Brain Teaser - Area enclosed by a parabola and a line

[CR Haeger]

I tried to work the following problem fully by hand and then with CAS help from the HP Prime. Eventually I got to "solutions" on the Prime but it took a lot of variables creation and substitutions to get there. I am sure most here will make quick work of this.

The problem:
* Find the area enclosed by a function f(x)=x^2 and a line g(x) that intersects and is normal to f(x) for all x>0. Solve for area at x=+1.0.

* Write the function h(x) in terms of x only which describes the area enclosed by f(x) and g(x).

* Find the minimum value of h(x)

   

Best, Carl

[Gerson W. Barbosa]

(09-11-2015 04:57 PM)CR Haeger Wrote:  I tried to work the following problem fully by hand and then with CAS help from the HP Prime. Eventually I got to "solutions" on the Prime but it took a lot of variables creation and substitutions to get there. I am sure most here will make quick work of this.

The problem:
* Find the area enclosed by a function f(x)=x^2 and a line g(x) that intersects and is normal to f(x) for all x>0. Solve for area at x=+1.0.

* Write the function h(x) in terms of x only which describes the area enclosed by f(x) and g(x).

* Find the minimum value of h(x)

* h(x) = (4x² + 1)³/(48x³)

* 4/3 at x = 1/2

I've fount the normal lines at x = 1, x = 2 and x = 3, then generalized for X. (Picture later if important).
I've used a quadratic solver to generalize the limits of integration.
Too lazy to do the remaining by hand:

http://www.wolframalpha.com/input/?i=int...%29%29%2CX

http://www.wolframalpha.com/input/?i=sol...%29%3D%3D0

Regards,

Gerson.

[Claudio L.]

My take:
h(x)=4/3*x^3+x
At x=1 it's 7/3 unless I got it all wrong.
Done by hand (no calculators were harmed during this teaser).

EDIT: I just realized my solution is wrong. Bad integration limit, Please disregard.

[Gerson W. Barbosa]

h(1) = 125/48

[CR Haeger]

(09-13-2015 12:17 AM)Gerson W. Barbosa Wrote:  

(09-11-2015 04:57 PM)CR Haeger Wrote:  I tried to work the following problem fully by hand and then with CAS help from the HP Prime. Eventually I got to "solutions" on the Prime but it took a lot of variables creation and substitutions to get there. I am sure most here will make quick work of this.

The problem:
* Find the area enclosed by a function f(x)=x^2 and a line g(x) that intersects and is normal to f(x) for all x>0. Solve for area at x=+1.0.

* Write the function h(x) in terms of x only which describes the area enclosed by f(x) and g(x).

* Find the minimum value of h(x)

* h(x) = (4x² + 1)³/(48x³)

* 4/3 at x = 1/2

I've fount the normal lines at x = 1, x = 2 and x = 3, then generalized for X. (Picture later if important).
I've used a quadratic solver to generalize the limits of integration.
Too lazy to do the remaining by hand:

http://www.wolframalpha.com/input/?i=int...%29%29%2CX

http://www.wolframalpha.com/input/?i=sol...%29%3D%3D0

Regards,

Gerson.

Good work!

I struggled (by hand) determining the general expression for the lower limit of integration. Tried solving x^2-m*x-b = 0 for x where m and b were slope, intercept of line g(x). Your expression seems better simplified.

Also, the HP Prime placed some absolute value limits on my h(x) equation... Ill post what I calculated soon.

[Thomas Klemm]

Let's assume that \(P=(u, u^2)\) is the intersection of the function \(f(x)=x^2\) and its normal line \(g(x)\).
Then from \(f'(x)=2x\) we can conclude that:

\(\begin{bmatrix}
x-u \\
y-u^2
\end{bmatrix}\cdot
\begin{bmatrix}
1 \\
2u
\end{bmatrix}=0
\)

Thus:
\(x-u+2u(y-u^2)=0\)

From the equation above we can find the formula for \(g(x)\):

\(y=\frac{u-x}{2u}+u^2\)

Thus we get for the difference \(g(x)-f(x)\):

\(y=\frac{u-x}{2u}+u^2-x^2=(u-x)(\frac{1}{2u}+u+x)\)

When we set both factors to 0 we can find the lower and upper limit of the integral.
We already know that the upper limit is: \(x=u\).
The lower limit is:

\(x=-\left (u+\frac{1}{2u}\right ) \)

Now let's pick the HP-15C and write two small programs.

