Differentials, Errors And Approximations

Given y = sin x and x changes from to.

Let so that

On differentiating y with respect to x, we get

We know

When, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and

∴ Δy = 0

Thus, there is approximately no change in y.

Given the radius of a sphere changes from 10 cm to 9.8 cm.

Let x be the radius of the sphere and Δx be the change in the value of x.

Hence, we have x = 10 and x + Δx = 9.8

⇒ 10 + Δx = 9.8

⇒ Δx = 9.8 – 10

∴ Δx = –0.2

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

When x = 10, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.2

⇒ ΔV = (400π)(–0.2)

∴ ΔV = –80π

Thus, the approximate decrease in the volume of the sphere is 80π cm³.

Given the radius of a circular plate initially is 10 cm and it increases by k%.

Let x be the radius of the circular plate, and Δx is the change in the value of x.

Hence, we have x = 10 and

∴ Δx = 0.1k

The area of a circular plate of radius x is given by

A = πx²

On differentiating A with respect to x, we get

We know

When x = 10, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.1k

⇒ ΔA = (20π)(0.1k)

∴ ΔA = 2kπ

Thus, the approximate increase in the area of the circular plate is 2kπ cm².

Given the error in the measurement of the edge of a cubical box is 1%.

Let x be the edge of the cubical box, and Δx is the error in the value of x.

Hence, we have

∴ Δx = 0.01x

The surface area of a cubical box of radius x is given by

S = 6x²

On differentiating A with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01x

⇒ ΔS = (12x)(0.01x)

∴ ΔS = 0.12x²

The percentage error is,

⇒ Error = 0.02 × 100%

∴ Error = 2%

Thus, the error in calculating the surface area of the cubical box is 2%.

Given the error in the measurement of the radius of a sphere is 0.1%.

Let x be the radius of the sphere and Δx be the error in the value of x.

Hence, we have

∴ Δx = 0.001x

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.001x

⇒ ΔV = (4πx²)(0.001x)

∴ ΔV = 0.004πx³

The percentage error is,

⇒ Error = 0.003 × 100%

∴ Error = 0.3%

Thus, the error in calculating the volume of the sphere is 0.3%.

Given pv^1.4 = constant and the decrease in v is.

Hence, we have

∴ Δv = –0.005v

We have pv^1.4 = constant

Taking log on both sides, we get

log(pv^1.4) = log(constant)

⇒ log p + log v^1.4 = 0 [∵ log(ab) = log a + log b]

⇒ log p + 1.4 log v = 0 [∵ log(a^m) = m log a]

On differentiating both sides with respect to v, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δv = –0.005v

⇒ Δp = (–1.4p)(–0.005)

∴ Δp = 0.007p

The percentage error is,

⇒ Error = 0.007 × 100%

∴ Error = 0.7%

Thus, the error in p corresponding to the decrease in v is 0.7%.

Given the height of a cone increases by k%.

Let x be the height of the cone and Δx be the change in the value of x.

Hence, we have

∴ Δx = 0.01kx

Let us assume the radius, the slant height and the semi-vertical angle of the cone to be r, l and α respectively as shown in the figure below.

From the above figure, using trigonometry, we have

∴ r = x tan(α)

We also have

∴ l = x sec(α)

(i) The total surface area of the cone is given by

S = πr² + πrl

From above, we have r = x tan(α) and l = x sec(α).

⇒ S = π(x tan(α))² + π(x tan(α))(x sec(α))

⇒ S = πx²tan²α + πx²tan(α)sec(α)

⇒ S = πx²tan(α)[tan(α) + sec(α)]

On differentiating S with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01kx

⇒ ΔS = (2πxtan(α)[tan(α) + sec(α)])(0.01kx)

∴ ΔS = 0.02kπx²tan(α)[tan(α) + sec(α)]

The percentage increase in S is,

⇒ Increase = 0.02k × 100%

∴ Increase = 2k%

Thus, the approximate increase in the total surface area of the cone is 2k%.

(ii) The volume of the cone is given by

From above, we have r = x tan(α).

On differentiating V with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01kx

⇒ ΔV = (πx²tan²α)(0.01kx)

∴ ΔV = 0.01kπx³tan²α

The percentage increase in V is,

⇒ Increase = 0.03k × 100%

∴ Increase = 3k%

Thus, the approximate increase in the volume of the cone is 3k%.

