Utilizing Understanding ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.3 as a study aid can enhance exam preparation.

## 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) = x1/3
Take x = 64,
x + ∆x = 66
⇒ ∆x = 2
When x = 64
then y = (64)1/3 = 4
Let ∆x = dx = 2
Now y = x1/3
∴ $$\frac{d y}{d x}$$ = $$\frac{1}{3}$$ 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) = x1/3
Take x = 27,
x + ∆x = 26
⇒ ∆x = – 1
When x = 27
then y = (27)1/3 = 3,
Let ∆x = dx = – 1
Now y = x1/3
∴ $$\frac{d y}{d x}$$ = $$\frac{1}{3}$$ 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) = x1/3
Take x = 27,
x + ∆x = 25
⇒ ∆x = – 2
when x = 27
then (27)1/3 = 3,
let dx = ∆x = – 2
Now y = x1/3
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{3}$$ 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) = x1/3
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 = x1/3
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{3}$$ 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) = x1/4
Take x = 81,
x + ∆x = 82
⇒ ∆x = 1
When x = 81,
then y = (81)1/4 = 3,
Let dx = ∆x = 1
Now y = x1/4
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{1}$$ 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) = x1/4
Take x = 16,
x + ∆x = 15
⇒ ∆x = – 1
When x = 16,
then y = (16)1/4 = 3,
Let dx = ∆x = – 1
Now y = x1/4
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{1}$$ 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) = x1/5
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 = x1/5
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{5}$$ 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 = x1/5 ………..(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) = x1/10
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 = x1/10
∴ $$\frac{d y}{d x}$$ = $$\frac{1}{10}$$ 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) = x3/2
When x = 4
then y = (4)3/2 = 8,
Let dx = ∆x = 0.032
Now y = x3/2
⇒ $$\frac{d y}{d x}$$ = $$\frac{3}{2}$$ x1/2
∴ $$\left.\frac{d y}{d x}\right]_{x=4}=\frac{3}{2}(4)^{1 / 2}=3$$
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 %.