Application of Derivatives NCERT Solutions
NCERT Exercise 6.4(2)
Question: 1. (xii) `(26.57)^(1/3)`
Solution:
Let `y= x^(1/3)`
`(dy)/(dx)=1/(3x)^(2/3)`
Let `x = 27` and `x+Delta x= 26.57`
` Delta x=(x+Delta x)-x`
` 26.57 -27 =-0.43`
`Delta y= (x+Delta x)^(1/3)-(x)^(1/3)`
`=(26.57)^(1/3)-(27)^(1/3)`
`=26.57^(1/3)-3`
`:. (2657)^(1/3)=3+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by`
` dy= (dy)/(dx).Delta x `
`= 1/(3x)^(2/3)xx (-0.43)`
`= (-0.43)/(3xx(27)^(2/3)) `
=` (-0.43)/(3xx9)`
`=-0.016`
`:. (26.57)^(1/3) `
` =3-0.016=2.984`
Thus approximate value of `(26.57)^(1/3)` is `28.94`
(xiii) `(81.5)^(1/4)`
Solution:
Let `y= x^(1/4)`
`(dy)/(dx)= 1/ 4x^(3/4)`
`:. Delta x= (x+Delta x)-x`
`=81.5-81 =0.5`
`Delta y=`
` (x+ Delta x)^(1/4) - (x) ^(1/4)`
`=(815)^(1/4)-(81)^(1/4)=(815)^(1/4)-3`
`:. (815)^(1/4) = 3+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy= (dy)/(dx).Delta x`
`= 1/(4x)^(3/4) xx (0.5)`
`=0.5/4xx(81)^ 3/4`
=`(0.5)/(4xx27)=0.0046`
`:. (815)^(1/4)=3+0.0046 = 3.0046`
Thus approximate value of `(81.5)^(1/4)` is `3.0046`
(xiv) `(3.968)^(3/2)`
Solution:
Let `y= x^(3/2)`
`(dy)/(dx)=3/2 x^(1/2)`
Let `x=4` and `x+Delta x=3.968`
`:. Delta x = (x+ Delta x)-x`
` =3.968-4 = -0.032`
`:. Delta x=(x+Delta x)-x`
`=3.968 -4=-0.032`
`Delta y= (x+ Delta x)^(3/2)-(x)^(3/2)`
`(3.968)^(3/2)-(4)^(3/2)=(3.968)^(3/2)-8`
`:. (3.968)^(3/2)= 8+Delta y`
Since `dy` is approximately equal to `Delta y` and is ginen by
`dy =(dy)/(dx) .Delta x`
`3/2 x^(1/2).(-0.032)`
` = 3/2xx(4)^(1/2)xx(-0.032)`
` 3/2 xx 2 xx (-0.032)= -0.096`
`:. (3.968)^(3/2) = 8-0.096 = 7.904`
Thus approximate value of `(3.968)^(3/2)` is `7.904`
(xv) `(32.15)^(1/5)`
Solution:
Let `y= x^(1/5)`
`= 1/(5x)^(4/5)`
Let `x= 32` and `x+Delta x = 32.15`
`:. Delta x= (x+ Delta x)-x `
`=32.15 - 32= 0.15`
` Delta y= (x+Delta x)^(1/5)-x^(1/5)`
`=(3215)^(1/5) -(32)^(1/5)`
`=(3215)^(1/5)-2`
`:. (3215)^(1/5)=2+ Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy = (dy)/(dx).Delta x`
`1/(5x)^(4/5) .015`
`(0.15)/(5(32)^(4/5))`
`= (0.15)/(5xx16)=0.0019`
`:. (3215)^(1/5)= 2+0.0019 = 2.0019`
Thus approximate value of `(3215)^(1/5)` is `2.0019`
Question: (2)Find the approximate value of `f(2.01)` Where `f(x)=4x^2+5x+2`
Solution:
Here `f(x)=4x^2+5x+2`
`f(x)= 8x+5`
Let `x=2` and `x+Delta x=2.01`
`Delta x= (x+Delta x)-x`
`= 2.01-2=0.01`
`f(2.01)=f(x+Delta x)`
Now `Delta = f(x+Delta x)-f(x)`
`:. f(x+Delta x) = f(x) + Delta y`
`= f(x) + f'(x).Delta x`
`(4x^2+5x+2) + (8x+5)(0.01)`
`=(4xx2^2+5 xx 2+2)+(8xx2+5)(0.01)`
`[16+10+2]+(21xx0.01)`
`=28 +0.21 =28.21`
Thus approximate value of `f(2.01)` is `28.21`
Question:(3) Find the approximate value of `f(5.001)` where `f(x)=x^2-7x^2+15`
Solution:
Here `f(x)=x^3-7x^2+15`
`f'(x) =3x^2-14x `
Let `x=5` and `x+Delta x = 5.001`
`:.Delta x=(x+Delta x)-x`
` 5.001-5=0.001`
`f(5.001)=f(x+Delta x)`
Now `Deltay=f(x+Deltax)-f(x)`
`:.f(x+Delta x)= f(x)+Delta y`
` f(x) + f'(x).Delta x `
`[:.Delta y=f'(x).Delta x]`
`=(x^3-7x^2+15)+(3x^2-14x(0.001)`
`=(5^3-7xx5^2+15)+(3xx5^2-14xx5^2-14xx5)[0.001]`
` [125-175+15]+[75-70][0.001]`
`-35 + 0.005=-34.995`
Thus approximately value of `f(5.001)` is `-34.995`
Question:(4) Find the approximate change in the valumee v of a cube of side `x ` metres caused byincreasing the side by `1%`
Solution:
Here side of cube is `x m`.