Let the error in measuring the radius of a sphere be k%.

Let x be the radius of the sphere and Δx be the error in the value of x.

Hence, we have

∴ Δx = 0.01kx

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01kx

⇒ ΔV = (4πx²)(0.01kx)

∴ ΔV = 0.04kπx³

The percentage error is,

⇒ Error = 0.03k × 100%

∴ Error = 3k%

Thus, the error in measuring the volume of the sphere is approximately three times the error in measuring its radius.

Let us assume that

Also, let x = 25 so that x + Δx = 25.02

⇒ 25 + Δx = 25.02

∴ Δx = 0.02

On differentiating f(x) with respect to x, we get

We know

When x = 25, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.02

⇒ Δf = (0.1)(0.02)

∴ Δf = 0.002

Now, we have f(25.02) = f(25) + Δf

⇒ f(25.02) = 5 + 0.002

∴ f(25.02) = 5.002

Thus,

Let us assume that

Also, let x = 4 so that x + Δx = 3.968

⇒ 4 + Δx = 3.968

∴ Δx = –0.032

On differentiating f(x) with respect to x, we get

We know

When x = 4, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.032

⇒ Δf = (3)(–0.032)

∴ Δf = –0.096

Now, we have f(3.968) = f(4) + Δf

⇒ f(3.968) = 2³– 0.096

⇒ f(3.968) = 8 – 0.096

∴ f(3.968) = 7.904

Thus, (3.968)^3/2 ≈ 7.904

Let us assume that

Also, let x = 0.008 so that x + Δx = 0.009

⇒ 0.008 + Δx = 0.009

∴ Δx = 0.001

On differentiating f(x) with respect to x, we get

We know

When x = 0.008, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.001

⇒ Δf = (8.3333)(0.001)

∴ Δf = 0.0083333

Now, we have f(0.009) = f(0.008) + Δf

⇒ f(0.009) = 0.2 + 0.0083333

∴ f(0.009) = 0.2083333

Thus, (0.009)^1/3 ≈ 0.2083333

Let us assume that f(x) = x⁵

Also, let x = 2 so that x + Δx = 1.999

⇒ 2 + Δx = 1.999

∴ Δx = –0.001

On differentiating f(x) with respect to x, we get

We know

When x = 2, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.001

⇒ Δf = (80)(–0.001)

∴ Δf = –0.08

Now, we have f(1.999) = f(2) + Δf

⇒ f(1.999) = 2⁵ – 0.08

⇒ f(1.999) = 32 – 0.08

∴ f(1.999) = 31.92

Thus, (1.999)⁵ ≈ 31.92

Let us assume that

Also, let x = 0.008 so that x + Δx = 0.007

⇒ 0.008 + Δx = 0.007

∴ Δx = –0.001

On differentiating f(x) with respect to x, we get

We know

When x = 0.008, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.001

⇒ Δf = (8.3333)(–0.001)

∴ Δf = –0.0083333

Now, we have f(0.007) = f(0.008) + Δf

⇒ f(0.007) = 0.2 – 0.0083333

∴ f(0.007) = 0.1916667

Thus, (0.007)^1/3 ≈ 0.1916667

Let us assume that

Also, let x = 0.09 so that x + Δx = 0.082

⇒ 0.09 + Δx = 0.082

∴ Δx = –0.008

On differentiating f(x) with respect to x, we get

We know

When x = 0.09, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.008

⇒ Δf = (1.6667)(–0.008)

∴ Δf = –0.013334

Now, we have f(0.082) = f(0.09) + Δf

⇒ f(0.082) = 0.3 – 0.013334

∴ f(0.082) = 0.286666

Thus,

Let us assume that

Also, let x = 400 so that x + Δx = 401

⇒ 400 + Δx = 401

∴ Δx = 1

On differentiating f(x) with respect to x, we get

We know

When x = 400, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 1

⇒ Δf = (0.025)(1)

∴ Δf = 0.025

Now, we have f(401) = f(400) + Δf

⇒ f(401) = 20 + 0.025

∴ f(401) = 20.025

Thus,

Let us assume that

Also, let x = 16 so that x + Δx = 15

⇒ 16 + Δx = 15

∴ Δx = –1

On differentiating f(x) with respect to x, we get

We know

When x = 16, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –1

⇒ Δf = (0.03125)(–1)