Program A
Calculates the limits to integrate program B.

Code:

LBL A
+
1/x
+
CHS
x<>y
STO 0
∫ B
RTN

Program B
Calculates the difference \(g(x)-f(x)\).

Code:

LBL B
RCL- 0
CHS
x<>y
RCL+ 0
RCL 0
RCL+ 0
1/x
+
*
RTN

We can test this for \(x = 1\):

1
ENTER
ENTER
ENTER
f A
2.6042


From the plot we can conclude that the minimum is somewhere between 0.1 and 1:



Here comes the tricky part. We have to force the solver to only search for positive values. The easiest way to do this is to add the ABS function to the program A:

Code:

LBL A
+
1/x
+
ABS
CHS
x<>y
ABS
STO 0
∫ B
RTN

Now we can try to solve for a solution:

0.1
ENTER
1
f SOLVE A
Error 8


This means that no solution was found. However we can still find the desired result in the stack:

z: 1.3339
y: 0.5354
x: 0.5085


Cheers
Thomas

[Bunuel66]

Sorry to restate what Gerson has already provided. That said I haven't used a quadratic solver ;-) Just a mix of hand computations and checks with a CAS software (the one included in Geogebra).

May I suggest: h(x)=[64x⁶+48x⁴+12x²+1]/48x³=(4x²+1)³/48x³

h(1)=125/48~2.604

With a minimum for h(x) at x=0.5 of 4/3

My 2 cents.

Regards

[CR Haeger]

Thomas - thanks for both elegant mathematical and HP15C programming solutions! I still need to figure out how you were able to determine, factor the lower integration limit expression.

Brunell66 - your solutions look correct as well at least for x>0. For x<0, should h(x) still be positive?

[Bunuel66]

Lower limit of integration:

Let P be a point on the curve. Coordinates of P are (x,x²).
Let M be a point on the straight line orthogonal to the tangent at P on the curve.

The vector T tangent at P has for coordinates (1,2x).
Then the vector PM is orthogonal to T.
With M of coordinates (xM, yM), if M is one the line one should have: (xM-x)+2x(yM-x²)=0.
This is just the dot product expressing the orthogonality.

When M is on the curve, yM=xM² then: (xM-x)+2x(xM²-x²)=0
As long as xM <> x: 1+2x(xM+x)=0
Solving for xM: xM=-x-1/2x=-(2x²+1)/2x

Regards

[Bunuel66]

Case x<0

If one understands the question as related to a positive area, it seems that:

h(x) becomes simply -h(x) for x<0 with the same minimum for x=-0.5

Seems to be reasonable as the problem has an axial symmetry relative to y.

Then the general solution would be just abs(h(x)) as the sign of h(x) is the same as the one of x.

Regards.

[eried]

Tried to use the geometry app in the Prime


of course it crashed and deleted all memory... F**k off. Besides is SOOOO nice to try to point an exact position with the cursor instead of typing the exact coordinate.

[CR Haeger]

Eried - sorry that the Prime crashed on you. I dont frequent the Geometry app very often.

