Brain Teaser  Area enclosed by a parabola and a line

09112015, 04:57 PM
(This post was last modified: 09122015 01:38 PM by CR Haeger.)
Post: #1




Brain Teaser  Area enclosed by a parabola and a line
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 

09132015, 12:17 AM
Post: #2




RE: Brain Teaser  Area enclosed by a parabola and a line
(09112015 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. * 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. 

09132015, 01:35 AM
(This post was last modified: 09132015 01:42 AM by Claudio L..)
Post: #3




RE: Brain Teaser  Area enclosed by a parabola and a line
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. 

09132015, 02:40 AM
Post: #4




RE: Brain Teaser  Area enclosed by a parabola and a line
h(1) = 125/48 

09132015, 01:19 PM
Post: #5




RE: Brain Teaser  Area enclosed by a parabola and a line
(09132015 12:17 AM)Gerson W. Barbosa Wrote:(09112015 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. Good work! I struggled (by hand) determining the general expression for the lower limit of integration. Tried solving x^2m*xb = 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. 

09132015, 04:23 PM
(This post was last modified: 09132015 04:26 PM by Thomas Klemm.)
Post: #6




RE: Brain Teaser  Area enclosed by a parabola and a line
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: \(\begin{bmatrix} xu \\ yu^2 \end{bmatrix}\cdot \begin{bmatrix} 1 \\ 2u \end{bmatrix}=0 \) Thus: \(xu+2u(yu^2)=0\) From the equation above we can find the formula for \(g(x)\): \(y=\frac{ux}{2u}+u^2\) Thus we get for the difference \(g(x)f(x)\): \(y=\frac{ux}{2u}+u^2x^2=(ux)(\frac{1}{2u}+u+x)\) 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 HP15C and write two small programs. Program A Calculates the limits to integrate program B. Code: LBL A Program B Calculates the difference \(g(x)f(x)\). Code: LBL B 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 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 

09132015, 10:06 PM
(This post was last modified: 09142015 09:49 AM by Bunuel66.)
Post: #7




RE: Brain Teaser  Area enclosed by a parabola and a line
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 

09142015, 03:57 PM
(This post was last modified: 09142015 04:06 PM by CR Haeger.)
Post: #8




RE: Brain Teaser  Area enclosed by a parabola and a line
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? 

09142015, 05:08 PM
(This post was last modified: 09142015 05:53 PM by Bunuel66.)
Post: #9




RE: Brain Teaser  Area enclosed by a parabola and a line
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: (xMx)+2x(yMx²)=0. This is just the dot product expressing the orthogonality. When M is on the curve, yM=xM² then: (xMx)+2x(xM²x²)=0 As long as xM <> x: 1+2x(xM+x)=0 Solving for xM: xM=x1/2x=(2x²+1)/2x Regards 

09142015, 05:51 PM
(This post was last modified: 09142015 07:31 PM by Bunuel66.)
Post: #10




RE: Brain Teaser  Area enclosed by a parabola and a line
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. 

09142015, 08:32 PM
Post: #11




RE: Brain Teaser  Area enclosed by a parabola and a line
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. My website: erwin.ried.cl 

09142015, 10:06 PM
(This post was last modified: 09142015 10:10 PM by CR Haeger.)
Post: #12




RE: Brain Teaser  Area enclosed by a parabola and a line  
09152015, 03:29 AM
(This post was last modified: 09182015 04:19 AM by Gerson W. Barbosa.)
Post: #13




RE: Brain Teaser  Area enclosed by a parabola and a line
(09132015 04:23 PM)Thomas Klemm Wrote: Thus: Things start to get complicated for the biquadratic parabola, if I've done it right: \(y=\frac{4u^{7}+ux}{4u^{3}}\) 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 

09152015, 11:56 AM
(This post was last modified: 09152015 12:00 PM by fhub.)
Post: #14




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 03:29 AM)Gerson W. Barbosa Wrote: Things start to get complicated for the biquadratic parabola, if I've done it right: 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=x01/(2*x0) and x2=x0 (of course). Now the area between the 2 curves is A(x0)=integral(1/(2*x0)*x+x0^2+1/2x^2,x,x01/(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*(41/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 

09152015, 01:30 PM
(This post was last modified: 09152015 01:45 PM by Gerson W. Barbosa.)
Post: #15




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 11:56 AM)fhub Wrote:(09152015 03:29 AM)Gerson W. Barbosa Wrote: Things start to get complicated for the biquadratic parabola, if I've done it right: 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. 

09152015, 02:44 PM
Post: #16




RE: Brain Teaser  Area enclosed by a parabola and a line
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. 

09152015, 02:59 PM
(This post was last modified: 09152015 03:21 PM by fhub.)
Post: #17




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 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 (xu)), 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 

09152015, 04:18 PM
(This post was last modified: 09152015 04:27 PM by Gerson W. Barbosa.)
Post: #18




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 02:59 PM)fhub Wrote: Well, I get a different result: 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 WA: solve 0.608258988304(1.24590156402x) = x^4 to 12 digits integrate 0.608258988304(1.24590156402x)  x^4, x=1.09201722595,0.74350488189 to 12 digits solve 0.608234446071(1.24595886588x) = x^4 to 12 digits integrate 0.608234446071(1.24595886588x)  x^4, x=1.09201232966,0.74351488189 to 12 digits solve 0.608209905131(1.24601617162x) = x^4 to 12 digits integrate 0.608209905131(1.24601617162x)  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. 

09152015, 05:00 PM
Post: #19




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 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 WA: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 

09152015, 05:08 PM
(This post was last modified: 09152015 05:10 PM by Gerson W. Barbosa.)
Post: #20




RE: Brain Teaser  Area enclosed by a parabola and a line
(09152015 05:00 PM)fhub Wrote:(09152015 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 WA: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! 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. 

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