Big right triangle, C at base of light pole, A at top, B at shadow tip
Then D at base of man
AC = 4
CD = 10 at start = y(t) = 10 -t
CB = x with velocity dx/dt
in general
x/4 = (x-y)/1.89 similar triangles
4x - 4y = 1.89 x
2.11 x = 4 y
2.11 (dx/dx) = 4(dy/dx)
2.11 (dx/dx*dx/dt) = 4 (dy/dx*dx/dt) chain rule
2.11 dx/dt = 4 dy/dt
dx/dt = (4/2.11)(-1m/s)
a) A young man measuring 1.89 m. walks at a velocity of 1m/s towards a lamppost which its light is at 4m from the ground. At what velocity does the young man's shadow decrease when he's 10 m away from the lamppost?
b) A young girl, standing at the edge of a dock, pulls a boat towards herself with a cable attached to the front of a boat at 0.3m from sea level. If the girl pulls the cable at a velocity of 1.6 m/s, her hands being at 2.5m from sea level, at what velocity does the boat approach the dock when it's 3m away?
2 answers
The second problem is similar.
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