Parking lot problem:
Characterize the angle of a vehicle in a parking lot such that if we consider the length of the car as unit 1 and width of vehicle as x, with x<1, such that the overall length required is exactly the same length when not parked at an angle.
For clarity, consider two parallel lines with a distance of unit 1. You could park a vehicle such that the entire front end lies on one of the parallel lines and the rear end of the vehicle lies on the other parallel line. There is some angle, 0 < theta < pi/2 for the varying widths, x, of the vehicle such that the right front corner of the vehicle touches one of the parallel lines and the left rear corner touches the other parallel line.
________________
| |...../\
| |......\ \
|_|____\ /__ angle theta in this corner
In assigning a line segment of length 1 starting from the right front corner of vehicle, you could assign new variables along the left side of vehicle, that are cut by the temporary line segment as r and s.
Then
sqrt(x^2+r^2) + sqrt(s^2-x^2) = 1, r+s=1, (r=1-s)
Short exercise would show x = sqrt(2s -1) -> s=(x^2+1)/2
The angle that is a complement of theta would thus be acos(x/((x^2+1)/2)) and so
theta=asin(2 x/(x^2+1))
Graphing this shows from x=0 to x=1, the angles starts off at x=0 at the point (x,theta)=(0,0) with a slope of 2 and tapers off at the point (1,pi/2) where the limit of the slope for x=1- is 1. Fascinating! What happens beyond 1 is immaterial to the problem. It looks similar to the dorsal fin of a shark. I may investigate the continuation of this curve as an extension of what occurs from x=0 to 1 and not the downward curve with a cusp at x=1. The slope of 2 at x=0 and slope of 1 at x=1 is interesting.
It turns out when x=1/sqrt(3) that this is a special value such that the right front of vehicle to left rear of vehicle forms a line segment that is perpendicular to the parallel lines. Theta for this case is pi/3. Intuitively, this can be accepted from familiarity with trigonometry.
One could also find out the effective new width formed by the vehicle with respect to the parallel lines.
From a little work, the width can be found to be:
sqrt(x^2)*sqrt((x^2-1)^2)/(1+x^2) + 2x/(1+x^2)
Being that 0 < x < 1, we can reduce further to x(3-x^2)/(1+x^2).
The graph of this function from x=0 to 1 starts at (0,0), goes higher than 1 then back down to the point (1,1). From a little calculus it can be found that the max x value is sqrt(2*sqrt(3)-3) ~0.68125 and with function value of sqrt(6*sqrt(3)-9) ~1.1799596. Unexpected from the initial problem but is sensible from the width function.
** In extending the shark fin curve mentioned above, one may consider the function, 2/(1+x^2) reflecting the derivative at the points x=0 and x=1 and becomes, 2 atan(x). Integrating from x=0 to inf on a site like wolfram alpha is unable to calculate it, giving the "does not converge" error, even when attempting to calculate 1/x integral(0,inf) 2 atan(x) but in this form, I find it to converge to pi. As integral [2 atan(x)] = 2 x tan^(-1)(x) - log(x^2 + 1) + constant, one may work with the form, integral [2 atan(x) b ] and see this is 2b x tan^(-1)(x) - 2b log(x^2 + 1) + constant, and for the limit x->inf, with b=1/x we see this turns into lim x->inf [2 atan(x) -2/x ln(x^2+1)] which converges to 2 atan(inf)=pi.
The function, 2 atan(x) is 0 at x=0, is pi/2 at x=1, and the limit of x->inf has the value pi. From this alone in knowing though some low values of x are far from pi, they are positive and always increasing which one could say for the span of the graph from x=0 to infinite, it would approximate a rectangle of infinite length and height of pi for which the area (integral of the function) from x=0 to inf would thus be x pi.
