ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.3

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.3

Question 1.
Using differentials, find the approximate values in the following (1 to 5) problems :
(i) \(\sqrt{26}\)
(ii) \(\sqrt{36.6}\) (NCERT)
(iii) \(\sqrt{49.5}\)
(iv) \(\sqrt{401}\) (NCERT)
Solution:
(i) Let us consider the function
y = √x = f(x)
Let x = 25 and x + ∆x = 26
∴ ∆x = 26 – 25 = 1
at x = 25 ;
y = \(\sqrt{25}\)= 5
Let dx = ∆x = 1
Since y = √x
∴ \(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x}}\)
at x = 25 ;
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{25}}\)
= \(\frac{1}{2 \times 5}=\frac{1}{10}\) = 0.1
∴ dy = ∆y
= \(\frac{d y}{d x}\) dx
= 0.1 × 1 = 0.1
Thus, \(\sqrt{26}\) = y + ∆y
= 5 + 0.1 = 5.1

(ii) Let y = √x ……….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we get
\(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
Take x = 36 ;
x + δx = 36.6
∴ δx = dx = 0.6
When x = 36
⇒ y = \(\sqrt{36}\) = 6
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{2 \sqrt{x}}\) δx
= \(\frac{1}{2 \times \sqrt{36}}\) × 0.6
= \(\frac{0.6}{12}=\frac{6}{120}\)
= \(\frac{1}{20}\) = 0.05
⇒ δy = 0.05 [∵ dy ≅ δy]
∴ from (1) ;
y + δy = \(\sqrt{x+\delta x}\) = \(\sqrt{36.6}\)
⇒ \(\sqrt{36.6}\) = 6 + 0.5 = 6.05.

(iii) Let y = f(x) = √x
Take x = 49,
y = \(\sqrt{49}\) = 7,
let dx = Δx = 0.5
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=49}=\frac{1}{2 \sqrt{49}}=\frac{1}{14}\)
∴ dy = \(\frac{d y}{d x}\) . dx
= \(\frac{1}{14}\) . (0.5)
= \(\frac{5}{140}=\frac{1}{28}\)
⇒ Δy = \(\frac{1}{28}\)
[∴ Δy ≅ dy]
∴ \(\sqrt{49.5}\) = y + Δy
= 7 + \(\frac{1}{28}\) = 7.036

(iv) Let y = f(x) = √x
Take x = 400, x + Δx = 401
∴ Δx = 1
When x = 400 than y = \(\sqrt{400}\) = 20
Let dx = Δx = 1
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=400}=\frac{1}{2 \sqrt{400}}=\frac{1}{40}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{40}\) × 1
= \(\frac{1}{40}\) = Δy
[∴ dy ≅ Δy]
Hence \(\sqrt{401}\) = y + Δy
= 20 + \(\frac{1}{40}\)
= 20 + 0.25 = 20.025.

Question 2.
(i) \(\sqrt{0.26}\)
(ii) \(\sqrt{0.24}\)
(iii) \(\sqrt{0.0037}\)
(iv) \(\sqrt{0.48}\) (NCERT)
Solution:
(i) Let y = f(x) = √x
Take x = 0.25,
x + Δx = 0.26
⇒ Δx = 0.01
When x = 0.25
then y = \(\sqrt{0.25}\) = 0.5
Let dx = Δx = 0.01
Now y = √x
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.25}=\frac{1}{2 \sqrt{0 \cdot 25}}\)
= \(\frac{1}{2 \times 0.5}\) = 1
∴ dy = \(\frac{d y}{d x}\) × dx
= 1 × 0.01 = 0.01
∴ \(\sqrt{0.26}\) = y + Δy
= 0.5 + 0.01 = 0.51.

(ii) Let y = f(x) = √x
Take x = 0.25,
x + Δx = 0.24
⇒ Δx = – 0.01
When x = 0.25
then y = \(\sqrt{0.25}\) = 0.5,
Let dx = Δx = – 0.01
Now y = √x
∴ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.25}=\frac{1}{2 \sqrt{0.25}}\)
= \(\frac{1}{2 \times 0.5}\) = 1
∴ dy = \(\frac{d y}{d x}\) dx
= – 1 × 0.01
= – 0.01 = Δy
∴ \(\sqrt{0.24}\) = y + Δy
= 0.5 – 0.01 = 0.49.

