A goat is tethered, but is otherwise free to graze in the space she can reach. The barn's front wall is tangent to the silo at the corner. The anchor to the tether is at the point of tangency. The silo has a circular base with radius of 10 feet. The tether is exactly long enough so it wraps around the silo exactly once - that is, the length of the tether equals the circumference of the silo. Calculate the grazing area of the goat. The barn wall is 65 feet long.I do not even know where to begin, please help amp; explain steps???
interesting. I am going to put the center of the silo at 0,0. This puts the spot where the silo meets the barn wall (and where the goat is tethered) at (10,0) the total length of the rope is 20 pi feet (equal to the circumference of the silo) now the goat wants to get to some spot and it stretches the tether as far as it can go. The cord wraps around the silo for some distance and then stretches out straight for some other distance. The length of rope in contact with the silo we will say is 10 t. The position of the point where the rope starts to lose contact is (10 cos t, 10 sin t) the remaining rope has length (20pi - 10t), and the position of the goat is at (10 cos t - (20pi - 10t) sin t, 10 sin t + (20pi-10 t) cos t) or x = 10 cos t - (20pi - 10t) sin t y = 10 sin t + (20pi-10 t) cos t area = integral of x dy/dt dt = integral of y dx/dt dt dy/dt = 10 cos t - (20pi - 10 t) sin t - 10 cos t = - (20pi - 10 t) sin t A = ∫ (10 cos t - (20pi - 10t) sin t)(10 t - 20pi) sin t dt t = 0 to 2pi ∫ -10(20pi-10t) sin t cos t + (20pi - 10t)^2 sin^2 t) dt lets do this in pieces ∫-10(20pi-10t) sin t cos t integrate by parts u = 20 pi - 10 t dv = sin t cos t du = -10 v = -1/4 cos 2t 10{(20 pi - 10 t) (1/4)cos 2t +10 ∫10/4 cos 2t dt} 10{(20 pi - 10 t) (1/4)cos 2t - 10/8 sin 2t | [0,2pi]} -50 pi ∫ (20pi - 10t)^2 sin^2 t) dt integrate by parts u = (20pi - 10t)^2 dv = sin^2 t du = -20(20pi - 10t) v = 1/2 (t - sin t cos t) 1/2 (t - sin t cos t)(20pi - 10t)^2 | [0,2pi] = 0 +∫10(20pi - 10t)(t-sin t cos t) dt = ∫-10(20pi - 10t) sin t cos t + 10(20pi - 10 t) t dt = -50 pi +100 pi t^2 - 100/3 t^3 -100 pi + 400 pi^2 - 800 / 3 pi^3 But the goat can't get into the silo... - 100 pi -200pi + 400 pi^2 - 800/3 pi^3
Need diagram to do this question , i can't guess the correct diagram !