///One may engage in mathematics, even to a level beyond their formal educational level without going on to be a power monger overpaid teacher, contributing to the debt of this nation while those in the private sector get far less, and this means also not having the great availability of young women (and/or men) for which to play around with as many teachers do in some way or another regardless of what the common person thinks and throwing out "I am a teacher and I do not do that" does not take away from this happening - it even occurs at the high school level and does not get enough attention besides the few cases making a national stir - believe me, what you hear about would be less than a percent of what goes on. I worked at a college to have learned of a lot of monkey business going on. I have seen right in front of my eyes young women getting all dressed up, loaded with perfume and hanging outside teachers offices and then once in, there is silence, putting your ear to the door and you just hear ruffling around, surely no tutoring session is going on and there's no talk about the assignments but maybe a whisper in the ear of making the grade better somehow. I've seen it with a tennis coach and an underage female and there's so much more that I do not understand when others find out they do nothing about it. I made a big stir while the wholesome tennis community did nothing. Seems women have a strong predilection to go after men of power, even if it's just as mundane as being a teacher. It disturbs me and society gets messed up when the female chooses in such a way that bypasses salient characteristics and moreover is often done knowing full well the man presently has a steady partner (perhaps wife). I've talked about this elsewhere such as at http://datingmateselection.blogspot.com/. Sad, a man seems to have to work the system in some way to obtain a lover and in standing on firm principles, the chances get cut dramatically. I wanted this page just to be about math problems in how they could relate to real life but I can't let go the greater social ramifications related to education where you have a person of authority amidst vulnerable students. This power structure extends into players of professional sports, well-known actors, movie directors and producers, cops, politicians, and politicians and could even be relatively low level ones such as at the local level. Going into a field does not imply the field is paramount to their intentions, it very well could be (or in a lesser degree) the ancillary perks related to sexuality. An honest person could reflect upon their decision of getting into a field and the purpose of moving up a ladder (if applicable) and in such honesty the sexual component could not be denied. /// **
PERCENT DIFFERENCE
It is commonly employed, the formula, abs(b-a)/((b+a)/2), for computing the "percent difference".
I propose two other forms effecting this that provides differing results.
1) abs(b-a)/sqrt(a*b) which could be referred to as the geometric percent difference, and
2) max(a,b)/sqrt(a*b)-1 which could be referred to as the geometric progression difference
Consider the case of a,b = 1,2. The conventional way gets 2/3 or 66.67%
In method 1), we get sqrt(2)/2 ~0.7071 or 70.71%.
In method 2) we get sqrt(2)-1 ~0.4142 or 41.42%. See that 41.42% over 1 gets to 1.4142 and 41.42% over 1.4142 gets to 2. **
AMAZING DISCOVERY!
In attempting to construct a function for a formula to compute a proposed real estate fee for selling homes, I wanted to merge all into a one-line function, one that could easily be written by 3 piece-wise functions. It took me part of two days to do it. I initially created a more difficult form but then realized I could make it a simpler form in judiciously adding two conversions of max functions. I could show the work-up but I lack the tools to easily show it here. You may verify with numerous values, work backwards from the formula I devised, or start from fresh.
The general form of the function would be a constant value from x values of -inf up until a critical x-value, and then a sloped line up/down until another critical x-value and from this point there would be a constant value to +inf. The form for this function could be termed "STRETCHED Z FUNCTION", not to be confused with the zeta function, as this would relate to a graph of a shape similar to the letter "Z" but upper and lower portions pulled apart so there is no more than one point for any vertical line, and that the upper and lower portions continue without bound (unless needing to be truncated per a particular need).
The formula, fixed to feed in values pertaining to the intended plot:
(t1+t2)/2+(t2-t1)/(2*(s2-s1))*(sqrt((s1-x)^2)-sqrt((s2-x)^2))
The value t1 is the constant value on the left side of critical points, t2 is the constant value on the right side of the critical points, s1 and s2 are the x-coordinates for the critical points (s1 and s2 may be interchanged in the formula but t1 and t2 cannot else the plot would flip [wrt itself-independent of location, or in more towards common math language, the flip would be about the line mid-way between the values of of t1 and t2]).
For example, try inputting this into an equation editor on-line or performing on a spreadsheet with sufficient points to see the graph:
(t1+t2)/2+(t2-t1)/(2*(s2-s1))*(sqrt((s1-x)^2)-sqrt((s2-x)^2)), t1=4,t2=10, s1=1,s2=5
** ** For cutting limbs/logs, what is the shortest length for which any length longer, you can cut in lengths between 12 and 15 inches? This problem I've considered for nearly 30 years. I finally got around to solving it. It turns out the solution is 48". You may see in a spreadsheet for a guide to setting up the problem to then extrapolate for any similar type of problem. 12*(x+1)=15*x After solving for x, multiply this intermediate by 12" to obtain 48". Assurance: 1) From 48" and higher up until 60" in length, cutting into 4 pieces evenly would have lengths varying from 12" to 15". 2) From 60" up to 75", cutting into 5 pieces evenly would have lengths varying from 12" to 15". And so on. 3) Now let's look at a piece 47" in total length: cutting into 3 pieces, the lengths would be over 15". Cutting into 4 pieces, the lengths would be under 12". **
________________
| |...../\
| |......\ \
|_|____\ /__ angle theta in this corner
In assigning a line segment of length 1 starting from the right front corner of vehicle, you could assign new variables along the left side of vehicle, that are cut by the temporary line segment as r and s.
