MHB Bordiba's questions at Yahoo Answers regarding roots, area and arc-length

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The discussion revolves around solving a calculus problem involving continuous functions that meet specific criteria regarding area and arc length. Two functions are proposed: an absolute value function and a quadratic function, both of which satisfy the conditions of being non-negative on the interval [0,1], having roots at 0 and 1, and enclosing an area of 1. The arc lengths of these functions are calculated, with the quadratic function yielding an approximate length of 3.249 and the absolute value function around 2.963. Additionally, a quartic function is explored, indicating that as the degree of the polynomial increases, the arc length initially decreases before eventually increasing again, with a notable turning point at the 10th degree polynomial.
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Here is the question:

Need help on calc 2 question?

I need a good explanation on how to solve this question
a) find at least two continuous functions F that satisfy each condition
i) F(x) ≥ 0 on [0,1]
ii) F(0)=0 and F(1)=0
iii) the area bounded by the graph of F and the x-axis for 0≤x≤1 equals 1

B) for each function in part (a) approximate the arc length of the graph of the function on the interval[0,1]

C) Can you find a function F that satisfies each condition on part (a) and whose graph has an arc-length of less than 3 on the interval [0,1]

I have posted a link there to this topic so the OP can see my work.
 
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Hello Bordiba,

a) I say we keep things as simple as possible. For our first function, let's try an absolute value function whose axis of symmetry is at $$x=\frac{1}{2}$$, and has the required roots and area. So let's begin with:

$$f(x)=A|2x-1|+B$$

With the required roots in mind, we may write:

$$f(0)=f(1)=A+B=0\,\therefore\,A=-B$$

With the required triangular area in mind, we find:

$$\frac{1}{2}(1-0)B=1\,\therefore\,B=2$$

and so we have:

$$f(x)=-2|2x-1|+2$$

Here is a plot of the function:

View attachment 1124

Next, let's try a quadratic function.

With the required roots in mind, we may write it as:

$$f(x)=Ax(x-1)=A\left(x^2-x \right)$$

With the required area in mind, we may write:

$$A\int_0^1 x^2-x\,dx=1$$

Applying the FTOC, we have:

$$A\left[\frac{1}{3}x^3-\frac{1}{2}x^2 \right]_0^1=1$$

$$A\left[2x^3-3x^2 \right]_0^1=6$$

$$A\left(2-3 \right)=6$$

$$A=-6$$

And so we have:

$$f(x)=6x(1-x)$$

Here is a plot of the function:

View attachment 1125

b) To find the arc-lengths, we need only apply the Pythagorean theorem for the first function:

$$s=2\sqrt{\left(\frac{1}{2} \right)^2+2^2}=\sqrt{17}$$

For the second function, we will use:

$$s=2\int_0^{\frac{1}{2}}\sqrt{1+(6(1-2x))^2}\,dx$$

Using the trigonometric substitution:

$$6(1-2x)=6-12x=\tan(\theta)\,\therefore\,-12\,dx=\sec^2(\theta)\,d\theta$$

we have:

$$s=\frac{1}{6}\int_0^{\tan^{-1}(6)}\sec^3(\theta)\,d\theta$$

With a bit of manipulation, we can get this in a form we can integrate directly:

$$\sec^3(\theta)=\frac{1}{2}\left(\sec^3(\theta)+ \sec^3(\theta) \right)=\frac{1}{2}\left(\sec^3(\theta)+ \sec(\theta)\left(\tan^2(\theta)+1 \right) \right)=$$

$$\frac{1}{2}\left(\sec^3(\theta)+ \sec(\theta)\tan^2(\theta)+\sec(\theta)\frac{\tan(\theta)+ \sec(\theta)}{\tan(\theta)+\sec(\theta)} \right)=$$

$$\frac{1}{2}\left(\left(\sec(\theta)\sec^2(\theta)+\sec(\theta)\tan^2(\theta) \right)+\left(\frac{\sec(\theta)\tan(\theta)+\sec^2(\theta)}{\sec(\theta)+\tan(\theta)} \right) \right)=$$

$$\frac{1}{2}\frac{d}{d\theta}\left(\sec(\theta)\tan(\theta)+\ln\left|\sec(\theta)+\tan(\theta) \right| \right)$$

So now we have:

$$s=\frac{1}{12}\int_0^{\tan^{-1}(6)}\,d \left(\sec( \theta)\tan( \theta)+\ln\left|\sec( \theta)+\tan( \theta) \right| \right)$$