(iii) Take x = 0.0036,
x + Δx = 0.0037
∴ Δx = 0.0001
When x = 0.0036
then y = \(\sqrt{0.0036}\) = 0.06
Let dx = Δx = 0.0001
Now, y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.0036}=\frac{1}{2 \times 0.06}=\frac{1}{0.12}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{0.12}\) × 0.0001
⇒ Δy = \(\frac{1}{10000} \times \frac{100}{12}=\frac{1}{1200}\)
∴ \(\sqrt{0.0037}\) = y + Δy
= 0.06 + \(\frac{1}{1200}\) = 0.0608.

(iv) Let y = f(x) = √x
Take x = 0.49,
x + ∆x = 0.48
⇒ ∆x = – 0.01
When x = 0.49
then y = \(\sqrt{0.49}\) = 0.7
Let dx = ∆x = – 0.01
Now y = √x
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}\)
⇒ \(\left.\frac{d y}{d x}\right]_{x=0.49}=\frac{1}{2 \sqrt{0 \cdot 49}}\)
= \(\frac{1}{2 \times 0.7}=\frac{1}{1.4}=\frac{5}{7}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{5}{7}\) × (- 0.01) = ∆y
[∵ ∆y ≅ dy]
∴ \(\sqrt{0.49}\) = y + ∆y
= 7 – \(\frac{0.05}{7}\) = 0.693.

 

Question 3.
(i) (66)^1/3
(ii) (26)^1/3 (NCERT)
(iii) (25)^1/3 (NCERT)
(iv) (26.57)^1/3 (NCERT)
Solution:
(i) Let y = f(x) = x^1/3
Take x = 64,
x + ∆x = 66
⇒ ∆x = 2
When x = 64
then y = (64)^1/3 = 4
Let ∆x = dx = 2
Now y = x^1/3
∴ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
⇒ \(\left.\frac{d y}{d x}\right]_{x=64}=\frac{1}{3}(64)^{-2 / 3}=\frac{1}{48}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{48}\) × 2
= \(\frac{1}{24}\) = ∆y
(∵ ∆y ≅ dy)
∴ (66)^1/3 = y + ∆y
= 4 + \(\frac{1}{24}\)
= \(\frac{97}{24}\) = 4.0416

(ii) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 26
⇒ ∆x = – 1
When x = 27
then y = (27)^1/3 = 3,
Let ∆x = dx = – 1
Now y = x^1/3
∴ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
∴ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{27}\) × (- 1)
= \(-\frac{1}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ (26)^1/3 = y + ∆y
= 3 – \(\frac{1}{27}\) = \(\frac{80}{27}\)

(iii) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 25
⇒ ∆x = – 2
when x = 27
then (27)^1/3 = 3,
let dx = ∆x = – 2
Now y = x^1/3
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
⇒ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{27}\) × (- 2)
= \(-\frac{2}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ (25)^1/3 = y + ∆y
= = 3 – \(\frac{1}{27}\)
= \(\frac{79}{27}\) = 2.926.

(iv) Let y = f(x) = x^1/3
Take x = 27,
x + ∆x = 26.57
⇒ ∆x = 26.57 – 27 = – 0.43
when x = 27
then (27)^1/3 = 3,
Let dx = ∆x = – 0.43
Now y = x^1/3
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{3}\) x^-2/3
∴ \(\left.\frac{d y}{d x}\right]_{x=27}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{27}\)
∴ dy = \(\frac{d y}{d x}\) × dx
= \(\frac{1}{27}\) × (- 0.43)
= \(-\frac{0.43}{27}\) = ∆y
[∵ ∆y ≅ dy]
∴ \(\sqrt[3]{26 \cdot 57}\) = y + ∆y
= = 3 – \(\frac{0.43}{27}\)
= 2.984.