Then `v=x^3`
`:. (dv)/(dx) =3x^2`
Let `Delta x` be change inside `=1%0fx=0.01x`
Now change in volume `Delta v =(dv)/(dx).Delta x`
`= (3x^2)(0.01x)= 0.03x^3m^3`
Question:(5)Find the approximate change in the surface area of a cube of side `x` metres caused by decreasing the side by `1%`
Solution:
Here side of cube is `x` m
Then `s=6x^2`
`:. (ds)/(dx)=12x`
Let `Delta x` be change in side `= 1% 0f x =- 0.01x`
Now Change in surface area `Delta v =(dv)/(dx).Delta x`
`= (12x) (-0.01 x) =-0.12x^2m^2`
Question:(6) If the radius of a sphere is measured as `7 m ` with an error of `0.02 m` then find the approximate error in calculating its volume..
Solution:
Let `r` be the radius of the sphere and`Delta r ` be the error in measuring the radius.
Then 1r=7m ` and ` Delta r =0.02 m`
Now volume of a sphare is given by
`v = 4/3 pir^3`
`(dv)/(dr)= 4/3 pi xx3r^2 =4pir^2`
`:. (dv) = (dv)/(dr).Delta r = ( 4pir^2). Delta r`
` =4pi xx7^2xx0.02= 3.92pi m^3`
Question: (7) If the radius of a sphere is measured is `9 m` with an error of `0.03 m , then find the approximate error in calculating its surface area.
Solution:
Let `r` be the radius of the sphare and `Delta r` be the error in measuring the radius.
Then `r=9m` and `Delta r =0.03m
Now surface area of a sphere is given by`
`s = 4pi r^2`
`(ds)/(dr) = 8pir`
`:. ds=(ds)/(dr).Delta r`
` =(8pir).Delta r`
`=8pixx9xx0.03`
`=2.16pi m^2`
Question: (8) If `f(x)=3x^2+15x+5`,
(A)` 47.66`
(B)`57.66`
(c) `67.66`
(D) `77.66`
Solution:
Here `f(x)= 3x^2+15x+5`
`f'(x)=6x+15`
Let `x=3 and ` x+Delta x =3.02`
`:. Delta r = (x+Deltax) -x 3.02 - 3=0.02`
`f(3,02) = f(x+Delta x)`
Now `Deltay = f(x+Delta x)-f(x)`
`:. f(x+Delta x)= f(x)+Delta y`
`=f(x)+ f'(x).Delta x `
`[&because ; Delta y = f'(x).Delta x]`
`=(3x^2+15x+5)+(6x+15).Delta x `
`[3xx3^2+15xx3 + 5]+ [6xx3+15] (0.02)`
`[27+45+5]+ [18+15] (0.02)`
`=77+33xx0.02`
`=77+0.66 = 77.66`
Thus answer is (D)
Question:(9) (The approximate change in the volume of a cube of side `x` metres caused by increasing the side by `3%` is
(A)`0.06x^3m^3`
(B)`0.6x^3m^3`
(C)`0.09 x^3 m^3`
(D)`0.9x^3m^3`
Solution:
Here side of cube is `x m`
Then ` v=x^3`
`(dv)/(dx) =3x^2`
Let `Delta x` be change in side ` = 3% of x=0.03x`
Now change in volume `Delta v =(dv)/(dx).Delta x`
`=(3x^2)(0.03x)=0.09x^3 m^3`
Thus answer is (c)
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