Need answers to calculus problems with clear solution?

1.A picture on a wall is 2ft high and 1ft above the eye鈥檚 observer. At what distance from the wall should the observer position in order to have a maximum subtended angle on the picture?





2.Approximate sin(3.14159) using differentials correct to 5 decimal places.Need answers to calculus problems with clear solution?
1. In the cross-section containing the human's eye and perpendicular to both wall and the floor, let A be the top of the picture, B the bottom of the picture, C below the picture but at the eye's level, and E the eye. Hence AB = 2ft. BC = 1 ft. Our task is to maximize angle AEB, which is bigger than 0 but smaller than pi/2. Let F = tan (angle AEB). When angle AEB is maximized, F is also maximized. Let BEC notes the angle BEC, and AEC notes the angle AEC. Also let h = CE, the distance of the eye from the wall. Hence:


3/h = tan AEC = tan (AEB + BEC)


= (F + tan(BEC)) / (1 - F tan(BEC))


= (F + 1/h)/(1 - F/h)


or 3(h - F) = Fh^2 + h


F = 2h / (h^2 + 3)


dF/dh = 0 when F is at maximum:


0 = dF/dh = { 2(h^2 + 3) - 2h*2h} / (h^2 + 3)^2


leads to h = 鈭?


So the observer position should be at 鈭? ft from the wall in order to have a maximum subtended angle on the picture.





2. Let y = pi - 3.14159


sin(3.14159) = sin(pi - 3.14159) = sin y


= y - y^3/3! + y^5/5! - .... %26lt;Taylor Expansion%26gt;


鈮?y = 0.0000026535....Need answers to calculus problems with clear solution?
Let me try to see if I may help you further on #2.


my first step is from: sin x = sin (pi - x)


my second step is from the Taylor Expansion of f = sin x, where the coefficients are: f'(0), f''(0)/2!, f'''(0)/3!, .....


my third step: y^3 is very very small and thus been ignored.

Report Abuse