Brain Teaser - Area enclosed by a parabola and a line
09-15-2015, 09:42 PM
Post: #21
 fhub Member Posts: 188 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-15-2015 05:08 PM)Gerson W. Barbosa Wrote:  I am not familiar with the Prime, but I think it can handle that, can't it?
Well, I've tried it first in Xcas, but I couldn't even solve the cubic equation x^3+u*x^2+u^2*x+1/(4*u^3)+u^3=0 for the left intersection point (and I've tried all kinds of solve functions).
Then I've tried it also in the HP-Prime emulator (in CAS mode), but with the same result: no solution.
So I gave up - I know why I stay with my good old Derive CAS!

Franz
09-16-2015, 02:06 AM (This post was last modified: 09-16-2015 02:41 PM by Gerson W. Barbosa.)
Post: #22
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-15-2015 02:44 PM)CR Haeger Wrote:  **or an old student looking at this again after a few decades of rust.

In this case Calculus Refresher, by A. A. Klaf might be handy. I bought this book last year in a local bookstore. But don't do like me, who kept it the shelf until today.
The equation of the normal line shouldn't be difficult to derive, but it's ready for use on page 70:

$y-y_{1}=-\frac{1}{\frac{dy}{dx}}(x-x_{1})$

For the case y = x^4, we have

$$y-y_{1}=-\frac{1}{4x_{1}^{3}}(x-x_{1})$$
$$(x_{1},y_{1})=(u,u^{4})$$
$$y-u^{4}=-\frac{1}{4u^{3}}(x-u)$$
$$y=u^{4}-\frac{x-u}{4u^{3}}$$
$$y=\frac{4u^{7}+u-x}{4u^{3}}$$

Thanks for posting this interesting problem!

Gerson.

Edited to fix a typo and to add a missing subscript (Thanks, Thomas!).
09-16-2015, 02:19 AM
Post: #23
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-13-2015 10:06 PM)Bunuel66 Wrote:  Sorry to restate what Gerson has already provided.

There's no need to be sorry. The different approaches followed by you and others are more important and valuable than the results themselves.
Thank you also for the follow-ups to your post.

Gerson.
09-16-2015, 05:12 AM
Post: #24
 Thomas Klemm Senior Member Posts: 1,536 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-16-2015 02:06 AM)Gerson W. Barbosa Wrote:  The equation of the normal line shouldn't be difficult to derive, but it's ready for use on page 70:

$y-y_{1}=-\frac{1}{\frac{dx}{dy}}(x-x_{1})$

There's a typo: $$dx$$ and $$dy$$ got switched. The correct formula is:

$y-y_{1}=-\frac{1}{\frac{dy}{dx}}(x-x_{1})$

But then this notation can be misleading as the derivative should be evaluated where $$x=x_1$$.
We could use $$\frac{dy}{dx}\Big|_{x=x_1}$$ however I prefer to use $$f'(x_1)$$.

Quote:For the case y = x^4, we have

$$y-y_{1}=-\frac{1}{4x^{3}}(x-x_{1})$$

The variable in the derivative should be $$x_1$$:

$$y-y_{1}=-\frac{1}{4x_1^{3}}(x-x_{1})$$

Quote:$$(x_{1},y_{1})=(u,u^{4})$$
$$y-u^{4}=-\frac{1}{4u^{3}}(x-u)$$

Now we're back on the same road.

Quote:Thanks for posting this interesting problem!

1+

Cheers
Thomas
09-16-2015, 03:09 PM (This post was last modified: 09-16-2015 03:10 PM by Gerson W. Barbosa.)
Post: #25
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-16-2015 05:12 AM)Thomas Klemm Wrote:
(09-16-2015 02:06 AM)Gerson W. Barbosa Wrote:  The equation of the normal line shouldn't be difficult to derive, but it's ready for use on page 70:

$y-y_{1}=-\frac{1}{\frac{dx}{dy}}(x-x_{1})$

There's a typo: $$dx$$ and $$dy$$ got switched. The correct formula is:

$y-y_{1}=-\frac{1}{\frac{dy}{dx}}(x-x_{1})$

But then this notation can be misleading as the derivative should be evaluated where $$x=x_1$$.
We could use $$\frac{dy}{dx}\Big|_{x=x_1}$$ however I prefer to use $$f'(x_1)$$.

