Mathematics Approximations For CBSE-NCERT
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### 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.
Q 3125045861

Use differential to approximate sqrt (36.6)
Class 12 Chapter 6 Example 21
Solution:

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
Solution:

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
Solution:

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
Solution:

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
Solution:

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.