Then
sqrt(x^2+r^2) + sqrt(s^2-x^2) = 1, r+s=1, (r=1-s)
Short exercise would show x = sqrt(2s -1) -> s=(x^2+1)/2
The angle that is a complement of theta would thus be acos(x/((x^2+1)/2)) and so
theta=asin(2 x/(x^2+1))
Graphing this shows from x=0 to x=1, the angles starts off at x=0 at the point (x,theta)=(0,0) with a slope of 2 and tapers off at the point (1,pi/2) where the limit of the slope for x=1- is 1. Fascinating! What happens beyond 1 is immaterial to the problem. It looks similar to the dorsal fin of a shark. I may investigate the continuation of this curve as an extension of what occurs from x=0 to 1 and not the downward curve with a cusp at x=1. The slope of 2 at x=0 and slope of 1 at x=1 is interesting.
It turns out when x=1/sqrt(3) that this is a special value such that the right front of vehicle to left rear of vehicle forms a line segment that is perpendicular to the parallel lines. Theta for this case is pi/3. Intuitively, this can be accepted from familiarity with trigonometry.
One could also find out the effective new width formed by the vehicle with respect to the parallel lines.
From a little work, the width can be found to be:
sqrt(x^2)*sqrt((x^2-1)^2)/(1+x^2) + 2x/(1+x^2)
Being that 0 < x < 1, we can reduce further to x(3-x^2)/(1+x^2).
The graph of this function from x=0 to 1 starts at (0,0), goes higher than 1 then back down to the point (1,1). From a little calculus it can be found that the max x value is sqrt(2*sqrt(3)-3) ~0.68125 and with function value of sqrt(6*sqrt(3)-9) ~1.1799596. Unexpected from the initial problem but is sensible from the width function.
** In extending the shark fin curve mentioned above, one may consider the function, 2/(1+x^2) reflecting the derivative at the points x=0 and x=1 and becomes, 2 atan(x). Integrating from x=0 to inf on a site like wolfram alpha is unable to calculate it, giving the "does not converge" error, even when attempting to calculate 1/x integral(0,inf) 2 atan(x) but in this form, I find it to converge to pi. As integral [2 atan(x)] = 2 x tan^(-1)(x) - log(x^2 + 1) + constant, one may work with the form, integral [2 atan(x) b ] and see this is 2b x tan^(-1)(x) - 2b log(x^2 + 1) + constant, and for the limit x->inf, with b=1/x we see this turns into lim x->inf [2 atan(x) -2/x ln(x^2+1)] which converges to 2 atan(inf)=pi.
The function, 2 atan(x) is 0 at x=0, is pi/2 at x=1, and the limit of x->inf has the value pi. From this alone in knowing though some low values of x are far from pi, they are positive and always increasing which one could say for the span of the graph from x=0 to infinite, it would approximate a rectangle of infinite length and height of pi for which the area (integral of the function) from x=0 to inf would thus be x pi.
