Fly Line Triangle
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Math, word problem derivative?
A child is flying a kite. If the kite is 90 feet above the child's hand level and the wind is blowing it on a horizontal course at 3 feet per second, how fast is the child paying out cord when 150 feet of the cord is out? (assume the cord forms a line)
ok so i drew out a triangle and said h=90, c= 150, db/dt= 3ft/sec and we have to find dc/dt
but we can get b by doing the pythagorean theorem so b = 120
h^2+ b^2= c^2
2h(dh/dt) + 2b(db/dt) = 2c(dc/dt)
substitution:
2(90)(dh/dt) + 2(120)(3) = 2(150)(dc/dt)
to find dc/dt can we say:
(dh/dt)^2 + (db/dt)^2 = (dc/dt)2
(dh/dt)^2 + 9 = (dc/dt)2
the square root((dh/dt)^2 + 9)= (dc/dt)
square root((dh/dt)^2)) + 3= (dc/dt)
(dh/dt) + 3 = (dc/dt)
so plugging in for dc/dt we get:
2(90)(dh/dt) + 2(120)(3) = 2(150)((dh/dt) + 3)
180(dh/dt) + 720= 300(dh/dt) + 900
-120(dh/dt)=180
(dh/dt)= -1.5..cause its going in the opp direction wright?
-1.5 + 3 = 1.5= dc/dt
1.5= dc/d
VERIFY PLEASSSSSSSSSE
h is a constant...it does not change...it stays at 90 since the kite is moving horizontally.
so,
90^2 + b^2 = c^2
d/dt(90^2 + b^2) = d/dt(c^2)
2b db/dt = 2c dc/dt
2(120)(3) = 2(150)(dc/dt)
dc/dt = 12/5 ft/s
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