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

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MarkFL

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Feb 24, 2012
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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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MarkFL

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Feb 24, 2012
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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 \(\displaystyle x=\frac{1}{2}\), and has the required roots and area. So let's begin with:

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

With the required roots in mind, we may write:

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

With the required triangular area in mind, we find:

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

and so we have:

\(\displaystyle f(x)=-2|2x-1|+2\)

Here is a plot of the function:

bordiba1.jpg

Next, lets try a quadratic function.

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

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

With the required area in mind, we may write:

\(\displaystyle A\int_0^1 x^2-x\,dx=1\)

Applying the FTOC, we have:

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

\(\displaystyle A\left[2x^3-3x^2 \right]_0^1=6\)

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

\(\displaystyle A=-6\)

And so we have:

\(\displaystyle f(x)=6x(1-x)\)

Here is a plot of the function:

bordiba2.jpg

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

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

For the second function, we will use:

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

Using the trigonometric substitution:

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

we have:

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

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

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

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

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

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

So now we have:

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

Applying the FTOC, we have:

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

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

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

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

With the root requirements in mind, we find:

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

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

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

Applying the FTOC, we have:

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

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

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

\(\displaystyle B=\frac{5}{4}\)

Thus, our function is:

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

Here is a plot of the function:

bordiba3.jpg

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

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

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Feb 24, 2012
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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:

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

Recall, we require:

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

Hence, the function is:

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

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

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

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

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

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

\(\displaystyle A(2-2(2n+1))=2(2n+1)\)

\(\displaystyle A(-2n)=2n+1\)

\(\displaystyle A=-\frac{2n+1}{2n}\)

And so the function becomes:

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

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

Now, computing the derivative, we find:

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

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

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

Using numeric integration, we find:

n$s(n)\approx$
13.24903
22.96314
32.90967
42.89612
52.89401
62.89589
72.89922
1002.97888
10002.99671
20002.99818

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