Here is my work in CAS and Function app:

   

   

and

   

[Gerson W. Barbosa]

(09-13-2015 04:23 PM)Thomas Klemm Wrote:  Thus:
\(x-u+2u(y-u^2)=0\)

From the equation above we can find the formula for \(g(x)\):

\(y=\frac{u-x}{2u}+u^2\)

Things start to get complicated for the biquadratic parabola, if I've done it right:

\(y=\frac{4u^{7}+u-x}{4u^{3}}\)

u: 0.760000000000 Area: 1.23320954839
u: 0.750000000000 Area: 1.23009174283
u: 0.740000000000 Area: 1.22972179352
u: 0.745000000000 Area: 1.22956112013
u: 0.744000000000 Area: 1.22953780108
u: 0.743750000000 Area: 1.22953629609
u: 0.743687500000 Area: 1.22953619031
u: 0.743671875000 Area: 1.22953618081
u: 0.743664062500 Area: 1.22953617857
u: 0.743660156250 Area: 1.22953617806
u: 0.743659179690 Area: 1.22953617801
u: 0.743658691410 Area: 1.22953617799
u: 0.743658203125 Area: 1.22953617797
u: 0.743656250000 Area: 1.22953617799
u: 0.743652798750 Area: 1.22953617828
u: 0.743652343750 Area: 1.22953617836
u: 0.743648437500 Area: 1.22953617913
u: 0.743640625000 Area: 1.22953618202
u: 0.743625000000 Area: 1.22953619273
u: 0.743500000000 Area: 1.22953652222
u: 0.743000000000 Area: 1.22954217016

Dull figures... Hopefully I'm mistaken :-)

Cheers,

Gerson.

Edit to include correct values. The table above is wrong, except for u=0.74, due to a silly error in my first Pascal program. I chose a numeric method because it's easy to extend the problem to x^6, x^8, x^10... The result is accurate to 8 digits, but more are possible (Three pairs of values are used to obtain the equation of the parabola that encompasses them, then its minimum is used as a new u).


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

[fhub]

(09-15-2015 03:29 AM)Gerson W. Barbosa Wrote:  Things start to get complicated for the biquadratic parabola, if I've done it right:

\(y=\frac{4u^{7}+u-x}{4u^{3}}\)

Hmmm, from where did you get this strange expression (u^7?)?

I'm also wondering about some very (unnecessarily) complicated solutions here, especially using 'numeric' solvers!?
This problem can be solved completely 'by hand' (or better 'by brain'), only for the definite integral it's maybe better to use a CAS just to avoid any mistakes and to get a simple (factored) result:

parabola: y=x^2
at special point x=x0 (>0) ==> y=x0^2 and slope k=2*x0
normal line: y=kn*x+d with normal slope kn=-1/k=-1/(2*x0)
entering the special point (x0,x0^2) and kn ==> d= x0^2+1/2
so the normal line is: y=-1/(2*x0)*x+x0^2+1/2

The 2 intersection points are simply the solutions of the quadratic equation
x^2=-1/(2*x0)*x+x0^2+1/2
with the 2 results
x1=-x0-1/(2*x0) and x2=x0 (of course).

Now the area between the 2 curves is
A(x0)=integral(-1/(2*x0)*x+x0^2+1/2-x^2,x,-x0-1/(2*x0),x0)
Although it's also easy to calculate this integral manually, it's indeed safer to use a CAS for it:
A(x0)=(4*x0^2+1)^3/(48*x0^3)

If you put now the x0^3 from the denominator under (...)^3 in the numerator, you get a much simpler result, for which it's very easy to calculate the minimum by hand:
A(x0)=1/48*(4*x0+1/x0)^3

A'(x0)=3/48*(4*x0+1/x0)^2*(4-1/x0^2) = 0 (for minimum)
In this expression only the last (...) can be 0, so you have immediately the answer:
x0=sqrt(1/4)=1/2 (x0=-1/2 excluded, because x0>0 was assumed)

Franz

[Gerson W. Barbosa]

(09-15-2015 11:56 AM)fhub Wrote:  

(09-15-2015 03:29 AM)Gerson W. Barbosa Wrote:  Things start to get complicated for the biquadratic parabola, if I've done it right:

\(y=\frac{4u^{7}+u-x}{4u^{3}}\)

Hmmm, from where did you get this strange expression (u^7?)?

Franz, I'm trying to extend the problem to the biquadratic parabola (y = x^4). The expression for lines normal to it at a point P(u, u^4) was obtained from Thomas Klemm's method above.

