OP Malhotra Class 12 Maths Solutions Chapter 11 Applications of Derivatives Ex 11(b)

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S Chand Class 12 ICSE Maths Solutions Chapter 11 Applications of Derivatives Ex 11(b)

Question 1.
Using differentials, find the approximate value of each of the following upto 3 places of decimal.
(i) \(\sqrt{37}\)
(ii) (15)^1/4
(iii) (255)^1/4
(iv) (81.5)^1/4
(v) (0.0037)^1/2
(vi) (0.6)^1/2
(vii) (26.57)^1/3
(viii) (0.009)^1/3
(ix) log_e = 4.04, it being given that
log₁₀4 = 0.6021 and log₁₀ e = 0.4343
(x) log_e = 10.02, it being given that
log_e 10 = 2.3026.
Solution:
(i) Let y = f(x) = \(\sqrt{x}\)
Take x = 36, x + ∆x = 37 so that
∆x = 37 – 36 = 1
When x = 36 then y = \(\sqrt{36}\) = 6
Let ∆x = dx = 1

(ii) Let y = f (x) = x^1/4
Take x = 16 ⇒ x + ∆x = 15
⇒ ∆x = – 1
when x = 16 then y = (16)^1/4 = 2
Let ∆x = dx = – 1

(iii) Let y = f (x) = x^1/4
Take x = 256 so that x + ∆x = 255
∴ ∆x = – 1
when x = 256 then y = (256)^1/4 = 4,
Let dx = ∆x = – 1

(iv) Let y = f (x) = x^1/4
Take x = 81, x + ∆x = 81.5 ⇒ ∆x = 0.5
when x = 81, then y = (81)^1/4 = 3,
Let dx = ∆x = 0.5

(v) Let y = f(x) = \(\sqrt{x}\) … (1)
Let x = 0.0036 and
let x + δx = 0.0037
∴ δx = 0.0037 – 0.0036 = 0.0001
∴ from (1) ; y + δy = \(\sqrt{x+\delta x}\)

(vi) Let y = f(x) = \(\sqrt{x}\) …(1)
Let x = 0.64 & let x + δx = 0.6
∴ δx = 0.6 – 0.64 = – 0.04
∴ from (1); y + δy = \(\sqrt{x+\delta x}\)

(vii) Let y = f (x) = x^1/3
Take x = 27, x + ∆x = 26.57
⇒ ∆x = 26.57 – 27 = – 0.43
When x = 27 then y = (27)^1/3 = 3,
Let dx = ∆x = – 0.43
Now y = x^1/3 ⇒ \(\frac { dy }{ dx }\) = \(\frac{1}{3} x^{-2 / 3}\)

(viii) Let y = f (x) = x^1/3
Take x = 0.008, x + ∆x = 0.009
⇒ ∆x = 0.001
when x = 0.008 then y = (0.008)^1/3 = 0.2
Let dx = ∆x = 0.001

(ix) Let us consider the function y = log_ex = f(x)
Let x = 4 and x + ∆x = 4.04
∴ ∆x = dx = 4.04 – 4 = 004
at x = 4 ; y = log₁₀ 4 = 0.6021

(x) Let us consider the function y = log_e x
Let x = 10 and x + ∆x = 10.02
∴ ∆x = 10.02 – 10 = 0.02
Let ∆x = dx = 0.02
at x = 10; y = log_e 10
since y = log_e x ⇒ \(\frac{d y}{d x}=\frac{1}{x}\)
∴ dy = ∆y = \(\frac { dy }{ dx }\)dx = \(\frac { 1 }{ x }\) x ∆x
at x = 10 ; dy = ∆y = \(\frac { 1 }{ 10 }\) x 0.02 = 0.002
Thus, log_e 10.02 = y + ∆y = log_e 10 + 0.002 = 2.3026 + 0.002 = 2.3046

Question 2.
If f (x) = 3x² + 15x + 5, then the approximate value of f(3.02) is
(a) 47.66
(b) 57.66
(c) 67.66
(d) 77.66
Solution:
Given f (x) = y = 3x² + 15x + 5
Take x = 3, x + ∆x = 3.02 ⇒ ∆x = 0.02
When x = 3 then y = 3.3² + 15.3 + 5
⇒ y = 27 + 45 + 5 = 77,
Let ∆x = dx = 0.02
Now = 3x² + 15x + 3