Quote:For the case y = x^4, we have

$$y-y_{1}=-\frac{1}{4x^{3}}(x-x_{1})$$

The variable in the derivative should be $$x_1$$:

$$y-y_{1}=-\frac{1}{4x_1^{3}}(x-x_{1})$$

Quote:$$(x_{1},y_{1})=(u,u^{4})$$
$$y-u^{4}=-\frac{1}{4u^{3}}(x-u)$$

Now we're back on the same road.

Thanks, Thomas, for pointing these out!
The example and the illustration in the book make it quite clear, but by looking at the formula alone that notation indeed seems to be misleading. In my hand-written solution above I used a capital X instead of x1, but that might be a bit confusing. I prefer your lower-case u as no subscript is required.
I'd revised the formula before posting, but I missed the dy and dx swap. Well, I typed that with the keyboard on my lap, the mouse sliding on top of a pile of books and the monitor far on on the floor, a totally anti-ergonomic way. (My wife reclaimed the sewing-machine stand I had been using for months, thus disturbing the progress of science )

Cheers,

Gerson.
09-16-2015, 03:55 PM
Post: #26
 Thomas Klemm Senior Member Posts: 1,536 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-16-2015 03:09 PM)Gerson W. Barbosa Wrote:  The example and the illustration in the book make it quite clear, but by looking at the formula alone that notation indeed seems to be misleading.

Katie will probably love that title.

Cheers
Thomas

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09-16-2015, 04:14 PM
Post: #27
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-16-2015 03:55 PM)Thomas Klemm Wrote:
(09-16-2015 03:09 PM)Gerson W. Barbosa Wrote:  The example and the illustration in the book make it quite clear, but by looking at the formula alone that notation indeed seems to be misleading.

Katie will probably love that title.

The Copyright date is 1944. Perhaps I should have suggested a more modern book, intended for technical persons.

Gerson.
09-18-2015, 05:51 PM
Post: #28
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
Just for the record, here are some numerical results for the cases y=x^4, y=6 and y=8. Notice the minimum areas occur for normal lines passing near y(3/4), y(5/6) and y(7/8), respectively. The values of u for minimum areas should be accurate to 7 or 8 digits.

y=x^4

u: 0.730000000000 Area: 1.232203807010
u: 0.740000000000 Area: 1.229721793524
u: 0.741869418802 Area: 1.229582549556
u: 0.743490000000 Area: 1.229543561785
u: 0.743514882370 Area: 1.229543552881
u: 0.743515349953 Area: 1.229543552884
u: 0.743543197140 Area: 1.229543564411
u: 0.744136760955 Area: 1.229549111764
u: 0.75000000000o Area: 1.230145031369

y=x^6

u: 0.810000000000 Area: 1.120523733964
u: 0.814713174965 Area: 1.119930602299
u: 0.814723174965 Area: 1.119930598819
u: 0.814724747205 Area: 1.119930598753
u: 0.814733174965 Area: 1.119930600634
u: 0.815000000000 Area: 1.119932604732
u: 0.820000000000 Area: 1.120664993006

y=x^8

u: 0.86790000000 Area: 1.24466886334
u: 0.86800000000 Area: 1.24466861630
u: 0.86796579708 Area: 1.24466859303
u: 0.86797579708 Area: 1.24466858824
u: 0.86798579708 Area: 1.24466859303
u: 0.86804996475 Area: 1.24466885164
u: 0.86810000000 Area: 1.24466932686

