When we are approximating a value, there occurs an error. In this section, we consider the error, which occurs by linear approximation, given by (4).

**Errors: Absolute
Error, Relative Error, and Percentage Error**

When we
are approximating a value, there occurs an error. In this section, we consider
the error, which occurs by linear approximation, given by (4). We shall
consider different types of errors. Taking *h*
=
*x* âˆ’ *x*_{0}
, we get *x* =
*x*_{0} +
*h* , then (5) becomes

*E *(* h*)* *=* f *(*x*_{0}* *+* h *)*
*âˆ’* f *(*x*_{0}* *)*
*âˆ’* f *â€²(*x*_{0}* *)*h *. ... (6)

Note
that *E*(0) =
0 and as we have alrea*dy* observed lim_{hâ†’0}
*E* ( *h*) = 0 follows from the continuity of *f* at *x*_{0}
. In addition, if *f* is
differentiable, then from (1), it follows that

Thus
when *f* is differentiable at *x*_{0} , then the above equation
shows that *E* ( *h*) actually approaches zero faster than *h* approaching zero. Now, we define

Suppose that certain quantity is to be determined. Itâ€™s exact
value is called the **actual value**. Some
times we obtain its **approximate value** through
some approximation process. In this case, we define

Absolute error = Actual value âˆ’ Approximate value.

So (6)
gives the absolute error that occurs by a linear approximation. Let us look at
an example illustrating the use of linear approximation.

** **

Use
linear approximation to find an approximate value of âˆš9.2 without using a calculator.

We need
to find an approximate value of âˆš9.2 using linear approximation. Now by (3), we
have *f *(*x*_{0}* *+* *Î”*x *)* *â‰ˆ* f *(*x*_{0}* *)*
*+* f *â€²(*x*_{0}* *)* *Î”*x *. To do this, we have to identify an appropriate function* f *, a point* x*_{0}* *and Î”*x* . Our choice should be such that the
right side of the above approximate equality, should be* *computable without the help of a calculator. So, we choose* f *(*x*)* *=* âˆšx
*,*
x*_{0}* *=* *9* *and* *Î”*x
*=* *0.2* *. Then,* f* â€²(*x*_{0})
= 1/2âˆš9 and hence

âˆš9.2 â‰ˆ *f* (9) +* f *â€²(9)(0.2) = 3 + 0.2/6 = 3.03333 .

Now if
we use a calculator, just to compare, we find âˆš9.2 = 3.03315 . We see that our
approximation is accurate to three decimal places and the error is 3.03315 âˆ’
3.03333 = âˆ’0.00018 . [Also note that one could choose *f (x)* = âˆš(1+* x*)* *, x_{0} = 8 and* *Î”*x
*= 0.2 . So the choice of* f *and x_{0}
are not necessarily unique].

So in
the above example, the absolute error is 3. 03315 3.03333 0.00018 . Note that
the absolute error says how much the error; but it does not say how good the
approximation is. For instance, let us consider two simple cases.

**Case 1 : **Suppose that the actual value of
something is** **5** **and its approximated value is**
**4** **, then** **the absolute error is 5 âˆ’ 4 = 1 .

**Case 2 : **Suppose that the actual value of
something is** **100** **and its approximated value is 95. In this case, the absolute error
is 100 âˆ’
95 =
5 . So the absolute error in the first case is smaller when compared to the
second case.

Among
these two approximations, which is a better approximation; and why? The
absolute error does not give a clear picture about whether an approximation is
a good one or not. On the other hand, if we calculate relative error or
percentage of error (defined below), it will be easy to see how good an
approximation is. If the actual value is zero, then we do know how close our
approximate answer is to the actual value. So if the actual value is not zero,
then we define,

**Definition 8.3**

If the actual value is not zero, then

Relative error = [ Actual value âˆ’ Approximate value ] / Actual value

Percentage error = Relative error Ã—100 .

**Note**:** **Absolute
error has unit of measurement where as relative error and percentage error are
units free.

Note
that, in the case of the above examples,

In the
first case

The
relative error = 1/5 = 0.2 ; and the percentage error = 1/5 Ã—100 = 20% .

In the
second case

The
relative error = 5/100 ; and the percentage error = 5/100 Ã—100 = 5%.

So the
second approximation is a better approximation than the first one. Note that,
in order to calculate the relative error or the percentage error one should
know the actual value of what we are approximating.

Let us
consider some examples.

** **

**Example 8.3**

Let us
assume that the shape of a soap bubble is a sphere. Use linear approximation to
approximate the increase in the surface area of a soap bubble as its radius
increases from 5 cm to 5.2 cm. Also, calculate the percentage error.

**Solution**

Recall
that surface area of a sphere with radius *r*
is given by *S* ( *r*) = 4*Ï€*
*r*^{2} . Note that even though
we can calculate the exact change using this formula, we shall try to
approximate the change using the linear approximation. So, using (4), we have

Change
in the surface area = *S*
(5.2) âˆ’
*S*(5) â‰ˆ *S*â€²(5)(0.2)

= 8*Ï€*(5)( 0 .2) =
8*Ï€* cm^{2}

Exact
calculation of the change in the surface gives

*S *(5.2)* *âˆ’* S*(5)* *=* *108.16*Ï€ -*
100* Ï€* = 8.16 cm^{2} .

Percentage
error =
relative error Ã— 100 = { 8.16*Ï€*âˆ’8*Ï€*
/ 8.16*Ï€* } Ã— 100 = 1.9607%

** **

**Example 8.4**

A right
circular cylinder has radius *r* =
10 cm. and height *h* =
20 cm. Suppose that the radius of the cylinder is increased from 10 cm to 10.1
cm and the height does not change. Estimate the change in the volume of the
cylinder. Also, calculate the relative error and percentage error.

**Solution**

Recall
that volume of a right circular cylinder is given by *V* = *Ï€*
*r*^{2}*h* , where *r* is the radius
and *h* is the height. So we have *V* ( *r*)
=
*Ï€**r*^{2}*h* = 20*Ï€**r*^{2}.

Thus the
estimate for the change in the volume is 40*Ï€*
cm^{3}.

Exact
calculation of the volume change gives

*V *(10.1)* *âˆ’*V *(10)* *=* *2040.2*Ï€** *âˆ’* *2000*Ï€** *=* *40.2*Ï€** *cm^{3}.

So
relative error = (40.2Ï€âˆ’40Ï€) / (40.2Ï€) =
1 / 201 = 0.00497 ; and hence

the
percentage error = relative error Ã— 100 = [1/201] Ã— 100 = 0.497% .

Tags : Linear Approximation | Mathematics , 12th Maths : UNIT 8 : Differentials and Partial Derivatives

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12th Maths : UNIT 8 : Differentials and Partial Derivatives : Errors: Absolute Error, Relative Error, and Percentage Error | Linear Approximation | Mathematics

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