use arc length formula to determine distance from arc PE to equator P?
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The short explanation is that the quickest way to get to the equator, of course, is to go due south.\n\nThere are a total of 90 degrees of latitude above the equator; you're 33'57" above the equator.\n\nA full "360 degrees" latitude constitutes the circumference.  From the radius value, calculate circumference:\n\nc = 2πr = 2π*3960\n\nNow, the fraction of the arc you have is between 33'57" and 0'0", and this is out of 360'.  \n\nConverting 33'57" to decimal (57/60 = 0.95), we get 33.95 degrees.\n\nCompared to the entire circumference, the ratio of this arc to the total circumference is 33.95/360 = 0.0943\n\nSo let's multiply that by the entire circumference:\n(0.0943*2π*3960) = \n\n2,346 miles\n\n\nI think it's right, hope it helps!
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the circumference of the earth\nat the equator is 24,902 mi / 40,076 km.
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It depends a lot on how high above the water your eyes are and how tall the distant object is.\n\nThe radius of the Earth at the equator is 6378135 meters.  Assume your eyes are 2 meters above sea-level, and the distant object is at sea-level.\n\nYou have a right triangle, where the longer leg is the radius of the Earth, and the hypotenuse is 6378315 + 2.  The shorter leg is the distance to the horizon.\n\nThe distance to the horizon = √(6378317^2 - 6378315^2) = 5051 meters, or about 3 miles.\n\n\nIf the distant object is 2 meters above sea-level, then it'll disappear when it's another 3 miles past the horizon, or 6 miles from you.
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Let L be the initial rope's length and R be the radius of the equator in metres.  Then \n\nL = 2piR\n\nAdding 1 to the length, we get:\n\nL+1 = 2pi(R+x)\n\nwhere x is the extra radius, i.e. the height at which the rope would float.  Combining these two equalities we get\n\n2piR + 1 = 2pi(R+x)\n2piR + 1 = 2piR + 2pi*x\n1 = 2pi*x\nx = 1/(2pi) = 0.159 or about 16 centimetres.\n\nThe cool part about this is, it's independent of the earth's radius!  The result is the same whether you start with the earth, a basketball or a marble.
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This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
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the arc length differential is da^2=dx^2+dy^2. divide RHS by dx^2 and square root both sides to get\n\nds=sqrt(1+(dy/dx)^2)\n\nso y(x)= x^5 => dy/dx=5x^4\n\nso to get arc length we integrate ds with respect to x with limits x= 0 to x= 1\n\nintegral from 0 to 1 of sqrt(1+25x^8) with respect to x\n\nsince a rounded answer is required i assume you must use a numerical scheme such as Simpson's rule to obtain answer but again the integrand is sqrt(1+25x^8) limits x= 0 to 1
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"Diameter: \n Equatorial 12,756.274 km \n Polar 12,713.504 km \n Mean 12,745.594 km"\n\n"At the equator the diameter of the earth is 7927 miles; at the poles, it is very close to 7900 miles. 27 miles, a difference of 0.0034 is just enough to give mapmakers a headache."\n\n"the earth's diameter is 7,912 English miles"
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the arc length differential is da^2=dx^2+dy^2. divide RHS by dx^2 and square root both sides to get\n\nds=sqrt(1+(dy/dx)^2)\n\nso y(x)=x^4 => dy/dx=4x^3\n\nso to get arc length we integrate ds with respect to x with limits x= 0 to x= 1\n\nintegral from 0 to 1 of sqrt(1+16x^6) with respect to x\n\nsince a rounded answer is required i assume you must use a numerical scheme such as Simpson's rule to obtain answer but again the integrand is sqrt(1+16x^6) limits x= 0 to 1
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If you find the latitude of the northernmost point and the latitude of a point directly south of that you can calculate it.  Multiply the difference in degrees by 60 to get the distance in nautical miles.  Then apply the conversion factors:  1 nautical mile = 1.152 statute miles and 1 statute mile = 1.609 kilometers.  Combined, that means to multiply the difference in longitude in degrees by 111.214.  (Use degrees and decimal degrees, not minutes and seconds).\n\nYou'll be pretty close with this method even though it assumes a spherical Earth.
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The way to figure this out is, \n\nS=segment length of arc \nA=angle subtended by arc\nR is radius of circle \n\nS/A=2*pi*R/(2*pi),\n\nso S=RA\n\nthen differentiate with respect to time t, \n\nand dS/dt=R*dA/dt,\nand dA/dt=w, the angular speed\n\ndS/dt=T, tangential velocity\n\nso T=R*w
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Imagine you owned a field which was 1 mile \nwide and 2 miles long.  If you'd like to\nknow how far you'd have to walk from one\ncorner diagonally to the opposite corner,\nthen it would be handy to know "square root\nof 5", since that's the number of miles \nthe diagonal is.