Question 3.
The volume of a sphere, radius r cm is \(\frac { 4 }{ 3 }\) πr³ cm³. Find the approximate decrease in volume of a sphere when the radius decreases from 3 to 2.98 cm.
Solution:
Given V = \(\frac { 4 }{ 3 }\)πr³
Differentiate both sides w.r.t. r; we have
\(\frac{d N}{d r}=\frac{4}{3} \pi\left(3 r^2\right)\) = 4π²
Since the radius r decrease from 3 to 2.98 cm
∴ r = 3; r + δr = 2.98
∴ δr = 2.98 – 3 = – 0.032
Thus dV = \(\frac { dV }{ dr }\) x δr [∵ dr≅δr]
= 4π²δr = 4π x 3² x (-0.02)
= – 0.72π; cm³
Hence the required decrease in volume of sphere be 0.72π cm³

Question 4.
Find the approximate change in 1/x when x = 1, δ δx = 0.2.
Solution:
Let y = \(\frac { 1 }{ x }\); given x = 1; δx = 0.2
∴ y + δy = \(\frac{1}{x+\delta x}\)
∴ \(\frac { dy }{ dx }\) = \(\frac { -1 }{ x² }\)
Thus δy = \(\frac { dy }{ dx }\) δx = \(\frac { -1 }{ x² }\)δx
= – \(\frac { 1 }{ 1² }\) x 0.2 = – 0.2
∴ approximate change in y = δy = – 0.2

Question 5.
A sphere of radius 10 cm shrinks to radius 9.8 cm. Find approximately the decrease in (i) volume, and (ii) surface area.
Solution:
Let x be the radius of sphere
Then V = volume of sphere = \(\frac { 4 }{ 3 }\) πr³
Here x = 10 and x + ∆x = 9.8
∴ ∆x = – 0.2 = dx
\(\frac { dV }{ dx }\) = \(\frac { 4 }{ 3 }\) x 3x² = 4πx²
∴ dV = \(\frac { dV }{ dx }\) dx = 4πx² x (- 0.2)
= – 0.8 πx²
at x = 10 ; dV = – 0.8 π x 10² = – 80 π
Thus, dV ≅ ∆AV = – 80 π
Thus, approximate decrease in volume = ∆V = 80 π cm³

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

Question 6.
A circular plate expands when heated from a radius of 5 cm to 5.06 cm. Find the approximate increase in area.
Solution:
Let A be the area of circle with radius r
Then A = πr²
∴ \(\frac { dA }{ dr }\) = 2πr
Given r = 5; r + δr = 5.06
⇒ δr = 0.06
Thus, dA = \(\frac { dA }{ dr }\)δr = 2πrδr
= 2π x 5 x 0.06 = 0.6π cm²
∴ approximate increase in area = 0.6π cm²

Question 7.
If the side of a cube is 10.01 cm, find approximately the volume of the cube.
Solution:
Let x be the side of cube
Then V = volume of cube = x³
Let x = 10 & x + δx = 10.01
∴ at x = 10; V = (10)³ = 1000
∴ \(\frac { dV }{ dx }\) = 3x²
∴ dV = \(\frac { dV }{ dx }\)dx ⇒ δV = \(\frac { dV }{ dx }\)δx
⇒ dV = 3x² x 8x = 3 x 10² x 0.01 = 3
Thus required volume = V + δV
= (10)³ + 3 = 1003 cm³

Question 8.
The shape of a bowl is such that V = h³ + 3h² + 11 h where V cm³ is the volume of water in the bowl and h cm the depth of the water. If, when h = 7, an additional small volume δV cm³ of water is poured into the bowl, prove that the level of the water rises approximately \(\frac { δV }{ 200 }\) cm.
Solution:
Given V = h³ + 3h² + 11 h
Differentiate both sides w.r.t. h; we have

Question 9.
A closed circular cylinder has height 16 cm, and radius r cm. The total surface area is A cm². Prove that \(\frac { dA }{ dr }\) = 4π(r + 8).
Hence, calculate an approximate increase in area if the radius increases from 4 to 4.02 cm, the height remaining constant (you may leave your answer in terms of n).
Solution:
Let h be the height and r be the radius of cylinder also, A be the total surface area of cylinder