Winter 18 1Consider the collection of functions that meet the following criteria

Question

Winter 18 1Consider the collection of functions that meet the following criteria: . f is continuous on the interval [-1,1] f bounds an area of 4 with the x axis on this interval. That is, Jf() dr There are lots of functions that do this. For example, the function f(z)-1-2 satisfies the first two conditions, but 1- dr-3. Adjusting, f(x) 31-z2) now meets all three criteria. t1, 0) Another example of a function that meets all three criteria is f(z) 2(1-1). We can show that this section of the parabola has arc length 6.5. This scction of the absolute value function has arc length 8.25. Your job is to find a function that satisfies the three criteria above. I will give extra points to the group whose function has the smallest arc length. Be creative. Maybe try rational functions? piecewise functions? trig functions? something else? To receive credit, you must show that your function satisfies the three criteria above. Then you must set up your arc length integral. You can use a calculator to evaluate this integral, but I will give bonus points if you can compute it by hand.

Explanation / Answer

equation of semicircle of radius 1 centered at origin above x axis is y=  (1-x2)

area of semicicle of radius 1 , centered at origin is (/2)

our required function f bounds an 4

let function is of form f(x)=k(1-x2)

area under f(x) is =[-1 to 1] k(1-x2) dx=k*(/2) = 4

=>k=(8/)

so our required function is f(x)=(8/)(1-x2)

so f(x) is continous , f(-1)=f(1)=0 , area bounded with x axis is 4

f(x)=(8/)(1-x2)

=>f '(x)=(8/)(-x/(1-x2))

arclength =[-1 to 1][1+(f '(x))2] dx

arclength =[-1 to 1][1+((8/)(-x/(1-x2)))2] dx

arclength =[-1 to 1][1+(64x2/2(1-x2))] dx

arclength = 5.83893..

by calculator

arclength =5.84

please rate if helpful. please comment if you have any doubt

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