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) = $$\sqrt { x }$$. Find the linear approximation at x = 27. Use the linear approximation to approximate $$\sqrt { 27.2 }$$
Solution:
x = 27
f(x) = $$\sqrt { 27 }$$ = 3
We need to find the value of $$\sqrt { 27.2 }$$
We know that
f(x0 + Δx) = f(x0) + f'(x0) Δx ∴ Approximate value of $$\sqrt { 27.2 }$$ = 3.0074 Question 2.
Use the linear approximation to find approximate value of
(i) (123)2/3
(ii) $$\sqrt { 15 }$$
(iii) $$\sqrt { 26 }$$
Solution:
(i) (123)2/3
Let x0 = 125, Δx = -2
f(x) = x2/3, f(x0) = 25
We know that
f(x0 + Δx) = f(x0) + f'(x0) Δx = 25 – 0.2666
(123)2/3 = 24.7334

(ii) $$\sqrt { 15 }$$ = (15)1/4
f(x) = x1/4, f(x0) = (16)1/4 = 2
We know that
f(x0 + Δx) = f(x0) + f’(x0) Δx = 2 – 0.03125
(15)1/4 = 1.96875

(iii) (26)1/3
f(x) = x1/3, f(x0) = (27)1/3 = 2, Δx = -1
We know that
f(x0 + Δx) = f(x0) + f’(x0) 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, x0 = 2
(ii) g(x) = $$\sqrt { x^2+9 }$$, x0 = -4
(iii) h(x) = $$\frac { x }{ x+1 }$$, x0 = 1
Solution:
(i) We know that the linear approximation
L(x) = f(x0) + f’(x0)(x – x0)
f(x) = x³ – 5x + 12
f'(x) = 3x² – 5
f'(x0) = f'(2) = 12 – 5 = 7
f(x0) = f(2) = 8 – 10 + 12 = 10
L(x) = 10 + 7 (x – 2)
= 10 + 7x – 14
= 7x – 4

(ii) g(x) = $$\sqrt { x^2+9 }$$, x0 = -4
g(x0) = g(14) = $$\sqrt {16+9 }$$ = 5 (iii) h(x) = $$\frac { x }{ x+1 }$$, x0 = 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) = 4πr2
S'(r) = 8πr
Change in surface area at r = 10 is
= S'(r) [10 – 9.8]
= 8π (10) (0.2) = 16π cm2
∴ Surface Area decreases by 16π cm2

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 = x1/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. 