第12类NCERT解决方案-数学第I部分–第6章导数的应用–练习6.4

问题1.使用微分，找到以下每个小数点后3位的近似值。

(i)√25.3

(ii)√49.5

(iii)√0.6

(iv)(0.009) ^1/3

(v)(0.999) ^1/10

(六)(15) ^1/4

(vii)(26) ^1/3

(viii)(255) ^1/4

(ix)(82) ^1/4

(x)(401) ^1/2

(西)(0.0037) ^1/2

(xii)(26.57) ^1/3

(xiii)(81.5) ^1/4

(十四)(3.968) ^3/2

(xv)(32.15) ^1/5

解决方案：

(i) √25.3

We take y = √x 

Let x = 25 and △x = 0.3

△y = √(x +△x) – √x

= √25.3 – √25

= √25.3 – 5

or √25.3 = 5 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{1}{2√x}(0.3)$

= $\frac{1}{2√25}(0.3)$

= 0.3/10

= 0.03

Hence, the approximate value of√25.3 = 5 + 0.03 = 5.03

(ii)√49.5

We take y = √x 

Let x = 49 and △x = 0.5

△y = √(x +△x) – √x

= √49.5 – √49

= √49.5 – 7

or √49.5 = 7 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{1}{2√x}(0.5)$

= $\frac{1}{2√49}(0.5)$

= 0.5/14

= 0.0357

Hence, the approximate value of √49.5 = 7 + 0.0357 = 7.0357

(iii) √0.6

We take y = √x 

Let x = 0.64 and △x = -0.04

△y = √(x +△x) – √x

= √0.6 – √0.64

= √0.6 – 0.8

or √0.6 = 0.8 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{1}{2√x}(-0.04)$

= $\frac{1}{2√0.64}(-0.04)$

= -0.04/1.6

= -0.025

Hence, the approximate value of √0.6 = 0.8 – 0.025 = 0.775

(iv) (0.009)^1/3

We take y = x^1/3 

Let x = 0.008 and △x = 0.001

△y = (x +△x)^1/3 – x^1/3

= (0.009)^1/3 – (0.008)^1/3

= (0.009)^1/3 – 0.2

or (0.009)^1/3 = 0.2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{0.001}{3x^{\frac{2}{3}}}$

= $\frac{0.001}{3(0.008)^{\frac{2}{3}}}$

= 0.0083

Hence, the approximate value of (0.009)^1/3 = 0.2 + 0.0083 = 0.2083

(v) (0.999)^1/10

We take y = x^1/10 

Let x = 1 and △x = -0.001

△y = (x +△x)^1/10 – x^1/10

= (1)^1/10 – (0.001)^1/10

= (1)^1/10 – 1

or (1)^1/10 = 1 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{-0.001}{10x^{\frac{9}{10}}}$

= $\frac{-0.001}{10(1)^{\frac{9}{10}}}$

= -0.0001

Hence, the approximate value of (0.999)^1/10 = 1 – 0.0001 = 0.9999

(vi) (15)^1/4

We take y = x^1/4

Let x = 16 and △x = -1

△y = (x +△x)^1/4 – x^1/4

= (15)^1/4 – (16)^1/4

= (15)^1/4 – 2

or (15)^1/4 = 2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{-1}{4x^{\frac{3}{4}}}$

= $\frac{-1}{4(16)^{\frac{3}{4}}}$

= -1/32

Hence, the approximate value of (15)^1/4 = 2 – 1/32 = 1.9687

(vii) (26)^1/3

We take y = x^1/3

Let x = 27 and △x = -1

△y = (x +△x)^1/3 – x^1/3

= (26)^1/3 – (27)^1/3

= (26)^1/3 – 3

or (26)^1/3 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{-1}{3x^{\frac{2}{3}}}$

= $\frac{-1}{3(27)^{\frac{2}{3}}}$

= -1/27

Hence, the approximate value of (26)^1/3 = 3 – 1/27 = 2.962

(viii) (255)^1/4

We take y = x^1/4

Let x = 256 and △x = -1

△y = (x +△x)^1/4 – x^1/4

= (255)^1/4 – (256)^1/4

= (255)^1/4 – 4

or (255)^1/4 = 4 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{-1}{4x^{\frac{3}{4}}}$