Question 4.
(i) (82)^1/4 (NCERT)
(ii) (15)^1/4 (NCERT)
(iii) (32.15)^1/5 (NCERT)
(iv) (31.9)^1/5
Solution:
(i) Let y = f(x) = x^1/4
Take x = 81,
x + ∆x = 82
⇒ ∆x = 1
When x = 81,
then y = (81)^1/4 = 3,
Let dx = ∆x = 1
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{1}\) x^-3/4
∴ \(\left.\frac{d y}{d x}\right]_{x=81}=\frac{1}{4}(81)^{-3 / 4}=\frac{1}{108}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{108}\) × 1
= \(\frac{1}{108}\)
∴ (82)^1/4 = y + ∆y
= 3 + \(\frac{1}{108}\)
= \(\frac{325}{108}\).

(ii) Let y = f(x) = x^1/4
Take x = 16,
x + ∆x = 15
⇒ ∆x = – 1
When x = 16,
then y = (16)^1/4 = 3,
Let dx = ∆x = – 1
Now y = x^1/4
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{1}\) x^-3/4
∴ \(\left.\frac{d y}{d x}\right]_{x=16}=\frac{1}{4}(16)^{-3 / 4}=\frac{1}{32}\)
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{32}\) × (- 1)
= \(-\frac{1}{32}\) = ∆y
[∵ dy ≅ ∆y]
∴ (15)^1/4 = y + ∆y
= 2 – \(\frac{1}{32}\)
= \(\frac{63}{32}\)
= 1.96875

(iii) Let y = f(x) = x^1/5
Take x = 32,
x + ∆x = 32.15
⇒ ∆x = 0.15
When x = 32,
then y = (32)^1/5 = 2,
Let dx = ∆x = 1
Now y = x^1/5
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{5}\) x^-4/5
∴ \(\left.\frac{d y}{d x}\right]_{x=32}=\frac{1}{5}(32)^{-4 / 5}=\frac{1}{80}\)
∴ dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{1}{80} \times \frac{15}{100}=\frac{3}{1600}\)
∴ (32.15)^1/5 = y + δy
= 2 + \(\frac{3}{1600}\)
= 2.00187

(iv) Let y = x^1/5 ………..(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{5} x^{\frac{1}{5}-1}=\frac{1}{5 x^{4 / 5}}\)
Take x = 32 ;
x + δx = 31.9
∴ δx = dx = 31.9 – 32 = – 0.1
When x = 32
∴ y = \((32)^{\frac{1}{5}}\)
= \(\left(2^5\right)^{\frac{1}{5}}\) = 2
∴ dy = \(\frac{d y}{d x}\) dx
= \(\frac{1}{5 x^{4 / 5}}\) δx
= \(\frac{1}{5 \times(32)^{4 / 5}}\) × (- 0.1)
= – \(\frac{0.1}{80}\)
⇒ δy = dy
= – \(\frac{0.1}{80}=-\frac{1}{800}\)
∴ from (1) ;
y + δy = (x + δx)^1/5
⇒ (31.9)^1/5 = y + δy
= 2 – \(\frac{1}{800}\)
= \(\frac{1599}{800}\)
= 1.99875.

Question 5.
(i) (0.999)^1/10 (NCERT)
(ii) (3.968)^3/2 (NCERT)
Solution:
(i) Let y = f(x) = x^1/10
Take x = 1,
x + ∆x = 0.999
⇒ ∆x = – 0.001
When x = 1
then y = (1)^1/10 = 1,
Let dx = ∆x = – 0.001
Now y = x^1/10
∴ \(\frac{d y}{d x}\) = \(\frac{1}{10}\) x^-9/10
⇒ \(\left.\frac{d y}{d x}\right]_{x=1}=\frac{1}{10}(1)^{-9 / 10}=\frac{1}{10}\)
Therefore dy = \(\frac{d y}{d x}\) ∆x
= \(\frac{1}{10}\) × (- 0.001)
= – 0.0001
= ∆y
[∵ ∆y ≅ dy]
∴ (0.999)^1/10 = y + ∆y
= 1 – 0.0001 = 0.9999.

