
Solution:
Let` r` be the radius of circle and A be the area of circle at any time `t`.
Then `(dr)/(dt)= 3 cm//s` and `r= 10 cm`.
Now `A = pir^2`
Differentiating both sides w.r.t. `t` ,we have
`(dA)/(dt) = 2pir(dr)/(dt) `
`=>(dA)/(dt)= 2pi xx 10 xx 3`
` = 60 pi cm^2//`
Thus rate of change of area is `60 pi cm^2//s.`
Question: (4) An edge of a variable cube is increasing at the rate of `3 cm//s`. How fast is the volume of the cube increasing when the edge is `10 cm` long?
Solution:
Let x be the length of an edge of cube and v be the volume of cube at any time `t`.
Then `(dx)/(dt)=3 cm //s and x =10 cm.`
Now ` v= x^3 => (dv)/(dt)`
` 3x^2 (dx)/(dt)`
`=>(dv)/(dt)`
` =3xx(10)^2 xx 3`
` = 900 cm^3//s`.
Thus rate of change of volume is `900 cm^3/s`
Question:(5) A stone is dropped into a quiet lake and waves move in circles at the speed of `5 cm//s`. At the instant when the radius of circular wave is 8cm, how fast is the enclosed area increasing?
Solution:
Let `x` be the radius of the circle and `A` be the area of circle at any time `t`.
Then `(dx)/(dt)= 5cm//s` and `x= 8 cm`
Now ` A = pi x^2 `
` => (dA)/(dt) =`
`2pi x (dx)/(dt)`
`=> (dA)/(dt)`
`= 2pi xx 8 xx 5`
` = 80 pi cm^2//s`
Thus rate of change of area is `80 pi cm^2//s`.
Question: (6) The radius of a circle is increasing at the rate of `0.7cm//s`. What is the rate of increase of its circumference?
Solution:
Let `r` be the radiius of circle and `c` be the circumference of circle at any time `t`
Then `(dr)/(dt)`
`= 0.7 cm//s`
Now ` c= 2 pir`
Differentiating both sides w.r.t. `t` ,we have
`(dc)/(dt) = 2pi(dr)/(dt) `
`=>(dc)/(dt)= 2pi xx 0.7`
`=1.4 pi cm// s`
Thus rate of change of circumference is `1. 4pi cm//s`
Question: (7) The length `x` of a rectangle is decreasing at the rate of `5cm//`minute and the width `y` is increasing at the rate of `4 cm//minute` When `x=8cm` and `y = 6 cm`, find the rates of change of (a)the perimeter and(b) the area of the rectangle.
Solution:
Here `x=8cm., y-6cm,(dx)/(dt)`
`=-5 cm//` minute and `(dy)/(dt)=4 cm//`minute.
Let `P` be the perimeter of rectangle and `A` be the area of rectangle at any time `t`
Then `P= 2(x+y)` and `A=xy`
Differentiating both sides w.r.t. `t` ,we have
`(dp)/(dt) = 2((dx)/(dt)+dy/dt) `
`:. (dP)/(dt)=2(-5+4)=-2cm //`Minute
Also `(dA)/(dt)=x (dy)/(dt)+y(dx)/(dt)`
`:. (dA)/(dt)= 8 xx 4 + 6 xx -5 `
`32 - 30 =2 cm^2//` minute.
Thus rate of change of perrimeter is `-2 cm// `minute and rate of change of area is `2 cm^2//` minute.
Question:(8) A baloon, Which always remains spherical on inflation, is being inflated by pumping in `900` cubic centimetres of gas per second. find the rate at which the radius of the ballon increases when the radius is `15` cm.
Solution:
Let `r` be the radious of spherical baloon and `v` be the volume of spherical balloon at any time `t`.
Then ` (dv)/(dt)= 900 cm^3 //` sec and `r=15 cm.