///One may engage in mathematics, even to a level beyond their formal educational level without going on to be a power monger overpaid teacher, contributing to the debt of this nation while those in the private sector get far less, and this means also not having the great availability of young women (and/or men) for which to play around with as many teachers do in some way or another regardless of what the common person thinks and throwing out "I am a teacher and I do not do that" does not take away from this happening - it even occurs at the high school level and does not get enough attention besides the few cases making a national stir - believe me, what you hear about would be less than a percent of what goes on. I worked at a college to have learned of a lot of monkey business going on. I have seen right in front of my eyes young women getting all dressed up, loaded with perfume and hanging outside teachers offices and then once in, there is silence, putting your ear to the door and you just hear ruffling around, surely no tutoring session is going on and there's no talk about the assignments but maybe a whisper in the ear of making the grade better somehow. I've seen it with a tennis coach and an underage female and there's so much more that I do not understand when others find out they do nothing about it. I made a big stir while the wholesome tennis community did nothing. Seems women have a strong predilection to go after men of power, even if it's just as mundane as being a teacher. It disturbs me and society gets messed up when the female chooses in such a way that bypasses salient characteristics and moreover is often done knowing full well the man presently has a steady partner (perhaps wife). I've talked about this elsewhere such as at http://datingmateselection.blogspot.com/. Sad, a man seems to have to work the system in some way to obtain a lover and in standing on firm principles, the chances get cut dramatically. I wanted this page just to be about math problems in how they could relate to real life but I can't let go the greater social ramifications related to education where you have a person of authority amidst vulnerable students. This power structure extends into players of professional sports, well-known actors, movie directors and producers, cops, politicians, and politicians and could even be relatively low level ones such as at the local level. Going into a field does not imply the field is paramount to their intentions, it very well could be (or in a lesser degree) the ancillary perks related to sexuality. An honest person could reflect upon their decision of getting into a field and the purpose of moving up a ladder (if applicable) and in such honesty the sexual component could not be denied. /// **
PERCENT DIFFERENCE
It is commonly employed, the formula, abs(b-a)/((b+a)/2), for computing the "percent difference".
I propose two other forms effecting this that provides differing results.
1) abs(b-a)/sqrt(a*b) which could be referred to as the geometric percent difference, and
2) max(a,b)/sqrt(a*b)-1 which could be referred to as the geometric progression difference
Consider the case of a,b = 1,2. The conventional way gets 2/3 or 66.67%
In method 1), we get sqrt(2)/2 ~0.7071 or 70.71%.
In method 2) we get sqrt(2)-1 ~0.4142 or 41.42%. See that 41.42% over 1 gets to 1.4142 and 41.42% over 1.4142 gets to 2. **
AMAZING DISCOVERY!
In attempting to construct a function for a formula to compute a proposed real estate fee for selling homes, I wanted to merge all into a one-line function, one that could easily be written by 3 piece-wise functions. It took me part of two days to do it. I initially created a more difficult form but then realized I could make it a simpler form in judiciously adding two conversions of max functions. I could show the work-up but I lack the tools to easily show it here. You may verify with numerous values, work backwards from the formula I devised, or start from fresh.
The general form of the function would be a constant value from x values of -inf up until a critical x-value, and then a sloped line up/down until another critical x-value and from this point there would be a constant value to +inf. The form for this function could be termed "STRETCHED Z FUNCTION", not to be confused with the zeta function, as this would relate to a graph of a shape similar to the letter "Z" but upper and lower portions pulled apart so there is no more than one point for any vertical line, and that the upper and lower portions continue without bound (unless needing to be truncated per a particular need).
The formula, fixed to feed in values pertaining to the intended plot:
(t1+t2)/2+(t2-t1)/(2*(s2-s1))*(sqrt((s1-x)^2)-sqrt((s2-x)^2))
The value t1 is the constant value on the left side of critical points, t2 is the constant value on the right side of the critical points, s1 and s2 are the x-coordinates for the critical points (s1 and s2 may be interchanged in the formula but t1 and t2 cannot else the plot would flip [wrt itself-independent of location, or in more towards common math language, the flip would be about the line mid-way between the values of of t1 and t2]).
For example, try inputting this into an equation editor on-line or performing on a spreadsheet with sufficient points to see the graph:
(t1+t2)/2+(t2-t1)/(2*(s2-s1))*(sqrt((s1-x)^2)-sqrt((s2-x)^2)), t1=4,t2=10, s1=1,s2=5
** ** For cutting limbs/logs, what is the shortest length for which any length longer, you can cut in lengths between 12 and 15 inches? This problem I've considered for nearly 30 years. I finally got around to solving it. It turns out the solution is 48". You may see in a spreadsheet for a guide to setting up the problem to then extrapolate for any similar type of problem. 12*(x+1)=15*x After solving for x, multiply this intermediate by 12" to obtain 48". Assurance: 1) From 48" and higher up until 60" in length, cutting into 4 pieces evenly would have lengths varying from 12" to 15". 2) From 60" up to 75", cutting into 5 pieces evenly would have lengths varying from 12" to 15". And so on. 3) Now let's look at a piece 47" in total length: cutting into 3 pieces, the lengths would be over 15". Cutting into 4 pieces, the lengths would be under 12". **
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