Answer to Question #268558 in Calculus for K11

Question

Problem A.1

The graph below is made of three line segments:

-1 1 2 3 4 5 6 7 8 9 10 11 12

1

2

3

4

y

x

f(x)

g(x)

h(x)

The segments correspond to the following three functions:

f(x) = x − 2, g(x) = p

4 − (x − 6)2 + 2, h(x) = x − 6

Find the total length L of the graph between x = 2 and x = 10.

Answer 1

(i) Length of f(x)

Given,

$f(x)=x-2, \quad for 2 \leq x \leq 4$

The length would be the length of hypotenuse of right angled isosceles triangle of side length =2 units

 $\begin{aligned} &\Rightarrow h^{2}=b^{2}+h^{2} \\ &\Rightarrow h^{2}=2^{2}+2^{2} \\ &\Rightarrow h^{2}=8 \\ &\Rightarrow h=2 \sqrt{2} \end{aligned}$

Length of graph contributed by $f(x)=2 \sqrt{2}$

(ii) Length of g(x)

Given,

$\begin{aligned} &g(x)=\sqrt{4-(x-6)^{2}}+2, \quad \text { for } 4 \leq x \leq 8 \\ &\Rightarrow g(x)-2=\sqrt{4-(x-6)^{2}} \\ &\Rightarrow(g(x)-2)^{2}=4-(x-6)^{2}\\ &\Rightarrow (x-6)^{2}+(g(x)-2)^{2}=2^2 \end{aligned}$

So, g(x) is a circle with centre at (6,2) and radius = 2 units

Length contributed by g(x) would be circumference of semicircle

Length contributed by g(x) = $\pi r=2\pi$

(iii) Length of $\mathrm{h}(\mathrm{x})$

Given,

$f(x)=x-6, \quad for \ 8 \leq x \leq 10$

The length would be the length of hypotenuse of right angled triangle of base = height =2 units

$\begin{aligned} &\Rightarrow h^{2}=b^{2}+h^{2} \\ &\Rightarrow h^{2}=2^{2}+2^{2} \\ &\Rightarrow h^{2}=8 \\ &\Rightarrow h=2 \sqrt{2} \end{aligned}$  

Length of graph contributed by $h(x)=2 \sqrt{2}$

Total length of graph

$\begin{aligned} &L=\text { length contributed by } f(x)+\text { length contributed by } g(x)+\text { length contributed by } h(x) \\ &\Rightarrow L=2 \sqrt{2}+2 \pi+2 \sqrt{2} \\ &\Rightarrow L=4 \sqrt{2}+2 \pi \end{aligned}$