= $\frac{-1}{4(256)^{\frac{3}{4}}}$

= -1/256

Hence, the approximate value of (255)^1/4 = 4 – 1/256 = 3.996

(ix) (82)^1/4

We take y = x^1/4

Let x = 81 and △x = 1

△y = (x +△x)^1/4 – x^1/4

= (82)^1/4 – (81)^1/4

= (82)^1/4 – 3

or (82)^1/4 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{1}{4x^{\frac{3}{4}}}$

= $\frac{1}{4(81)^{\frac{3}{4}}}$

= 1/108

Hence, the approximate value of (82)^1/4 = 3 + 1/108 = 3.009

(x) (401)^1/2

We take y = x^1/2

Let x = 400 and △x = 1

△y = (x +△x)^1/2 – x^1/2

= (401)^1/2 – (400)^1/2

= (401)^1/2 – 20

or (401)^1/2 = 20 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{1}{2\sqrt{x}}$

= $\frac{1}{2\sqrt{400}}$

= 1/40

Hence, the approximate value of (401)^1/2 = 20 + 1/40 = 20.025

(xi) (0.0037)^1/2

We take y = x^1/2

Let x = 0.0036 and △x = 0.0001

△y = (x +△x)^1/2 – x^1/2

= (0.0037)^1/2 – (0.0036)^1/2

= (0.0037)^1/2 – 0.06

or (0.0037)^1/2 = 0.06 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{0.0001}{2\sqrt{x}}$

= $\frac{0.0001}{2\sqrt{0.0036}}$

= 0.0001/0.12

Hence, the approximate value of (0.0037)^1/2 = 0.06 + 0.0001/0.12 = 0.0608

(xii) (26.57)^1/3

We take y = x^1/3

Let x = 27 and △x = -0.43

△y = (x +△x)^1/3 – x^1/3

= (26.57)^1/3 – (27)^1/3

= (26.57)^1/3 – 3

or (26.57)^1/3 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{-0.43}{3x^{\frac{2}{3}}}$

= $\frac{-0.43}{3(27)^{\frac{2}{3}}}$

= -0.43/27

Hence, the approximate value of (26.57)^1/3 = 3 – 0.43/27 = 2.984

(xiii) (81.5)^1/4

We take y = x^1/4

Let x = 81 and △x = 0.5

△y = (x +△x)^1/4 – x^1/4

= (81.5)^1/4 – (81)^1/4

= (81.5)^1/4 – 3

or (81.5)^1/4 = 3 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{0.5}{4x^{\frac{3}{4}}}$

= $\frac{0.5}{4(81)^{\frac{3}{4}}}$

= 0.5/108

Hence, the approximate value of (81.5)^1/4 = 3 + 0.5/108 = 3.004

(xiv) (3.968)^3/2

We take y = x^3/2

Let x = 4 and △x = -0.032

△y = (x +△x)^3/2 – x^3/2

= (3.968)^3/2 – (4)^3/2

= (3.968)^3/2 – 8

or (3.968)^3/2 = 8 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{(-0.032)(3)\sqrt{x}}{2}$

= $\frac{(-0.032)(3)\sqrt{4}}{2}$

= -0.096

Hence, the approximate value of (3.968)^3/2 = 8 – 0.096 = 7.904

(xv) (32.15)^1/5

We take y = x^1/5

Let x = 32 and △x = 0.15

△y = (x +△x)^1/5 – x^1/5

= (32.15)^1/5 – (32)^1/5

= (32.15)^1/5 – 2

or (32.15)^1/5 = 2 + △y 

Here, dy is approximately equal to △y 

dy = (dy/dx)△x

= $\frac{0.15}{5x^{\frac{4}{5}}}$

= $\frac{0.15}{5(32)^{\frac{4}{5}}}$

= 0.15/80

Hence, the approximate value of (32.15)^1/5 = 2 + 0.15/80 = 2.0018

问题2。找到f(2.01)的近似值，其中f(x)= 4x ² + 5x + 2。

解决方案：

Find the approximate value of f(2.01)

So, f(2.01) = f(2 + 0.01)

x = 2 and △x = 0.01      

Given that f(x) = y = 4x² + 5x + 2       -(1)

On differentiating we get    

dy/dx = 8x + 5

dy = (8x + 5)dx = (8x + 5)△x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(2.01) = y + Δy        -(2)

f(2 + .01) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(2.01) = (4x² + 5x + 2) + (8x + 5)△x

= (4(2)² + 5(2) + 2) + (8(2) + 5)(0.01)