∴ Δf = –0.03125

Now, we have f(15) = f(16) + Δf

⇒ f(15) = 2 – 0.03125

∴ f(15) = 1.96875

Thus, (15)^1/4 ≈ 1.96875

Let us assume that

Also, let x = 256 so that x + Δx = 255

⇒ 256 + Δx = 255

∴ Δx = –1

On differentiating f(x) with respect to x, we get

We know

When x = 256, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –1

⇒ Δf = (0.00390625)(–1)

∴ Δf = –0.00390625

Now, we have f(255) = f(256) + Δf

⇒ f(255) = 4 – 0.00390625

∴ f(255) = 3.99609375

Thus, (255)^1/4 ≈ 3.99609375

Let us assume that

Also, let x = 2 so that x + Δx = 2.002

⇒ 2 + Δx = 2.002

∴ Δx = 0.002

On differentiating f(x) with respect to x, we get

We know

When x = 2, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.002

⇒ Δf = (–0.25)(0.002)

∴ Δf = –0.0005

Now, we have f(2.002) = f(2) + Δf

⇒ f(2.002) = 0.25 – 0.0005

∴ f(2.002) = 0.2495

Thus,

log_e4.04, it being given that log₁₀4 = 0.6021 and log₁₀e = 0.4343

Let us assume that f(x) = log_ex

Also, let x = 4 so that x + Δx = 4.04

⇒ 4 + Δx = 4.04

∴ Δx = 0.04

On differentiating f(x) with respect to x, we get

We know

When x = 4, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.04

⇒ Δf = (0.25)(0.04)

∴ Δf = 0.01

Now, we have f(4.04) = f(4) + Δf

⇒ f(4.04) = log_e4 + 0.01

⇒ f(4.04) = 1.3863689 + 0.01

∴ f(4.04) = 1.3963689

Thus, log_e4.04 ≈ 1.3963689

log_e10.02, it being given that log_e10 = 2.3026

Let us assume that f(x) = log_ex

Also, let x = 10 so that x + Δx = 10.02

⇒ 10 + Δx = 10.02

∴ Δx = 0.02

On differentiating f(x) with respect to x, we get

We know

When x = 10, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.02

⇒ Δf = (0.1)(0.02)

∴ Δf = 0.002

Now, we have f(10.02) = f(10) + Δf

⇒ f(10.02) = log_e10 + 0.002

⇒ f(10.02) = 2.3026 + 0.002

∴ f(10.02) = 2.3046

Thus, log_e10.02 ≈ 2.3046

log₁₀10.1, it being given that log₁₀e = 0.4343

Let us assume that f(x) = log₁₀x

Also, let x = 10 so that x + Δx = 10.1

⇒ 10 + Δx = 10.1

∴ Δx = 0.1

On differentiating f(x) with respect to x, we get

We know

When x = 10, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.1

⇒ Δf = (0.04343)(0.1)

∴ Δf = 0.004343

Now, we have f(10.1) = f(10) + Δf

⇒ f(10.1) = log₁₀10 + 0.004343

⇒ f(10.1) = 1 + 0.004343 [∵ log_aa = 1]

∴ f(10.1) = 1.004343

Thus, log₁₀10.1 ≈ 1.004343

cos 61°, it being given that sin 60° = 0.86603 and 1° = 0.01745 radian

Let us assume that f(x) = cos x

Also, let x = 60° so that x + Δx = 61°

⇒ 60° + Δx = 61°

∴ Δx = 1° = 0.01745 radian

On differentiating f(x) with respect to x, we get

We know

When x = 60°, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01745

⇒ Δf = (–0.86603)(0.01745)

∴ Δf = –0.0151122

Now, we have f(61°) = f(60°) + Δf

⇒ f(61°) = cos(60°) – 0.0151122

⇒ f(61°) = 0.5 – 0.0151122

∴ f(61°) = 0.4848878

Thus, cos 61° ≈ 0.4848878

Let us assume that

Also, let x = 25 so that x + Δx = 25.1

⇒ 25 + Δx = 25.1

∴ Δx = 0.1

On differentiating f(x) with respect to x, we get

When x = 25, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.1

⇒ Δf = (–0.004)(0.1)

