Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.1 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1

Question 1.

Let f(x) = \(\sqrt[3] { x }\). Find the linear approximation at x = 27. Use the linear approximation to approximate \(\sqrt[3] { 27.2 }\)

Solution:

x = 27

f(x) = \(\sqrt[3] { 27 }\) = 3

We need to find the value of \(\sqrt[3] { 27.2 }\)

We know that

f(x_{0} + Δx) = f(x_{0}) + f'(x_{0}) Δx

∴ Approximate value of \(\sqrt[3] { 27.2 }\) = 3.0074

Question 2.

Use the linear approximation to find approximate value of

(i) (123)^{2/3}

(ii) \(\sqrt[4] { 15 }\)

(iii) \(\sqrt[4] { 26 }\)

Solution:

(i) (123)^{2/3}

Let x_{0} = 125, Δx = -2

f(x) = x^{2/3}, f(x_{0}) = 25

We know that

f(x_{0} + Δx) = f(x_{0}) + f'(x_{0}) Δx

= 25 – 0.2666

(123)^{2/3} = 24.7334

(ii) \(\sqrt[4] { 15 }\) = (15)^{1/4}

f(x) = x^{1/4}, f(x_{0}) = (16)^{1/4} = 2

We know that

f(x_{0} + Δx) = f(x_{0}) + f’(x_{0}) Δx

= 2 – 0.03125

(15)^{1/4} = 1.96875

(iii) (26)^{1/3}

f(x) = x^{1/3}, f(x_{0}) = (27)^{1/3} = 2, Δx = -1

We know that

f(x_{0} + Δx) = f(x_{0}) + f’(x_{0}) Δx

= 3 – 0.370

(26)^{1/3} = 2.963

Question 3.

Find a linear approximation for the following functions at the indicated points

(i) f(x) = x³ – 5x + 12, x_{0} = 2

(ii) g(x) = \(\sqrt { x^2+9 }\), x_{0} = -4

(iii) h(x) = \(\frac { x }{ x+1 }\), x_{0} = 1

Solution:

(i) We know that the linear approximation

L(x) = f(x_{0}) + f’(x_{0})(x – x_{0})

f(x) = x³ – 5x + 12

f'(x) = 3x² – 5

f'(x_{0}) = f'(2) = 12 – 5 = 7

f(x_{0}) = f(2) = 8 – 10 + 12 = 10

L(x) = 10 + 7 (x – 2)

= 10 + 7x – 14

= 7x – 4

(ii) g(x) = \(\sqrt { x^2+9 }\), x_{0} = -4

g(x_{0}) = g(14) = \(\sqrt {16+9 }\) = 5

(iii) h(x) = \(\frac { x }{ x+1 }\), x_{0} = 1

Question 4.

The radius of a circular plate is measured as 12.65. cm instead of the actual length 12.5 cm find the following in calculating the area of the circular plate:

(i) Absolute error

(ii) Relative error

(iii) Percentage error

Solution:

Actual radius of the circular plate = 12.5 cm

Measured radius of the circular plate = 12.65

dr = 12.65 – 12.5

= 0.15

A = π r²

dA = 2π r dv

Change in Area

A(12.65) – A(12.5) = dA

= 2π × 12.5 × 0.15

= 3.75 π

Exact calculation of the Area changes gives

A(12.65) – A(12.5) = π(12.65)² – π(12.5)²

= 160.0225 π – 156.25 π

= 3.7725 π

Absolute error = 3.7725 π – 3.75 π

= 0.0225 π cm²

Relative error

= \(\frac { 3.7725π-3.75π }{ 3.7725π }\)

= \(\frac { 0.0225π }{ 3.7725π }\)

= 0.00596

= 0.006 cm²

Percentage error = Relative error × 100

= 0.006 × 100

= 0.6% .

Question 5.

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:

(i) Change in the volume.

(ii) Change in the surface area

Solution:

(i) Given r = 10

dr = 10 – 9.8 = 0.2

Volume v = \(\frac { 4 }{ 3 }\)πr³

dv = \(\frac { 4 }{ 3 }\).3πr²dv

Change in volume

v(10) – v(9.8) = 4π(10)²(0.2)

= 80π cm³

(ii) Surface area of the sphere

S(r) = 4πr^{2}

S'(r) = 8πr

Change in surface area at r = 10 is

= S'(r) [10 – 9.8]

= 8π (10) (0.2) = 16π cm^{2}

∴ Surface Area decreases by 16π cm^{2}

Question 6.

The time T, taken for a complete oscillation of a single pendulum with length l, is given by the equation T = 2π\(\sqrt{\frac { l }{ g }}\) where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of l.

Solution:

Given T = 2π\(\sqrt{\frac { l }{ g }}\)

On taking log both sides, we get

log T = log 2 + log π + \(\frac { 1 }{ 2 }\) log l – \(\frac { 1 }{ 2 }\) log g

On differentiating both sides w. r. to l, we get

So, the percentage error in T is 1%

Question 7.

Show that the percentage error in the nth root of a number is approximately \(\frac { 1 }{ n }\) times the percentage error in the number.

Solution:

Let x be the number

Let y = x^{1/n}

log y = \(\frac { 1 }{ n }\) log x

Taking differentiate on both sides, we have

= \(\frac { 1 }{ n }\) times the percentage error in the number.