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The Distance Formula\nThe Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Here's how we get from the one to the other:\n\n \n\nSuppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The points look like this:\n\n\n  \n\n\n plotted points\n\n  \n\n \n\nYou can draw in the lines that form a right-angled triangle, using these points as two of the corners:\n\n\n  \n\n\n right triangle\n\n  \n\n \n\nIt's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values:\n\n\n  \n\n\n lengths\n   \n\nThen use the Pythagorean Theorem to find the length of the third side (which is the hypoteneuse of the right triangle):\n\n      c2 = a2 + b2\n\n...so: Copyright © Elizabeth Stapel 2000-2006 All Rights Reserved\n\n      c = 5\n\nThis format always holds true. Given two points, you can always plot them, draw the right triangle, and then find the length of the hypoteneuse. The length of the hypoteneuse is the distance between the two points. Since this format always works, it can be turned into a formula:\n\n      Distance Formula:\n\n      Given the two points (x1, y1) and (x2, y2), the distance between these points is given by the formula:\n\n            Distance Formula\n\nDon't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you.\n\n    * Find the distance between the points (–2, –3) and (–4, 4).\n\n      Just plug them in to the Distance Formula:\n\n            d = sqrt(53)\n\n      Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.\n\nThe commonest mistake made when using the Formula is to accidentally mismatch the x-values and y-values. Don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly.\n\nAlso, don't get careless with the square-root symbol. If you get in the habit of omitting the square root and then "remembering" to put it back in when you check your answers in the back of the book, then you'll forget the square root on the test, and you'll miss easy points.\n\nYou also don't want to be careless with the squaring inside the Formula. Remember that you simplify inside the parentheses before you square, not after, and remember that the square in on everything inside the parentheses, including the minus sign, so the square of a negative is a positive.\n\nIn other words, if you do each step completely, instead of sloppily or in your head, then you're much more likely to get the right answers.\n\nBy the way, it is almost always better to leave the answer in "exact" form (the square root "sqrt(53)" above). Rounding is usually reserved for the last step of word problems. If you're not sure which format is preferred, do both, like this:\n\n      d = sqrt(53), or about 7.28\n\nNow YOU try it!\n\nVery often you will encounter the Distance Formula in veiled forms. That is, the problem will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember the Formula. For instance:\n\n    * Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.\n\n      The radius is the distance between the center and any point on the circle, so find the distance:\n\n            d = sqrt[ (2 + 1)^2 + (-3 + 2)^2 ] = sqrt(10)\n\n      Then the radius is sqrt(10), or about 3.16, rounded to two decimal places.\n\n    * Find all points (4, y) that are 10 units from the point (–2, –1).\n\n      I'll plug the two points and the distance into the Distance Formula:\n\n            10 = sqrt[(-2 - 4)^2 + (-1 - y)^2] = [y^2 + 2y + 37]\n\n
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According to www.about.com the circumference of the earth measured at the equator is 24,901.55 miles. Measuring through the poles is a slightly shorter distance at 24,859.82 miles.
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This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
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write the equation of the distance from a point to the center,\nthat way it should work\nalso writing in english might get you more answers\ns
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All these can be answered using the basic formula for distance of:\nDistance = rate x time\nD = rt\n\na) Ann's rate is 3 m/s and her time is 2 minutes, but we need that in seconds.\nr = 3 m/s\nt = 2 min x 60 sec/min = 120 sec\n\nThen it is easy to calculate the distance as:\nD = 3 x 120 \n= Ann will have travelled 360 metres after 2 minutes\n\nb) Ann and Ben are closing at a rate of 10 m/s (3+7)\nr = 10 m/s\nt = 120 seconds (2 minutes)\nD = 10 x 120 = 1200\nSo they have travelled 1200 meters (360m for Ann, 840m for Ben).\nSubtracting 6000 - 1200, you get 4800m.\nSo they are 4800 meters apart after 2 minutes.\n\nc) Again, they cover 10 m/s together, so\nt = D/r\nt = 6000 m / 10 m/s\nt = 600 seconds\nt = 10 minutes\nIn 10 minutes, Ann would travel:\nDistance from Point P to M = 3 m/s x 600 sec\nD = 1800 metres from P to M\n\nd) Distance = 6000, rate = 7 m/s\nt = 6000 / 7\nt = 857 1/7 seconds\n\nNow convert this to minutes and seconds:\n857 / 60 = 14 r. 17\nSo it took Ben 14 minutes 17 seconds (actually 17.142857... seconds), but they said to the nearest second.\n\ne) To determine the ratio of scales, you need to get them to the same units.\n2 cm : 5 km\n1 cm : 2.5 km\n1 cm : 2500 m\n1 cm : 250000 cm\n(i) 1:250000 scale\n\nFrom above we know that 1 cm is equivalent to 2.5 km.  Write these as similar ratios.\n1 / 2.5 km = x / 6 km\nx = 6 / 2.5\nx = 2.4 cm\n(ii) 6km would be represented as 2.4 cm on the map.\n\nIn summary:\n(a) 360 m\n(b) 4800 m\n(c) 1800 m\n(d) 14 minutes 17 seconds\n(e:i) 1:250000\n(e:ii) 2.4 cm
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Let r=the earth's radius and h=your height.\n\nDraw a right triangle ABC, where AC = length r, BC = length r+h, and <A is the right angle and angle C is very small.  C is the center of the earth, and BA is tangent to the earth.  BA is the distance from your head to the horizon.  By Pythagoreas' theorem, \n\nlength(BA) = sqrt((r+h)^2 - r^2)\n= sqrt(2rh + h^2).