Gerson.
09-18-2015, 08:20 PM
Post: #29
 Thomas Klemm Senior Member Posts: 1,536 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
We start with a slightly different definition of $$h(x)=g(x)-f(x)$$ but now consider it as a function of two variables $$u$$ and $$x$$:
$h(u,x)=\frac{u-x}{2u}+u^2-x^2$
From this we define the indefinite integral:
$H(u,x)=\int h(u,x)\, dx$
For each $$u$$ we find the lower limit $$a(u)$$ and the upper limit $$b(u)$$ as roots of $$h(u,x)=0$$:
\begin{align*} a(u) &= -(u+\frac{1}{2u})\\ b(u) &= u \end{align*}
We evaluate the anti-derivative at these limits, calculate the derivative and set it equal to 0:
$\frac{\mathrm{d} }{\mathrm{d} u}(H(u,b)-H(u,a))=0$
To calculate the total differential we use partial differentials:
$\frac{\mathrm{d} }{\mathrm{d} u}H(u,x)=\frac{\partial H}{\partial u}+\frac{\partial H}{\partial x}\frac{\partial x}{\partial u}$
From the definition of $$H(u,x)$$ we find:
$\frac{\partial H(u,x)}{\partial x}=h(u,x)$
However this function is 0 when evaluated at the lower or the upper limit. Thus this term vanishes.
This leaves only the first term:
$\frac{\partial}{\partial u}H(u,x)=\frac{\partial}{\partial u}\int h(u,x)\,dx$
But we can switch the order of integration and differentiation:
\begin{align*} \frac{\partial}{\partial u}\int h(u,x)\,dx &=\int \frac{\partial}{\partial u}h(u,x)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{u-x}{2u}+u^2-x^2)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{1}{2}-\frac{x}{2u}+u^2-x^2)\,dx \\ &=\int\frac{x}{2u^2}+2u\,dx \\ &=\frac{x^2}{4u^2}+2ux \end{align*}
We evaluate this expression at the upper and lower limit and find the difference. But we can do that for both terms separately:
\begin{align*}
\frac{b^2-a^2}{4u^2} + 2u(b-a) &= (b-a)\left(\frac{b+a}{4u^2}+2u\right)\\
&= \left(2u+\frac{1}{2u}\right)\left(2u-\frac{1}{8u^3}\right) \\
&= \frac{4u^2+1}{2u} \cdot \frac{16u^4-1}{8u^3} \\
&= \frac{(4u^2+1)(16u^4-1)}{16u^4} \\
&= \frac{(4u^2+1)(4u^2+1)(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(2u+1)(2u-1)}{16u^4} = 0 \\
\end{align*}
We conclude that $$2u-1=0$$ which leads to the solution $$u=\tfrac{1}{2}$$.

Kind regards
Thomas
09-21-2015, 05:53 PM (This post was last modified: 09-21-2015 09:00 PM by fhub.)
Post: #30
 fhub Member Posts: 188 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-18-2015 05:51 PM)Gerson W. Barbosa Wrote:  y=x^4
u: 0.743514882370 Area: 1.229543552881

y=x^6
u: 0.814724747205 Area: 1.119930598753

y=x^8
u: 0.86797579708 Area: 1.24466858824
Hi Gerson,

since your Area result for x^8 looked rather illogical for me (IMO the area should continuously decrease for higher exponents n), I've written a small program (in TurboPascal) for this problem.
The left intersection point is of course calculated numerically (no exact value possible for n>4), but the integral is calculated exactly (i.e. no numerical method), and the results really confirm my assumption of continuously decreasing areas:
in fact for n->inf the results are: u->1 and A->0