Now `v=4/3 pi r^3`
Differentiating both sides w.r.t. `t` ,we have
`(dv)/(dt) = 4/3 pi xx 3r^2 (dr)/(dt)`
`900=4pi r^2 (dr)/(dt) `
`:.(dr)/(dt) = (900)/ (4pi r^2)`
` = (900)/(4pi xx (15)^2)`
` 1/picm//` Sec.
Thus rate of change of radius is `1/ pi cm//` sec.
Question: (9) A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is `10` cm.
Solution:
Let `r` be the radius of spherical ballon and `v` be its volum.
Then `r=10` cm
Now ` v= 4/3 pi r^2`
Differentiating both sides w.r.t. `t` ,we have
`(dv)/(dt) = 4/3 pi xx 3r^2`
`=> (dv)/(dr) = 4pi x^2`
` (dv)/(dr) = 4pi xx (10)^2`
`= 400 pi cm^3// `cm.
Thus rate of change of volume is `400 pi cm^3// `cm
Question: (10) A ladder `5m` long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of `2 cm//` s. How fast is its ladder 4m away from the wall? Fig Page no 424
Solution:
Let the foot of the ladder be `x` m from the wall and let the ladder reach at a height of `y` m on the wall at any time `t`.Then `OA=xm, OB=y m` and`AB = 5m `
`(dx/dt)= 2 cm //`sec
`= 2/100 m//`sec
In `Delta OAB`, by pythagoras theorem
` x^2+ y^2 = 25`
Differentiating both sides w.r.t. `t` ,we have
`2x(dx)/(dt)+2y(dy)/dt =0`
` => 2x xx 2/(100)+ 2y (dy)/dt=0`
`2y (dy)/dt = - x/ 25 `
`=> (dy)/dt =-x/(50 y)`
Now x= 4, Putting value of `x` in(i), we have
`(4)^2 +y^2 = 25 `
`=> y^2 =25 - 16 - 9 `
`=> y=3`
Putting value of `x` and `y` in (ii), we have
`(dy)/dt =- 4/(50xx3)`
`=-2/75 m//` sec
`(dy)/ (dt)`
` =- 2/75 xx 100 cm//`sec
`=- 8/3 cm// `sec.
Thus rate of decrease of height on the wall is 8/3 cm//` sec
Question: (11) A particle move along the curve `6y=x^2+2`. Find the point on the curve at which the `y` coordinate is changing `8` times as fast as the `x` -coordinate.
Solution:
Here `6y=x^2+2` and `(dy)/ dt = 8 (dx)/dt`
Differentiating both sides w, r. t. `t` we have
`6 (dy)/dt = 3x^2 (dx)/dt`
`:.6 xx 8 (dx)/dt `
` 3x^2 (dx)/ dt`
`=> 3x^2 = 48`
` => x^2 =16 `
` => x^2 = +- 4`
When `x=4`
`6y=(4)^3 + 2 `
`=>6y = 66`
`=> y=11`
When `x= -4`
`6y=(-4)^3 +2`
`=> 6y=-64`
`=> y= (-64)/6 =(-32)/3`
Thus required points are `(4, 11)` and `(-4,(-32)/3)`.
Question: (12) Theradius of an air bubble is increasing at the rate of `1/2 cm//` s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
Let `r` be the radius of bubble and `v` be the volume of bubble at any time `t`.
Then ` (dr)/(dt)=1/2 cm//` s and `r=1cm`
now `v= 4/3 pi r^2`
Differentiating both sides w.r.t. `t` , we have
` (dv)/(dt)= 4/3 pi xx 3r^2 (dr)/(dt)`
` => (dv)/(dt)`
`= 4 pi r^2 xx 1/2`
`=>(dv)/(dt) = 2pi r^2`
`:. (dv)/(dt) = 2pi xx (1)^2`
` = 2pi cm^3//`s
Thus rate of change of volume is `2pi cm^3//`s
Question: (13) A balloon, which always remain spherical, has a variable diameter `3/2 (2x+1)` Find the rate of change of its volume with respect to `x`.
Solution:
Let `r` be the the radius of sphrical balloon and `v` be the volume of sphrical balloon.