∴ Δf = –0.0004

Now, we have f(25.1) = f(25) + Δf

⇒ f(25.1) = 0.2 – 0.0004

∴ f(15) = 0.1996

Thus,

Let us assume that f(x) = sin x

Let so that

On differentiating f(x) with respect to x, we get

We know

When, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and

∴ Δf = 0

Now, we have

Thus,

Let us assume that f(x) = cos x

Let so that

On differentiating f(x) with respect to x, we get

We know

When, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and

⇒ Δf = (–0.86603)(–0.0873)

∴ Δf = 0.07560442

Now, we have

Thus,

Let us assume that

Also, let x = 81 so that x + Δx = 80

⇒ 81 + Δx = 80

∴ Δx = –1

On differentiating f(x) with respect to x, we get

We know

When x = 81, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –1

⇒ Δf = (0.00926)(–1)

∴ Δf = –0.00926

Now, we have f(80) = f(81) + Δf

⇒ f(80) = 3 – 0.00926

∴ f(80) = 2.99074

Thus, (80)^1/4 ≈ 2.99074

Let us assume that

Also, let x = 27 so that x + Δx = 29

⇒ 27 + Δx = 29

∴ Δx = 2

On differentiating f(x) with respect to x, we get

We know

When x = 27, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 2

⇒ Δf = (0.03704)(2)

∴ Δf = 0.07408

Now, we have f(29) = f(27) + Δf

⇒ f(29) = 3 + 0.07408

∴ f(29) = 3.07408

Thus, (29)^1/3 ≈ 3.07408

Let us assume that

Also, let x = 64 so that x + Δx = 66

⇒ 64 + Δx = 66

∴ Δx = 2

On differentiating f(x) with respect to x, we get

We know

When x = 64, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 2

⇒ Δf = (0.02083)(2)

∴ Δf = 0.04166

Now, we have f(66) = f(64) + Δf

⇒ f(66) = 4 + 0.04166

∴ f(66) = 4.04166

Thus, (66)^1/3 ≈ 4.04166

Let us assume that

Also, let x = 25 so that x + Δx = 26

⇒ 25 + Δx = 26

∴ Δx = 1

On differentiating f(x) with respect to x, we get

We know

When x = 25, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 1

⇒ Δf = (0.1)(1)

∴ Δf = 0.1

Now, we have f(26) = f(25) + Δf

⇒ f(26) = 5 + 0.1

∴ f(26) = 5.1

Thus,

Let us assume that

Also, let x = 36 so that x + Δx = 37

⇒ 36 + Δx = 37

∴ Δx = 1

On differentiating f(x) with respect to x, we get

We know

When x = 36, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 1

⇒ Δf = (0.08333)(1)

∴ Δf = 0.08333

Now, we have f(37) = f(36) + Δf

⇒ f(37) = 6 + 0.08333

∴ f(37) = 6.08333

Thus,

Let us assume that

Also, let x = 0.49 so that x + Δx = 0.48

⇒ 0.49 + Δx = 0.48

∴ Δx = –0.01

On differentiating f(x) with respect to x, we get

We know

When x = 0.49, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.01

⇒ Δf = (0.7143)(–0.01)

∴ Δf = –0.007143

Now, we have f(0.48) = f(0.49) + Δf

⇒ f(0.48) = 0.7 – 0.007143

∴ f(0.48) = 0.692857

Thus,

Let us assume that

Also, let x = 81 so that x + Δx = 82

⇒ 81 + Δx = 82

∴ Δx = 1

On differentiating f(x) with respect to x, we get

We know

When x = 81, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 1

⇒ Δf = (0.00926)(1)

∴ Δf = 0.00926

Now, we have f(82) = f(81) + Δf

⇒ f(82) = 3 + 0.00926

∴ f(82) = 3.00926

Thus, (82)^1/4 ≈ 3.00926

Let us assume that

Also, let so that

On differentiating f(x) with respect to x, we get

We know

When, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and

∴ Δf = 0.0104166

Now, we have

Thus,

Let us assume that

Also, let x = 32 so that x + Δx = 33

⇒ 32 + Δx = 33

∴ Δx = 1

On differentiating f(x) with respect to x, we get

We know

When x = 32, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 1

⇒ Δf = (0.0125)(1)

