Here we will use differentials to approximate values of certain quantities.
Let f: D ---- R, DCR, be a given function and y = f(x). Let ∆ x be a small increment in x, so that ∆ y will be the corresponding increment in y then,
∆ y is given by the formula, ∆ y = f(x + ∆ x) - f(x).
i) The differential of x is denoted by dx and it is defined by dx = ∆ x
ii) The differential of y, denoted by dy, is defined by dy = f'(x) dx or dy = (dy/dx) ∆x
Approximate Value of irrationals
For finding the approximate value of irrationals, first we have the take the integral part or bigger number as 'x' and the decimal part or smaller number as ∆x. Here, we take dy = ∆y and for evaluating dy use the formula dy = (dy/dx) ∆x.
For example: Use differentials to approximate (25)1/3
(25)1/3 = (27 + (-2))1/3
Take x = 27, which is a perfect cube and ∆x = -2
y + ∆y = (x + ∆x)1/3
∆y = (x + ∆x)1/3 - x1/3
= (27 + (-2))1/3 - (27)1/3
∆y = (25)1/3 - 3 ……………………..(i)
∆y = dy = (dy/dx) ∆x
Equation (i) becomes -0.074 = (25)1/3 - 3
-0.074 + 3 = (25)1/3
Hence (25)1/3= 2.926
Approximate value of a function
In this case a function f(x) will be given and we have to find the value of the function at a given decimal number. Here also, we take the integral part as 'x' and decimal part as ∆x. The formula is, f(x+∆x)=∆y + f(x), where ∆y=f'(x)∆x
For example: Find the approximate value of f(3.02) where f(x)=3x2+5x+3
Let x = 3 and ∆x = 0.02,
f(x) = 3x2 + 5x + 3
f'(x) = 6x + 5
∆y = f'(x) ∆x
= (6x + 5) (0.02)
= (6 x 3 + 5) (0.03)
= 2 3x 0.03
f (3 + .02) = 0.46 + f(3)
=0.46 + [3(3)2+5(3)+3]
=0.46 + 45
f(3.02) = 45.46
Here we learn to find the approximate error in volume, surface area etc caused by the error in taking radius.
For example: If the radius of a sphere is measured as 9m with an error of 0.03m, then find the approximate in calculating its surface area.
Solution: r=9m and ∆r=0.03m
Thus the approximate error in calculating the volume is 9.72∏m3
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