5. Approximations

Approximations:

To find a very small change or variation of a quantity, we can use derivatives to give the approximate value of it. The approximate value is represented by delta △.

Suppose change in the value of x, dx = x then,

dy/dx = △x = x.

Since the change in x, dx ≈ x therefore, dy ≈ y.

Example :

Approximate 25.5

using differential.

Solution:

Let us consider y = x

, where x = 25 and ∆x = 0.5. Then,

∆y = f(x+∆x) - f(x)= x+△x-x

• ∆y =25.5-25
• ∆y =25.5–5
• 25.5=△y+5

Since dy is approximately equal to ∆y, therefore

dy=dydx△x=12x(0.5)=0.05

Therefore, the approximate value of 25.5=5+0.05=5.05

Example :

Find the approximate value of the function f(3.02), where f(x) is given as 3x²+5x+3.

Solution:

Given that, f(x) = 3x²+5x+3

Assume x = 3, and ∆x = 0.02.

Hence, we can write the given function as:

f (3. 02) = f (x + ∆x) = 3(x + ∆x)² + 5(x + ∆x) + 3

We know that,

∆y = f (x + ∆x) – f (x). 

The above expression can be written as

f (x + ∆x) = f (x) + ∆y

As, dx = ∆x, it can be approximately written as f (x) + f ′(x) ∆x

Hence, f (3.02) ≈ (3x² + 5x + 3) + (6x + 5) ∆x

Now, substitute the values of x and ∆x, we get

= (3(3)² + 5(3) + 3) + (6(3) + 5) (0.02) 

Now, simplify it to get the approximate value

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

= 45 + 0.46 

= 45.46

Therefore, the approximate value of f(3.02) is 45.46.