∴ Δf = 0.0125

Now, we have f(33) = f(32) + Δf

⇒ f(33) = 2 + 0.0125

∴ f(33) = 2.0125

Thus, (33)^1/5 ≈ 2.0125

Let us assume that

Also, let x = 36 so that x + Δx = 36.6

⇒ 36 + Δx = 36.6

∴ Δx = 0.6

On differentiating f(x) with respect to x, we get

We know

When x = 36, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.6

⇒ Δf = (0.0833333)(0.6)

∴ Δf = 0.05

Now, we have f(36.6) = f(36) + Δf

⇒ f(36.6) = 6 + 0.05

∴ f(36.6) = 6.05

Thus,

Let us assume that

Also, let x = 27 so that x + Δx = 25

⇒ 27 + Δx = 25

∴ Δx = –2

On differentiating f(x) with respect to x, we get

We know

When x = 27, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 2

⇒ Δf = (0.03704)(–2)

∴ Δf = –0.07408

Now, we have f(25) = f(27) + Δf

⇒ f(25) = 3 – 0.07408

∴ f(25) = 2.92592

Thus, (25)^1/3 ≈ 2.92592

Let us assume that

Also, let x = 49 so that x + Δx = 49.5

⇒ 49 + Δx = 49.5

∴ Δx = 0.5

On differentiating f(x) with respect to x, we get

We know

When x = 49, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.5

⇒ Δf = (0.0714286)(0.5)

∴ Δf = 0.0357143

Now, we have f(49.5) = f(49) + Δf

⇒ f(49.5) = 7 + 0.0357143

∴ f(49.5) = 7.0357143

Thus,

Given f(x) = 4x² + 5x + 2

Let x = 2 so that x + Δx = 2.01

⇒ 2 + Δx = 2.01

∴ Δx = 0.01

On differentiating f(x) with respect to x, we get

We know and derivative of a constant is 0.

When x = 2, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01

⇒ Δf = (21)(0.01)

∴ Δf = 0.21

Now, we have f(2.01) = f(2) + Δf

⇒ f(2.01) = 4(2)² + 5(2) + 2 + 0.21

⇒ f(2.01) = 16 + 10 + 2 + 0.21

∴ f(2.01) = 28.21

Thus, f(2.01) = 28.21

Given f(x) = x³ – 7x² + 15

Let x = 5 so that x + Δx = 5.001

⇒ 5 + Δx = 5.001

∴ Δx = 0.001

On differentiating f(x) with respect to x, we get

We know and derivative of a constant is 0.

When x = 5, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.001

⇒ Δf = (5)(0.001)

∴ Δf = 0.005

Now, we have f(5.001) = f(5) + Δf

⇒ f(5.001) = 5³ – 7(5)² + 15 + 0.005

⇒ f(5.001) = 125 – 175 + 15 + 0.005

⇒ f(5.001) = –35 + 0.005

∴ f(5.001) = –34.995

Thus, f(5.001) = –34.995

Let us assume that f(x) = log₁₀x

Also, let x = 1000 so that x + Δx = 1005

⇒ 1000 + Δx = 1005

∴ Δx = 5

On differentiating f(x) with respect to x, we get

We know

When x = 1000, we have

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 5

⇒ Δf = (0.0004343)(5)

∴ Δf = 0.0021715

Now, we have f(1005) = f(1000) + Δf

⇒ f(1005) = log₁₀1000 + 0.0021715

⇒ f(1005) = log₁₀10³ + 0.0021715

⇒ f(1005) = 3 × log₁₀10 + 0.0021715

⇒ f(1005) = 3 + 0.0021715 [∵ log_aa = 1]

∴ f(1005) = 3.0021715

Thus, log₁₀1005 = 3.0021715

Given the radius of a sphere is measured as 9 cm with an error of 0.03 m = 3 cm.

Let x be the radius of the sphere and Δx be the error in measuring the value of x.

Hence, we have x = 9 and Δx = 3

The surface area of a sphere of radius x is given by

S = 4πx²

On differentiating S with respect to x, we get

We know

When x = 9, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 3

⇒ ΔS = (72π)(3)

∴ ΔS = 216π

Thus, the approximate error in calculating the surface area of the sphere is 216π cm².

Given a cube whose side x is decreased by 1%.

Let Δx be the change in the value of x.