Gerson.

[CR Haeger]

Franz - thank you for your post and straightforward solutions. I posted this thread in part, to see how others approached this "simple" problem by hand or with the assistance of a calculator. What I found so far, is that relying heavily on the HP Prime (CAS) still takes some problem setup, managing variables and managing syntax. For current algebra/calculus students** I can see where CAS can be helpful but can also complicate things. I also relearned that the HP15C is a great machine, especially in the hands of capable users.

Gerson - that is an interesting problem extension. I wondered how variations on f(x)=x^2 might be handled.

Best,
Carl

**or an old student looking at this again after a few decades of rust.

[fhub]

(09-15-2015 01:30 PM)Gerson W. Barbosa Wrote:  Franz, I'm trying to extend the problem to the biquadratic parabola (y = x^4). The expression for lines normal to it at a point P(u, u^4) was obtained from Thomas Klemm's method above.

Aaah ok, but that's in fact a very complicated calculation!

You get the left intersection point from the cubic equation x^3+u*x^2+u^2*x+1/(4*u^3)+u^3=0 (which is the quartic polynomial divided by (x-u)), and then you have to calculate the definite integral from this complicated left point to u.
And finally solve the terrible equation A'(u)=0 ...

Well, I get a different result:
u=0.743514881896 with A=1.22954355289

Franz

[Gerson W. Barbosa]

(09-15-2015 02:59 PM)fhub Wrote:  Well, I get a different result:
u=0.743514881896 with A=1.22954355289

I think I can blame it on my Simpson 3/8 integrator in Pascal (too many iterations required!). Your result is indeed a minimum, as checked in W|A:

solve 0.608258988304(1.24590156402-x) = x^4 to 12 digits

integrate 0.608258988304(1.24590156402-x) - x^4, x=-1.09201722595,0.74350488189 to 12 digits


solve 0.608234446071(1.24595886588-x) = x^4 to 12 digits

integrate 0.608234446071(1.24595886588-x) - x^4, x=-1.09201232966,0.74351488189 to 12 digits

solve 0.608209905131(1.24601617162-x) = x^4 to 12 digits

integrate 0.608209905131(1.24601617162-x) - x^4, x=-1.09200743451,0.74352488189 to 12 digits


u = 0.743504881896 Area = 1.22954355433

u = 0.743514881896 Area = 1.22954355289

u = 0.743524881896 Area = 1.22954355432


Gerson.

[fhub]

(09-15-2015 04:18 PM)Gerson W. Barbosa Wrote:  I think I can blame it on my Simpson 3/8 integrator in Pascal (too many iterations required!). Your result is indeed a minimum, as checked in W|A:

I don't think that you can solve any intermediate steps approximately, at least I have done the complete calculation with exact terms, only the last equation A'(u)=0 has to be done approximately of course. And these intermediate expressions (left intersection point, the formula for A(u) after the definite integration) are really HUGE!

For all this I've used my good old Derive 4.11 (the DOS version because I'm familiar with it since more than 20 years)!

Franz

[Gerson W. Barbosa]

(09-15-2015 05:00 PM)fhub Wrote:  

(09-15-2015 04:18 PM)Gerson W. Barbosa Wrote:  I think I can blame it on my Simpson 3/8 integrator in Pascal (too many iterations required!). Your result is indeed a minimum, as checked in W|A:

I don't think that you can solve any intermediate steps approximately, at least I have done the complete calculation with exact terms, only the last equation A'(u)=0 has to be done approximately of course. And these intermediate expressions (left intersection point, the formula for A(u) after the definite integration) are really HUGE!

For all this I've used my good old Derive 4.11 (the DOS version because I'm familiar with it since more than 20 years)!

Franz

Wolfram Alpha would not allow me to copy and past the very expression for the left integration limit (unless of course I subscribed to their service), so I didn't even try. I am not familiar with the Prime, but I think it can handle that, can't it?

Gerson.