Then `r =3/4 (2x+1)` and `v=4/3 pir^3`
Differentiating both sides w.r.t. `x` ,we have
`(dr)/(dx)=3/4 xx2 =3/2`
Also `(dv)/dx`
` =4/3 pi xx3r^2 (dr)/dx `
=>`(dv)/dx`
` = 4pi r^2 3/2 `
` = 6pir^2`
`:. (dv)/dx=6pi[3/4(2x+1)]^2`
` = 6pi xx 9/16 (2x + 1)^2`
`=27/8 pi(2x+1)^2`
Thus rate of change of volume is `27/8 pi (2x +1)^2`
Question: (14) Sand is pouring from a pipe at the rate of `12 cm^3//` s. The falling sand forma, a cone on the ground in such a way that the height of the cone is always one -sixth of the radius of the base. How fast is the height of the sand cane increasing when the height is `4` cm?
Solution:
Let `r` be the radius , h be the height and ` v`be the volume of sand cone at any time `t`.
Then ` (dv)/dt = 12 cm^3//` s and ` h= 1/6 r `
`=>r =6h`
Now ` v= 1/3 pi r^2 h`
`:. v= 1/3 pi xx (6h)^2.h`
`= 12pih^3`
Differentiating both sides w.r.t. `x` ,we have
`(dv)/(dt)=12 pi xx 3h^2 (dh)/dt`
`=> (dv)/dt =36 pih^2 (dh)/dt`
` :. 12 = 36 pi h^2 (dh)/dt `
`=> (dh)/dt =1/(3pi h^2)`
When `h=4 cm`
`dh/dt = 1/(3pi xx (4)^2)`
` 1/48 pi cm//` sec
Thus rate of change of height is `1/48 pi cm//` sec.
Question: (15) The total cost `C (x)` in Rupees associated with the production of `x` units of an item is given by
`C(x)= 0.007x^3 - 0.003x^2 + 15 x + 4000`
Find the marginal cost when `17` units are produced.
Solution:
Here `C(x) = 0. 007x^3 -0.003x^2 ` `+ 15 x + 4000` and `x =17`
We know that marginal cost is rate of change of otal cost.
`:.`Marginal cost `= (dc)/dx`
now `C(x)= 0.007 x^2 - 0.003x^2 + 15x + 4000`
Differentiating both sides w.r.t. `x` ,we have
`(dC)/(dx)=0.021x^2- 0.006x+15`
` = 0. 021xx(17)^2 - 0.006 xx 17 +15`
Thus marginal cost is Rs. `20.967` when `17 units are produced.
Question: (16) The total revenue in Rupees received from the sale of `x` units of a product is given by
` R(x) = 13x^2 + 26x + 15`
Find the marginal revenue when `x=7`
Solution:
Here `R(x) = 13x^2+ 26 x + 15 and x =7`
We know that marginal revenue is rate of change of total revenue.
`:.` Marginal revenue =` (dR)/dx `
Now ` R(x) = 13x^2 +26 x +15`
Differentiating both sides w.r.t. `x` ,we have
`(dR)/(dx)=26x+26 `
`= 182+26=208`
Thus marginal revenue is `Rs 208 ` When `7` units are produced.
Choose the correct answer in the Exercise `17` and `18`
(17)The rate of change of the area of a circle with respect to its radius `r` at `r= 6cm is
(i)`10 pi`
(ii)` 12 pi`
(iii)`8 pi`
(iii)` 11pi`
Solution:
Let `A` be the area of the circle
Now `A = pir^2`
`:.(dA)/dr =2pi r`
`(dA)/dr =2pi xx 6 = 12pi`
`(:. r = 6 cm)`
Thus answer is (ii)
Question: 18. The total revenue in Rupees received from the sale of `x` units of a product is given by `R(x)=3x^2+36x+5.` The marginal revenue , when `x= 15` is
(a)116
(b)96
(c)90
(d)126
Solution:
Here `R(x)= 3x^2+36x+5`
`(dR)/dx`
` =6x+36`
`(dR)/dx =6xx15+36=126`
Thus answer is (d)
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