= 16 + 10 + 2 + 0.16 + 0.05

= 28.21

问题3.找到f(5.001)的近似值，其中f(x)= x ³ – 7x ² + 15。

解决方案：

Find the approximate value of f(5.001)

So, f(5.001) = f(5 + .001)

x = 5 and △x = 0.01, 

Given that f(x) = y = x³ – 7x² + 15        -(1)

On differentiating we get    

dy/dx = 3x² – 14x 

dy = (3x² – 14x) dx = (3x² – 14x) △x

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(5.001) = y + Δy        -(2)

f(5 + .001) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(5.001) = (x³ – 7x² + 15) + (3x² – 14x) △x

= ((5)³ – 7(5)² + 15) + (3(5)² – 14(5)) (0.001)

= 125 – 175 + 15 + 0.075 + 0.07

= -34.995

问题4.求出因边增加1％而引起的边数x米的立方体的近似变化。

解决方案：

Volume of cube = x³

V = x³

On differentiating we get    

dV/dx = 3x²

Given that Δx = 0.0

dV = $(\frac{dV}{dx}).Δx=3x.Δx$

dV = (3x²)(-0.01x)

= 0.03x³m³ 

Thus, the approximate change in volume is 0.03x³m³ 

问题5：找出因边减少1％而引起的边数x米的立方体的表面积的近似变化。

解决方案：

Surface area of cube = 6x²

S = 6x²

On differentiating we get    

dS/dx = 12x

Given that Δx = -0.01[-ve sign because of decrease]

dS = $(\frac{dS}{dx}).Δx=12x.Δx$

dS = (12x)(-0.01x)

= -0.12x²m²    

Thus, the approximate change in volume is -0.12x²m²

问题6.如果球体的半径测量为7m且误差为0.02m，则在计算其值时找到近似误差。

解决方案：

Let r be the radius of the sphere and △r be the error in measuring the radius, 

then r = 7m and △r = 0.02m. 

Now the volume V of the sphere is given by,

v = 4/3πr² 

On differentiating we get    

$\frac{dv}{dr}=4πr^2$

Therefore, dv = (4πr²)Δr = 4π(7)².(0.02)

= 12.30m³

Thus, the approximate error in calculating the volume is 12.30m³

问题7.如果球体的半径测量为9m且误差为0.03m，则在计算其表面积时可以找到近似误差。

解决方案：

Let r be the radius of the sphere and Δr be the error in measuring the radius. 

Then r = 9m and Δr = 0.03m.

Now the surface area S of the sphere is given by,

S = 4πr²

On differentiating we get    

dS/dr = 8πr

Therefore, dS = (8πr)Δr

= 8π(9).(0.03)

= 2.16πm²                     

Thus, the approximate error in measuring the surface area is 2.16πm².

问题8.如果f(x)= 3x ² + 15x + 5，则f(3.02)的近似值为

(A)47.66(B)57.66(C)67.66(D)77.66

解决方案：

Given: f(x) = y = 3x² + 15x + 5        -(1)

f(3.02) = f(3 + 0.2)

So, x = 3 and Δx = 0.02   

On differentiating eq(1) we get    

f'(x) = y = 6x + 15

dy = (6x + 15) dx = (6x + 15)Δx

Now changing x = x + Δx and y = y +Δy in eq(1)

f(x + Δx) = y + Δy

f(3.02) = y + Δy        -(2)

f(3 + 0.2) = y + Δy

since 

then Δy = dy and Δx = dx

From eq(2), we get

f(3.02) = (3x² + 15x + 5) + (6x + 15)Δx

f(3.02) = (3x² + 15x + 5) + (6x + 15)(0.02)

= (3(3)² + 15(3) + 5) + (6(3)+ 15)(0.02)

= 27 + 45 + 5 + 0.36 + 0.3

= 77.66

Hence the correct option is 77.66

问题9：由于边增加3％而引起的边x米的立方体的体积的近似变化是

(A)0.06X ³米^{3(B)0.6X 3}米^{3(C)0.09X 3}米^{3(d)0.9X 3}米³

解决方案：

We have

Volume of cube = v = x³

On differentiating we get

$\frac{dv}{dr}=3x^2$

Given that the side increasing 3%, so Δx = 0.03x

$dv=(\frac{dv}{dr}).Δx=3x^2.(0.03x)$

$dv=x^2(0.03x)=0.09x^3m^3$

Thus, the approximate change is 0.09x³m³ 

The correct option is (C)