A 13 ft ladder is leaning against a wall. If the top of the
ladder slips down the wall at a rate of 2 ft/s, how fast will
the foot be moving away from the wall when the top is 5 ft
above the ground?
L=13
"\\frac{dy}{dt}=2"
y=5
we are to find "\\frac{dx}{dt}" when "\\frac{dy}{dt}=2" and y=5
"x^2+y^2=13^2" ......................(i)
"\\therefore 2x\\frac{dx}{dt}+2y\\frac{dy}{dt}=0" ............(ii)
from(i) "x=\\sqrt{13^2-5^2}=12"
from (ii);
"2(12)\\frac{dx}{dt}+2(5)(2)=0"
"\\frac{dx}{dt}=-\\frac{5}{6}"
"=-\\frac{5}{6}ft\/s"
Comments
Leave a comment