A ladder 25 feet long is leaning against the wall of a building. Initially, the

Question

A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder at time t seconds is given by the parametric equations (7+2t,0). The location of the top of the ladder will be given by parametric equations (0,y(t)). The formula for y(t)= ???? The domain of t values for y(t) ranges from ?? to ?? Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a= ??

Explanation / Answer

By the Pythagorean Theorem (7+2t)^2 + y^2 = 25^2 where y is the height of the top of the ladder. So y = sqrt(625-(7+2t)^2 ) When t=0, y=sqrt(625-49)=24. When t=2, y=sqrt(625-121)=22.45. When t=4, y=sqrt(625-225)=20. When t=6, y=sqrt(625-361)=16.25. When t=8, y=sqrt(625-529)=9.80. When t=9, y=sqrt(625-625)=0. On [0,2] the average velocity is (24-22.45)/2=0.775 ft/sec. On [2,4] the average velocity is (22.45-20)/2=1.225 ft/sec. On [6,8] the average velocity is (16.25-9.80)/2=3.225 ft/sec. On [8,9] the average velocity is (9.80-0)/1 = 9.80 ft/sec.

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