A cow is tied to a silo with diameter 30-ft. The rope's length is 100-ft. Find the area the cow can graze on.This one's a tricky one! I've been trying to solve this question for a long time, so any help would be appreciated. Thanks.-John
L = length of rope R = radius of silo w = portion of rope that wraps around silo s = path of cow when the rope is taut, describes an involute A_inv = area of involute A_semi = pi*L^2/2 A_tot = A_semi + A_inv dA_0 = area of infinitesimal triangle in involute with height = (L - w) and base = ds dA_0 = 1/2*(L - w)*ds A_inv = 2*int{dA_0} ... factor of 2 to account for symmetry A_inv = 2*int{1/2*(L - w)*ds} A_inv = int{(L - w)*ds} by similarity of triangles, ds/(L - w) = dw/R ds = (L - w)dw/R A_inv = 1/R*int{(L - w)^2*dw} A_inv = 1/R*int{(L^2 - 2Lw + w^2)dw} the limits of w are 0 to W, where W is length of rope wrapped around silo when the cow just crosses the x-axis solve for W tan(pi - W/R) = (L - W)/R tan(pi - W/15) = (100 - W)/15 tan(pi - W/15) = 100/15 - W/15 and W = 26.59 ft A_inv = 1/R*(L^2*w - Lw^2 + w^3/3) A_inv = 1/15*(100^2*26.59 - 100*26.59^2 + 26.59^3/3) A_inv = 13430.9 ft^2 A_tot = pi*100^2/2 + 13430.9 A_tot = 29138.9 ft^2
Edited when I saw that I had used the diameter, not the radius! Mathteacher and I are doing the same calculations, but with slightly different parameter. Paramaterize the cow's extreme position (i.e., the rope is taunt) by the angle theta, clock-wise, where the rope leaves the silo (tangent everywhere, except at angle theta=0, where it forms a half-circle). Rope is tied at the ground at (15, 0). At theta, 15*theta of rope is along the silo and 100-15*theta is left. (x, y) = 15(cos(theta), sin(theta)) + (100-15theta)(-sin(theta, cos(theta)) = (15cos(theta)-(100-15theta)sin(theta), 15sin(theta)+(100-15theta)cos(theta)) It hits area covered by going clock-wise, as soon as the y value hits 0 again. Call this, Theta. It must be found numerically, since it is known that this type of answer is transcendental. Area = 2 Integral [0 to Theta] (15sin(theta)+(100-15theta)cos(theta))*(... d(theta) This seems to be doable, as checked with wolframalpha, which gave an elementary result. Hope it helps and is correct! I'll try to add to it.
You could divide the region into two parts. The first part is the first quarter revolution where he can move without the rope wrapping around the pole. A1 = π/2 * L? The second part is when the rope starts wrapping around the pole. A2 = area of grazing outside of the pole. We can find A2 using an elemental analysis.
At some levels, yes. I did high school physics and first year university physics without knowing calculus. But this was a long time ago. I know that some physics problems certainly involve calculus. I don't know whether you will need it these days as early as your first level course. Ask a few people who have taken the course at your university or check out the textbook or problem set if you can get your hands on it in advance? In any event, if you intend to go into some further courses, then calculus will come into things sooner or later.
(edited version) Now clearly the cow can graze on half a disc with a radius of 100-ft. The question must be to determine the area where the silo bends the rope. Let us model this part of the setup by a vector function r(t) where t denotes the length of the rope that touches the silo. The rest of the rope will then be tangent to the silos circle at that point. Let us first find an expression for this angle, a(t). a(t) = t/r where r denotes the radius of the silo, ie.
