Answer to Question #268572 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.

Problem A.2

Let f(x), g(x) and h(x) be the functions from Problem A.1. Find the derivative λ

(x) of the

following function with respect to x:

λ(x) = f(x) · g(x) + f(x) · h(x) − g(x) · h(x)

*Please give specific answers to both Problem A.1 & A.2

Answer 1

A.1

(I) length of f(x)

Given

$f(x)=x-2, for \ 2\leq x\leq4$

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

$\implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2}$

$\implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2}$

length of graph contributed by

$f(x)=2\sqrt{2}$

$f(x)=2\sqrt{2}$

(ii) Length of g(x)

Given

$g(x)=2\sqrt{4-(x-6)^2+2}, \ for \ 4\leq x \leq8$

$g(x)=2\sqrt{4-(x-6)^2+2}, \ for \ 4\leq x \leq8$

$\implies g(x)-2=\sqrt{4-(x-6)^2}\\\implies(g(x)-2)^2=4-(x-6)^2\\\implies(x-6)^2+(g(x)-2)^2=2^2$

$\implies g(x)-2=\sqrt{4-(x-6)^2}\\\implies(g(x)-2)^2=4-(x-6)^2\\\implies(x-6)^2+(g(x)-2)^2=2^2$

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)=$\pi r=2\pi$

(III) Length of h(x)

Given,

$f(x)=x-6, for\ 8\leq x\leq10$

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

$\implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2}$

$\implies h^2=b^2+h^2\\\implies h^2=2^2+2^2\\\implies h^2=8\\h=2\sqrt{2}$

length of graph contributed by

h(x)=$2\sqrt{2}$

$2\sqrt{2}$

Total length of the graph

$L=length\ contributed \ by \ f(x)+length\ contributed \ by \ g(x)+length\ contributed \ by \ h(x)$ $\implies L=2\sqrt{2}+2\pi+2\sqrt{2}\\\implies L=4\sqrt{2}+2\pi$

$\implies L=2\sqrt{2}+2\pi+2\sqrt{2}\\\implies L=4\sqrt{2}+2\pi$$\implies L=2\sqrt{2}+2\pi+2\sqrt{2}\\\implies L=4\sqrt{2}+2\pi$$\implies L=2\sqrt{2}+2\pi+2\sqrt{2}\\\implies L=4\sqrt{2}+2\pi$

A.2

λ(x) = f(x) · g(x) + f(x) · h(x) − g(x) · h(x)

$\lambda (x)=((x-2)(2\sqrt{4-(x-6)^2+2})+((x-2)(x-6))-((x-6)(2\sqrt{4-(x-6)^2+2})\\$

$\lambda (x)=\frac{d}{dx}((x-2)(2\sqrt{4-(x-6)^2+2})+\frac{d}{dx}((x-2)(x-6))-\frac{d}{dx}((x-6)(2\sqrt{4-(x-6)^2+2})\\$

$\lambda'(x)=\frac{2(-2x^2+20x-42)}{\sqrt{-x^2+12x-30}}+(2x-8)+\frac{2(-2x^2+24x-66)}{\sqrt{-x^2+12x-30}}$

$\lambda'(x)=\frac{\left(-8x^2+88x-216\right)\sqrt{-x^2-30+12x}}{-x^2-30+12x}+2x-8$