Here's the list for n=2..20:
Code:
 n= 2   u=0,500000000   A=1,333333333 n= 4   u=0,743514882   A=1,229543553 n= 6   u=0,814724746   A=1,119930599 n= 8   u=0,850322007   A=1,034513042 n=10   u=0,872567780   A=0,966429043 n=12   u=0,888129582   A=0,910591797 n=14   u=0,899777205   A=0,863713409 n=16   u=0,908898125   A=0,823611190 n=18   u=0,916276056   A=0,788780129 n=20   u=0,922392221   A=0,758146096
And here a list from 100 to 1000:
Code:
 n= 100   u=0,976938065   A=0,381166646 n= 200   u=0,986757839   A=0,275172584 n= 300   u=0,990499224   A=0,226444741 n= 400   u=0,992515499   A=0,196933121 n= 500   u=0,993789394   A=0,176609553 n= 600   u=0,994672548   A=0,161517493 n= 700   u=0,995323449   A=0,149737975 n= 800   u=0,995824495   A=0,140212382 n= 900   u=0,996222942   A=0,132302458 n=1000   u=0,996547911   A=0,125597430

Franz
09-21-2015, 06:40 PM (This post was last modified: 09-21-2015 06:53 PM by Gerson W. Barbosa.)
Post: #31
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-21-2015 05:53 PM)fhub Wrote:  since your Area result for x^8 looked rather illogical for me (IMO the area should continuously decrease for higher exponents n), I've written a small program (in TurboPascal) for this problem.
The left intersection point is of course calculated numerically (no exact value possible for n>4), but the integral is calculated exactly (i.e. no numerical method), and the results really confirm my assumption of continuously decreasing areas:
in fact for n->inf the results are: u->1 and A->0

Here's the list for n=2..20:
Code:
 n= 2   u=0,500000000   A=1,333333330 n= 4   u=0,743514882   A=1,229543550 n= 6   u=0,814724746   A=1,119930600 n= 8   u=0,850322007   A=1,034513040 n=10   u=0,872567780   A=0,966429043 n=12   u=0,888129582   A=0,910591797 n=14   u=0,899777205   A=0,863713409 n=16   u=0,908898125   A=0,823611190 n=18   u=0,916276056   A=0,788780129 n=20   u=0,922392221   A=0,758146096

Hello Franz,

Thanks again for pointing out yet another wrong result of mine. Thanks also for providing solutions to this plentiful of cases.
Since my results for y=x^6 are correct within the accuracy I stated, I surely have made a mistake in the function submitted to the numerical integration and solver for case y=x^8 (I have yet to find it out).
Yes, there's no need to use a numerical integrator as the functions are quite easy to integrate [ integral (x^n)dx = x^(n+1)/(n+1) + k, n ≠ -1 ], only the left integration limits being the complicated parts.
I've found that Derive 4.11 is installed in my HP-200LX and tried some numerical integrations (very slow on the 200LX, BTW). But I haven't explored it much as it is not so intuitive and didn't find a manual.

Best regards,

Gerson.
09-21-2015, 09:48 PM (This post was last modified: 09-21-2015 09:53 PM by Gerson W. Barbosa.)
Post: #32
 Gerson W. Barbosa Senior Member Posts: 1,452 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
(09-21-2015 05:53 PM)fhub Wrote:
(09-18-2015 05:51 PM)Gerson W. Barbosa Wrote:  y=x^4
u: 0.743514882370 Area: 1.229543552881

y=x^6
u: 0.814724747205 Area: 1.119930598753

y=x^8
u: 0.86797579708 Area: 1.24466858824

... your Area result for x^8 looked rather illogical for me (IMO the area should continuously decrease for higher exponents n) ...

The following might have been closer if my function Y had been Y := u*u*u*u*u*u*u*u - (x-u)/(8*u*u*u*u*u*u*u) - x*x*x*x*x*x*x*x rather than Y := u*u*u*u*u*u*u*u - (x-u)/(6*u*u*u*u*u*u*u) - x*x*x*x*x*x*x*x. In FreePascal the yellow is brighter and the blue is fainter, but this is not an excuse as I ought to have mistrusted the increase in the area.

y=x^8

u: 0.840000000000 Area: 1.038962719953
u: 0.849000000000 Area: 1.034585573337
u: 0.849999511578 Area: 1.034517355546
u: 0.850000000000 Area: 1.034517342489
u: 0.850322146435 Area: 1.034513041971
u: 0.851000000000 Area: 1.034532095727
u: 0.860000000000 Area: 1.038377424828
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