`star` Approximations

`\color{green} ✍️` Let `f : D →R,` `D ⊂ R,` be a given function and let `y = f (x).`

`\color{green} ✍️` Let `Δx` denote a small increment in `x.` Recall that the increment in `y` corresponding to the increment in `x,` denoted by `Δy,` is given by `color{blue}{Δy = f (x + Δx) – f (x).}`

We define the following

(i) The differential of `x,` denoted by `dx,` is defined by `dx = Δx.`

(ii) The differential of `y,` denoted by `dy,` is defined by `dy = f′(x) dx` or

`color{red}{dy = ( (dy)/(dx) ) Δx}`

`\color{green} ✍️` In case `dx = Δx` is relatively small when compared with `x,`

`dy` is a good approximation of `Δy` and we denote it by `dy ≈ Δy.`

`=>` For geometrical meaning of `Δx, Δy, dx` and `dy,` one may refer to Fig.

`\color{green} ✍️` Let `Δx` denote a small increment in `x.` Recall that the increment in `y` corresponding to the increment in `x,` denoted by `Δy,` is given by `color{blue}{Δy = f (x + Δx) – f (x).}`

We define the following

(i) The differential of `x,` denoted by `dx,` is defined by `dx = Δx.`

(ii) The differential of `y,` denoted by `dy,` is defined by `dy = f′(x) dx` or

`color{red}{dy = ( (dy)/(dx) ) Δx}`

`\color{green} ✍️` In case `dx = Δx` is relatively small when compared with `x,`

`dy` is a good approximation of `Δy` and we denote it by `dy ≈ Δy.`

`=>` For geometrical meaning of `Δx, Δy, dx` and `dy,` one may refer to Fig.

Q 3125045861

Use differential to approximate `sqrt (36.6)`

Class 12 Chapter 6 Example 21

Class 12 Chapter 6 Example 21

Take `y = sqrt x` let x=36 and let `Delta x = 0.6` Then

`Delta y = sqrt(x+ Δ x) - sqrt x = sqrt (36.6) - sqrt(36) = sqrt (36.6) - 6`

or `sqrt (36.6) = 6 + Δ y`

Now dy is approximately equal to Δy and is given by

`dy = ( (dy)/(dx) ) Δ x = 1/(2 sqrt x) (0.6) = 1/(2 sqrt 36) (0.6) = 0.05` (as `y = sqrt x )` )

Thus, the approximate value of `sqrt 36.6` is 6 + 0.05 = 6.05.

Q 3165045865

Use differential to approximate ` (25)^(1/3)`

Class 12 Chapter 6 Example 22

Class 12 Chapter 6 Example 22

Let ` y = x^2/3` . Let x = 27 and let Δx = – 2. Then

` Δy = (x+Δ x )^(1/3) - x^(1/3) = (25)^(1/3) - (27)^(1/3) = (25)^(1/3) - 3`

or `(25)^(1/3) = 3+ Δ y`

Now dy is approximately equal to Δy and is given by

`dy = ( (dy)/(dx) ) Δ x = 1/(3x^(2/3)) (-2) ` ( as ` y = x^(1/3) `)

` = 1/(3 ( (27)^(1/3))^2) = (-2)/(27) = -0.074`

Thus, the approximate value of ` (25)^(1/3) ` is given by

`3 + (– 0. 074) = 2.926`

Q 3175045866

Find the approximate value of f (3.02), where `f (x) = 3x^2 + 5x + 3`.

Class 12 Chapter 6 Example 23

Class 12 Chapter 6 Example 23

Let x = 3 and Δx = 0.02. Then

`f (3. 02) = f (x + Δx) = 3(x + Δx)^2 + 5(x + Δx) + 3`

Note that Δy = f (x + Δx) – f (x). Therefore

f (x + Δx) = f (x) + Δy

≈ f (x) + f ′(x) Δx (as dx = Δx)

or `f (3.02) ≈ (3x^2 + 5x + 3) + (6x + 5) Δx`

`= (3(3)^2 + 5(3) + 3) + (6(3) + 5) (0.02)` (as x = 3, Δx = 0.02)

= (27 + 15 + 3) + (18 + 5) (0.02)

= 45 + 0.46 = 45.46

Hence, approximate value of f (3.02) is 45.46.

Q 3105045868

Find the approximate change in the volume V of a cube of side x meters

caused by increasing the side by 2%.

Class 12 Chapter 6 Example 24

caused by increasing the side by 2%.

Class 12 Chapter 6 Example 24

Note that

`V = x^3`

or `dV = ( (dV)/(dx) ) Δ x = (3x^2) Δ x`

`= ( 3 x^2) (0.02x) = 0.06 x^3 m^3` (as 2% of x is 0.02x)

Thus, the approximate change in volume is `0.06 x^3 m^3` .

Q 3125145961

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm,

then find the approximate error in calculating its volume.

Class 12 Chapter 6 Example 25

then find the approximate error in calculating its volume.

Class 12 Chapter 6 Example 25

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then r = 9 cm and Δr = 0.03 cm. Now, the volume V of the sphere is given by

`V = 4/3 pi r^3`

or ` (dV)/(dr) = 4 pi r^2`

Therefore `dV = ( (dV)/(dr) ) Δ r = (4 pi r^2) Δ r`

`= 4π(9)^2 (0.03) = 9.72π cm^3`

Thus, the approximate error in calculating the volume is `9.72π cm^3`.