Chapter 8

1. Find the linear approximation for

\sqrt{1 + X}, x \geq - 1 a t x_{0} = 3

. Use the linear approximation to estimate f (3.2) .
Solution
We know from (4), that L(x) =

f (x_{0}) + f' (x_{0}) (x - x_{0})

. We have

x_{0} = 3 △ x = 0.2

and hence

f (3) = \sqrt{1 + 3}

= . Also,

f' (x) = \frac{1}{2 \sqrt{1 + x}} a n d h e n c e f' (3) = \frac{1}{2 \sqrt{1 + 3}} . T h u s, L (x) = 2 + \frac{1}{4} (x - 3) = \frac{x}{4} + \frac{x}{5} g i v e s t h e r e q u i r e d l i n e a r a p p r o x i m a t i o n . N o w, f (3.2) = \sqrt{4.2} ≃ L (3, 2) = \frac{3.2}{4} + \frac{5}{4}

=
Actually, if we use a calculator to calculate we get

\sqrt{4.2}

= .

2. 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) =

{πr}^{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 (), we have
Change in the surface area = S(5.2) − S(5)
= 8

π

()()
=

{πcm}^{2}

Exact calculation of the change in the surface gives
S(5.2) − S(5) =

π

- 100

π

{πcm}^{2}

.
Percentage error = relative error × 100 =

\frac{8.16 π - 8 π}{8.16 π} x 100

= %

3. Let g(x) =

x^{2}

+ sin x . Calculate the differential dg .
Solution
Note that g is differentiable and g'(x) = .
Thus dg = ()dx .

4. Percentage error= ×100

5. error =Actual value – value

6. All the rules for limits (limit theorems) for functions of also hold true for .

Chapter 8

Gap-fill exercise