(ii) Let y = f(x) = x^3/2
When x = 4
then y = (4)^3/2 = 8,
Let dx = ∆x = 0.032
Now y = x^3/2
⇒ \(\frac{d y}{d x}\) = \(\frac{3}{2}\) x^1/2
∴ \(\left.\frac{d y}{d x}\right]_{x=4}=\frac{3}{2}(4)^{1 / 2}=3\)
Therefore dy = \(\frac{d y}{d x}\) ∆x
= – 3 × 0.0032
= – 0.096
= ∆y
[∵ dy ≅ ∆y]
∴ (3.968)^3/2 = y + ∆y
= 8 – 0.096 = 7.904.

Question 6.
Find the approximate value of (1999)⁵. (NCERT Exemplar)
Solution:
Let us consider the function y = x⁵ = f(x)
Let x = 2
and x + ∆x = 1.999
∴ ∆x = 1.999 – 2 = – 0.001
Let dx = ∆x = – 0.001
at x = 2 ;
y = x⁵
= 2⁵ = 32
Since y = x⁵
⇒ \(\frac{d y}{d x}\) = 5x⁴
at x = 2 ;
\(\frac{d y}{d x}\) = 5 × 2⁴ = 80
∴ ∆y = dy
= \(\frac{d y}{d x}\) dx
= 80 × (- 0.001)
= – 0.08
Thus, (1.999)⁵ = y + ∆y
= 32 – 0.08 = 31.92.

Question 7.
(i) If f (x) = x⁴ – 10, then find the approximate value of f(2, 1). (NCERT)
(ii) Find the approximate value of f (3.02), where f(x) = 3x² + 5x + 3.
Solution:
(i) Given f(x) = x⁴ – 10 = y
Let x = 2 and x + δx = 2.1
∴ δx = 2 . 1 – 2 . 0
= 0 . 1 = dx
When x = 2
∴ y = f(2)
= 2⁴ – 10
= 16 – 10 = 6
∴ f'(x) = \(\frac{d y}{d x}\) = 4x³
Thus dy = δy
= \(\frac{d y}{d x}\) δx
= 4x³ . δx
= 4 × 2³ × 0.1 = 3.2
∴ f (x + δx) = y + δy
= (x + δx)⁴ – 10
= (2 . 1)⁴ – 10
= f(2 . 1)
⇒ f(2.1) = 6 + 3.2 = 9.2

(ii) Given f(x) = 3x² + 5x + 3
Take x = 3,
x + ∆x = 3.02
⇒ ∆x = 0.02
When x = 3
then y = 3.3² + 15 . 3 + 3
⇒ y = 27 + 15 + 3 = 45,
Let ∆x = dx = 0.02
Now y = 3x² + 5x + 3
∴ \(\frac{d y}{d x}\) = 6x + 5
⇒ \(\left.\frac{d y}{d x}\right]_{x=3}\) = 6 × 3 + 5 = 23
∴ dy = \(\frac{d y}{d x}\) ∆x
= 23 × 0.02
= 0.46 = ∆y
[∵ ∆y ≅ dy]
∴ f(3.02) = y + ∆y
= 45 + 0.46 = ∆y.

Question 7 (old).
(ii) Find the approximate value of f(5.001), where f(x) = x³ – 7x² + 15. (NCERT)
Solution:
Given f(x) = x³ – 7x² + 15
⇒ f'(x) = 3x² – 14x
Take x = 5 and ∆x = 0.001
then f(5.001) = f(x + ∆x)
= (x + ∆x)³ – 7 (x + ∆x)² + 15
Now ∆y = f (x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆y
⇒ f(x + ∆x) = f (x) + f'(x) ∆x
[∵ dy = ∆y = \(\frac{d y}{d x}\) ∆x = f'(x) ∆x]
⇒ f(5.001) = f(5) + f'(5) × 0.001
= [5³ – 7.5² + 15] + [3.5² – 14 × 5] × 0.01
= – 35 + 0.005 = – 34.995

Question 8.
If the radius of a circle increases from 5 cm to 5-1 cm, find the increase in area.
Solution:
Let r be the radius of circle and A be the area of circle
Then A = πr² ………….(1)
Diff. w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr dr
Since radius r increases from 5 cm to 5.1 cm.
∴ r = 5 and r + δr = 5.1
⇒ δr = dr = 0.1
When r = 5 ; A = π (5)² = 25π
∴ dA = \(\frac{d A}{d r}\) dr
= (2πr) δr
= 2π × 5 × 0.1 = π
⇒ δA = π cm²
[∵ dA ≅ δA]
Thus approximate increase in area = δA = π cm²
and A + δA = 25 π + π = 26 π cm²