Applying the FTOC, we have:

$$s=\frac{1}{12}\left[\sec( \theta)\tan( \theta)+\ln\left|\sec( \theta)+\tan( \theta) \right| \right]_0^{\tan^{-1}(6)}=$$

$$\frac{1}{12}\left(6\sqrt{37}+\ln\left(\sqrt{37}+6 \right) \right)\approx3.249029586202852$$

c) Let's try a quartic function this time, since the area seem to be decreasing as the order increases. We may try:

$$f(x)=A\left(x-\frac{1}{2} \right)^4+B$$

With the root requirements in mind, we find:

$$f(0)=f(1)=\frac{A}{16}+B=0\,\therefore\,A=-16B$$

Now, with the requirement on the area in mind, we may write:

$$B\int_0^1 1-16\left(x-\frac{1}{2} \right)^4\,dx=1$$

Applying the FTOC, we have:

$$B\left[x-\frac{16}{5}\left(x-\frac{1}{2} \right)^5 \right]_0^1=1$$

$$B\left[5x-16\left(x-\frac{1}{2} \right)^5 \right]_0^1=5$$

$$B\left(\left(5-\frac{1}{2} \right)-\left(0+\frac{1}{2} \right) \right)=5$$

$$B=\frac{5}{4}$$

Thus, our function is:

$$f(x)=-20\left(x-\frac{1}{2} \right)^4+\frac{5}{4}=\frac{5}{4}\left(1-(2x-1)^4 \right)$$

Here is a plot of the function:

View attachment 1126

Using numeric integration, we find the arc-length is:

$$s=\int_0^1\sqrt{1+100(2x-1)^6}\,dx\approx
2.96314204075918$$
 

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I wanted to post a bit more about part c) of this problem. I realized I said that as the degree increases, the arc-length decreases, yet as I reflected on this, I realized that as the order approaches infinity, the function becomes square and so the arc-length must approach 3. So, it occurred to me that perhaps the arc-length does in fact decrease at first, but then at some point begins increasing again.

So, let's define our function as:

$$f(x)=A(2x-1)^{2n}+B$$

Recall, we require:

$$f(0)=f(1)=A+B=0\,\therefore\,B=-A$$

Hence, the function is:

$$f(x)=A(2x-1)^{2n}-A$$

Now, to determine $A$, we will require the given area to be $1$:

$$A\int_0^1 (2x-1)^{2n}-1\,dx=1$$

$$A\left[\frac{1}{2(2n+1)}(2x-1)^{2n+1}-x \right]_0^1=1$$

$$A\left[(2x-1)^{2n+1}-2(2n+1)x \right]_0^1=2(2n+1)$$

$$A\left((1-2(2n+1))-(-1-0) \right)=2(2n+1)$$

$$A(2-2(2n+1))=2(2n+1)$$

$$A(-2n)=2n+1$$

$$A=-\frac{2n+1}{2n}$$

And so the function becomes:

$$f(x)=-\frac{2n+1}{2n}(2x-1)^{2n}+\frac{2n+1}{2n}$$

$$f(x)=\frac{2n+1}{2n}\left(1-(2x-1)^{2n} \right)$$

Now, computing the derivative, we find:

$$f'(x)=-2(2n+1)(2x-1)^{2n-1}$$

And so, the arc-length as a function of $n$ is:

$$s(n)=\int_0^1\sqrt{1+\left(2(2n+1)(2x-1)^{2n-1} \right)^2}\,dx$$

Using numeric integration, we find:

[TABLE="class: grid, width: 250, align: center"]
[TR]
[TD]n[/TD]
[TD]$s(n)\approx$[/TD]
[/TR]
[TR]
[TD]1[/TD]
[TD]3.24903[/TD]
[/TR]
[TR]
[TD]2[/TD]
[TD]2.96314[/TD]
[/TR]
[TR]
[TD]3[/TD]
[TD]2.90967[/TD]
[/TR]
[TR]
[TD]4[/TD]
[TD]2.89612[/TD]
[/TR]
[TR]
[TD]5[/TD]
[TD]2.89401[/TD]
[/TR]
[TR]
[TD]6[/TD]
[TD]2.89589[/TD]
[/TR]
[TR]
[TD]7[/TD]
[TD]2.89922[/TD]
[/TR]
[TR]
[TD]100[/TD]
[TD]2.97888[/TD]
[/TR]
[TR]
[TD]1000[/TD]
[TD]2.99671[/TD]
[/TR]
[TR]
[TD]2000[/TD]
[TD]2.99818[/TD]
[/TR]
[/TABLE]

It seems the turning point is at $n=5$, representing a 10th degree polynomial.
 
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