Ant puzzle for math inclined

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ehanes7612

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Say you have a cylinder with an ant on the outside curved wall, a distance (d) from the top rim of the cylinder (which is open air) There is a piece of sugar on the inside of the cylinder wall also a distance (d) from the top. What is the shortest distance the ant has to walk to get to the sugar. Bonus points if you can show the equation. Assume (d) is the shortest distance to the top rim of the cylinder You dont need to know anymore information
 
Draw a line between the ant and the sugar around the circumference. Draw a line from the half way point of this line to the rim of the cylinder (This would be a right angle since the ant and sugar are the same distance (d) from the rim.). Call the intersection with the rim point e. A line connecting the the ant with point e and then the sugar is the shortest distance.
 
On second thought, I want to change my answer.

Draw a line between the ant and the sugar around the circumference. Create a right angle triangle using this line as the base and the sugar as the pointy bit. The top would have to be folded over the rim to reach the sugar. The shortest distance is the hypotenuse of the triangle. I'm not sure the hypotenuse would intersect the rim at the mid point of the base line (point e in my previous answer). I suspect not.
 
the hypotenuse would intersect the rim at the mid point because the distance (d) is the same for both the sugar and ant...most of what you said is in the right direction..not sure what you mean by drawing a line around the circumference..if it helps...make the cylinder a flat (2 dimensional) object
 
Since you didn't specify exactly where the sugar is on the inside, I'll assume it's just on the inside of the ant's location on the outside.

Therefore, the shortest distance is 2d plus the glass thickness. Straight up, straight down. Done.
 
Since you didn't specify exactly where the sugar is on the inside, I'll assume it's just on the inside of the ant's location on the outside.

Therefore, the shortest distance is 2d plus the glass thickness. Straight up, straight down. Done.

Sorry, your solution is only one case (but I get your reasoning)


You dont know where the sugar is in relation to the ant other than the distance to the rim and on the inside (this is why I wrote "this is the only information you need" ..your solution is what mathematicians/physicists call a trivial solution and not general enough,..on a test , the student would be marked wrong for this type of answer..the professor may respond "you are not allowed to take advantage of the ambiguity in the question to make oversimplified assumptions (bad (lazy) science) , next time create a solution that addresses all cases "

and yes, I have seen professors do this
 
total distance walked =d + sq root of (a sq + d sq).
where distance a is the horizontal distance between the ant and the sugar.

???????

you are on the right track


it's actually...2 x sq root (a^2 +d^2) (plus the thickness of cylinder to be extremely accurate) ...if a = 0 as in Ray's case..then you get 2d
 
so the cylinder is an example of extrinsic curvature..it can be easily made into a flat object. A more interesting case is that of intrinsic curvature..such as a sphere..which cant be made into a flat object. So , if you really like trig and you understand why we use trig functions... figure out this example

Suppose the ant is outside a hemispherical bowl and the sugar piece is inside the bowl directly across on the other side (inside the bowl)..what is the shortest distance (neglect the thickness of the bowl rim)?..you can set your variable(s) any way you want

if you get this you are well on your way to understanding General relativity
 
so the cylinder is an example of extrinsic curvature..it can be easily made into a flat object. A more interesting case is that of intrinsic curvature..such as a sphere..which cant be made into a flat object. So , if you really like trig and you understand why we use trig functions... figure out this example

Suppose the ant is outside a hemispherical bowl and the sugar piece is inside the bowl directly across on the other side (inside the bowl)..what is the shortest distance (neglect the thickness of the bowl rim)?..you can set your variable(s) any way you want

if you get this you are well on your way to understanding General relativity

Isnt it basically the same formula but the distances are expressed as 'arc' lengths - 2PiR(C/360) were r=radius and c=circumference of the bowl???
 
Isnt it basically the same formula but the distances are expressed as 'arc' lengths - 2PiR(C/360) were r=radius and c=circumference of the bowl???

that would work, but C is the arc angle of the 'pie' ..you need a ratio from 0 to 1 to measure the arc length and that is multiplied by the circumference..circumference is always 2*pie*R

there is a more formal way in physics involving sin (angle) (which is a ratio of zero to one) because many times integration is involved , and you need an independent variable (angle typically expressed as theta or phi) in the equation..you also need sin(angle) for symmetry arguments that are inevitable in physics..so for example..since it's a hemispherical sphere ..the arch angle from the ant to the rim would be expressed as R*sin^2(theta)*pie/2 ..it's useful for more complicated derivations..not so much for this one
 

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