Question 9.
If the radius of a spherical balloon shrinks from 10 cm to 9-8 cm, find the approximate decrease in its
(i) volume
(ii) surface area.
Solution:
Let x be the radius of sphere
Then V = volume of sphere = \(\frac{4}{3}\) πx³
Here x = 10 and x + ∆x = 9-8
∴ ∆x = – 0.2 = dx
\(\frac{d V}{d x}\) = \(\frac{4 \pi}{3}\) × 3x²
= 4πx²
∴ dV = \(\frac{d V}{d x}\) dx
= 4πx² × (- 0.2)
= – 0.8 πx²
at x = 10 ;
dV = – 0.8 π × 10² = – 80 π
Thus, dV = ∆V = – 80 π
Thus, approximate decrease in volume = ∆V = 80 π cm³.

(ii) Let S = surface area of spherical balloon = 4πx²
∴ \(\frac{d S}{d x}\) = 8πx
∴ dS = \(\frac{d S}{d x}\) dx
= 8πx dx
= 8π × 10 × (- 0.2)
⇒ dS = – 16π = ∆S
Thus approximate decrease in surface area = ∆S = 16π cm²

Question 10.
(i) Find the approximate change in the volume of a cube of side x metres caused by increasing the side by 1%.
(ii) Find the approximate change in the surface area of a cube of side x metres by decreasing the side by 1%.
Solution:
(i) Given the length of each side of cube be x m.
Then volume of cube = V = x³
∴ \(\frac{d V}{d x}\) = 3x²
Let ∆x be the change in x and the corresponding change in V be ∆V.
∴ ∆V = \(\frac{d V}{d x}\) ∆x
[∵ ∆V = dV ; ∆x = dx]
= 3x² ∆x
also it is given that \(\frac{\Delta x}{x}\) × 100 = 1
⇒ ∆x = \(\frac{x}{100}\) = 0.01 x
∴ ∆V = 3x² × 0.01x
= 0.03 x³ m³

(ii) Let x be the side of cube
then S = surface area of cube = 6x²
Diff. both sides w.r.t. x, we have,
∴ \(\frac{d S}{d x}\) = 12x dx
∴ dS = change in surface area
= \(\frac{d S}{d x}\) . ∆x
∴ dS = 12x . ∆x [∵ dx = ∆x]
= 12x × (- 0.01 x)
[∆x = 1 % decreasing of x = – 0.01 x]
= – 0.12 x² m²
Hence the approximate change in surface area is 0.12 x² m².

Question 11.
(i) If the radius of a sphere is measured as 9 m with an error of 0-03 m, then find the approximate error in calculating its volume. (NCERT)
(ii) Find the percentage error in computing the surface area of a cubical box if an error of 1% is made in measuring the lengths of the edges of the box.
Solution:
(i) Let r be the radius of sphere
then error in r = ∆r = 0 03 m ;
r = 9 cm
Let V = volume of sphere = \(\frac{4 \pi}{3}\) r³
∴ Log V = log \(\frac{4 \pi}{3}\) + 3 log r
∴ \(\frac{1}{V}\) ∆V = 0 + \(\frac{3}{r}\) ∆r
∴ ∆V = \(\frac{3}{r}\) ∆r × \(\frac{4}{3}\) πr³
= 4πr² ∆r
= 4π × 9² × 0.03
= 9.72 πm²

(ii) Let x be the length of the edge of the cube then S = surface area of cubical base = 6x²
Taking logarithm on both sides, we get
log S = log 6 + 2 log x ;
on differentiating
⇒ \(\frac{d S}{S}\) × 100 = 2 (\(\frac{dx}{x}\) × 100) …………….(1)
Given, \(\frac{dx}{x}\) × 100 = 1
∴ from (1) ; we have
\(\frac{d S}{S}\) × 100 = 2 × 1 = 2
∴ error in calculating surface area is 2 %.