Hence, we have

∴ Δx = –0.01x

The surface area of a cube of radius x is given by

S = 6x²

On differentiating A with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.01x

⇒ ΔS = (12x)(–0.01x)

∴ ΔS = –0.12x²

Thus, the approximate change in the surface area of the cube is 0.12x² m².

Given the radius of a sphere is measured as 7 m with an error of 0.02 m.

Let x be the radius of the sphere and Δx be the error in measuring the value of x.

Hence, we have x = 7 and Δx = 0.02

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

When x = 7, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.02

⇒ ΔV = (196π)(0.02)

∴ ΔV = 3.92π

Thus, the approximate error in calculating the volume of the sphere is 3.92π m³.

Given a cube whose side x is increased by 1%.

Let Δx be the change in the value of x.

Hence, we have

∴ Δx = 0.01x

The volume of a cube of radius x is given by

V = x³

On differentiating A with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.01x

⇒ ΔV = (3x²)(0.01x)

∴ ΔV = 0.03x³

Thus, the approximate change in the volume of the cube is 0.03x³ m³.

given (ΔL/L)×100 = 2 (if we let the length of pendulum is L)

we all know the formula of period of a pendulum is T=2π×√(l/g)

By the formula of approximation in derivation ,we get-

=1%

given that

% Error in measuring the edge of a cube [(ΔL/L)×100] is = a (if L is edge of the cube)

We have to find out (ΔA/A)×100 = ? (IF let the surface of the cube is A)

By the formula of approximation of derivation we get,

=2×a

=2a

given % error in measuring the radius of a sphere Δr/r×100 =k (if let r is radius)

Find out : (Δv/v)×100 = ?

We know by the formula of the volume of the sphere

So,

=3×k

=3k%

let height of a cylinder=h=radius of that cylinder=r

% error in height Δh/h×100 = a (given)

Volume of cylinder= v =(1/3)×πr²h

We have given that h=r

Then

So,

Finally

=3×a

=3a%

we know that the area of a equilateral traiangle is =A =(√3/4)×a²

Where a= side of equilateral triangle

So by the formula of approximation of derivation, we get,

=2×k

=2k% ans

let y=f(x)=logx

Let x=4,

X+Δx=4.01,

Δx=0.01,

For x=4,

Y=log4=1.3868,

y=logx

Δy=dy

Δy=0.0025

So, log(4.01)=y+Δy

=1.3893

we know that volume of sphere = v = (4/3)×πr³ (r is radius of sphere)

r = 100mm

=4πr²Δr

Δr = (98-100)

=-2

Δv=4π(100)²×(-2)

Δv = - 80,000π mm³ans

given that the radius is half then the height of the cone so

Let h = 2r (where r is radius and h is height of the cone)

Volume of the cone = v

(because h = 2r )

Δv = 2πr²Δr

So finally ,

= 3×λ

= 3λ%

let pv^1/4=k (constant)

Pv^1/4=k

P=k.v^-1/4

log(p) = log(k.v^-1/4)

log(p) = log(k) – (1/4)log(v)

given y=xⁿ

Δy = n.x^n-1.Δx = x

So finally ratio is = n:1

f(x) = x^1/5

F’(x) = (1/5).x^-4/5

F(a+h) = f(a) + h×f’(a)

Now

Let a = 32 & h=1

=2.0125

given that circumference is = C = 2πr = 28 cm

That’s mean r=14/π

ΔC = 2πΔr = 0.01

Δr = (0.01/2π)

We all know that area of a circle is = A =πr²

ΔA= 2πr×dr

So finally,

= 1/14

by the formula of differentiation we all know that-

……….eq(1)

Ify=x² then

,so put the value of in eq(1),we get-

Δy=2×10×(0.1)

Δy=2

given that

Y=logx then y’=1/x

Δy=?

X=3

Δx=0.03

By putting the values of above in the formula we get

Δy=0.01

given that

(if let r is radius)

(if let A is area of circle)

We know that the area of a circle(A)=πr² then

dA=2πr×dr

now

we know that if there is a little approximation in variables then,

=2×a

=2a

given that

(if let r is a radius of a sphere)

We know that

Then, dv = (4πr²)×dr

Finally

=3×a

=3a%

given that cube edge error %[(Δx/x)×100]=a

Volume %=?

Let the edge of cube is x,

Volume; v=x³

Then, dv=3x².dx

So finally

=3×a

=3a