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Math challenging question
1. A wall is 1.8m high and 1.2m from a building. Find the length of the shortest ladder that will touch the building, the top of the wall, and the ground beyond the wall. 2. Show that the maximum value of y= a sinx +b cosx is (a^2 +b^2)^(1/2) and the minimum value is - (a^2 +b^2)^(1/2).
最佳解答:
1. First draw a sketch of the wall 1.8 m high at 1.2 from the building wall, for example, see: http://i263.photobucket.com/albums/ii157/mathmate/math/ladder.jpg where t=angle of elevation of ladder, L=total length. By simple trigonometry for the height where the ladder touches the building, we can deduce that Lsin(t)=1.8+1.2tan(t), or L=1.8/sin(t)+1.2/cos(t) differentiate w.r.t. t and equate to zero for a minimum: we get dL/dt=1.2tan(t)/cos(t)-1.8cot(t)/sin(t)=0 and note that the function is singular at t=0 and t=pi/2. Multiply by cot(t) to get 1.2tan(t)-1.8cot(t)^2 tan(t)=(1.8/1.2)^(1/3) tan(t)=1.1447142... t=0.852770877... See the plots of dL/dt w.r.t. t http://i263.photobucket.com/albums/ii157/mathmate/math/ladder1.png http://i263.photobucket.com/albums/ii157/mathmate/math/ladder2.jpg Substitute in the L(t) expression above to get L(0.852...) L=1.8/sin(0.852770877)+1.2/cos(0.852770877) =4.214 m. 2008-05-19 05:19:24 補充： 2. y=asin(x)+bcos(x) dy/dx=acos(x)-bsin(x)=0 (at max. or min.) from which tan(x)=a/b, sin(x)=a/sqrt(a^2+b^2), cos(x)=b/sqrt(a^2+b^2) y =asin(x)+bcos(x) =(a^2+b^2)/sqrt(a^2+b^2) =+/- sqrt(a^2+b^2)
其他解答:63D0B758E2D502CC
Math challenging question
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發問:1. A wall is 1.8m high and 1.2m from a building. Find the length of the shortest ladder that will touch the building, the top of the wall, and the ground beyond the wall. 2. Show that the maximum value of y= a sinx +b cosx is (a^2 +b^2)^(1/2) and the minimum value is - (a^2 +b^2)^(1/2).
最佳解答:
1. First draw a sketch of the wall 1.8 m high at 1.2 from the building wall, for example, see: http://i263.photobucket.com/albums/ii157/mathmate/math/ladder.jpg where t=angle of elevation of ladder, L=total length. By simple trigonometry for the height where the ladder touches the building, we can deduce that Lsin(t)=1.8+1.2tan(t), or L=1.8/sin(t)+1.2/cos(t) differentiate w.r.t. t and equate to zero for a minimum: we get dL/dt=1.2tan(t)/cos(t)-1.8cot(t)/sin(t)=0 and note that the function is singular at t=0 and t=pi/2. Multiply by cot(t) to get 1.2tan(t)-1.8cot(t)^2 tan(t)=(1.8/1.2)^(1/3) tan(t)=1.1447142... t=0.852770877... See the plots of dL/dt w.r.t. t http://i263.photobucket.com/albums/ii157/mathmate/math/ladder1.png http://i263.photobucket.com/albums/ii157/mathmate/math/ladder2.jpg Substitute in the L(t) expression above to get L(0.852...) L=1.8/sin(0.852770877)+1.2/cos(0.852770877) =4.214 m. 2008-05-19 05:19:24 補充： 2. y=asin(x)+bcos(x) dy/dx=acos(x)-bsin(x)=0 (at max. or min.) from which tan(x)=a/b, sin(x)=a/sqrt(a^2+b^2), cos(x)=b/sqrt(a^2+b^2) y =asin(x)+bcos(x) =(a^2+b^2)/sqrt(a^2+b^2) =+/- sqrt(a^2+b^2)
其他解答:63D0B758E2D502CC
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