(I) length of f(x)
Given
f ( x ) = x − 2 , f o r 2 ≤ x ≤ 4 f(x)=x-2, for \ 2\leq x\leq4 f ( x ) = x − 2 , f or 2 ≤ x ≤ 4
the length would be the length of hypotenuse and right angled isosceles triangle of side length = 2 units
⟹ h 2 = b 2 + h 2 ⟹ h 2 = 2 2 + 2 2 ⟹ h 2 = 8 h = 2 2 \implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2} ⟹ h 2 = b 2 + h 2 ⟹ h 2 = 2 2 + 2 2 ⟹ h 2 = 8 h = 2 2
length of graph contributed by
f ( x ) = 2 2 f(x)=2\sqrt{2} f ( x ) = 2 2
(ii) Length of g(x)
Given
g ( x ) = 2 4 − ( x − 6 ) 2 + 2 , f o r 4 ≤ x ≤ 8 g(x)=2\sqrt{4-(x-6)^2+2}, \ for \ 4\leq x \leq8 g ( x ) = 2 4 − ( x − 6 ) 2 + 2 , f or 4 ≤ x ≤ 8
⟹ g ( x ) − 2 = 4 − ( x − 6 ) 2 ⟹ ( g ( x ) − 2 ) 2 = 4 − ( x − 6 ) 2 ⟹ ( x − 6 ) 2 + ( g ( x ) − 2 ) 2 = 2 2 \implies g(x)-2=\sqrt{4-(x-6)^2}\\\implies(g(x)-2)^2=4-(x-6)^2\\\implies(x-6)^2+(g(x)-2)^2=2^2 ⟹ g ( x ) − 2 = 4 − ( x − 6 ) 2 ⟹ ( g ( x ) − 2 ) 2 = 4 − ( x − 6 ) 2 ⟹ ( x − 6 ) 2 + ( g ( x ) − 2 ) 2 = 2 2
so, g(x) is a circle with center at (6,2) and radius = 2 units
length contributed by g(x) would be circumference of semicircle
length contributed by g(x)=π r = 2 π \pi r=2\pi π r = 2 π
(III) Length of h(x)
Given,
f ( x ) = x − 6 , f o r 8 ≤ x ≤ 10 f(x)=x-6, for\ 8\leq x\leq10 f ( x ) = x − 6 , f or 8 ≤ x ≤ 10
the length would be the length of hypotenuse of right angled triangle of base =height=2 units
⟹ h 2 = b 2 + h 2 ⟹ h 2 = 2 2 + 2 2 ⟹ h 2 = 8 h = 2 2 \implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2} ⟹ h 2 = b 2 + h 2 ⟹ h 2 = 2 2 + 2 2 ⟹ h 2 = 8 h = 2 2
length of graph contributed by
h(x)=2 2 2\sqrt{2} 2 2
Total length of the graph
L = l e n g t h c o n t r i b u t e d b y f ( x ) + l e n g t h c o n t r i b u t e d b y g ( x ) + l e n g t h c o n t r i b u t e d b y h ( x ) L=length\ contributed \ by \ f(x)+length\ contributed \ by \ g(x)+length\ contributed \ by \ h(x) L = l e n g t h co n t r ib u t e d b y f ( x ) + l e n g t h co n t r ib u t e d b y g ( x ) + l e n g t h co n t r ib u t e d b y h ( x ) ⟹ L = 2 2 + 2 π + 2 2 ⟹ L = 4 2 + 2 π \implies L=2\sqrt{2}+2\pi+2\sqrt{2}\\\implies L=4\sqrt{2}+2\pi ⟹ L = 2 2 + 2 π + 2 2 ⟹ L = 4 2 + 2 π
Comments