If the boat is x meters from the dock, then the length of rope is given by
r^2 = x^2 + 1
2r r' = 2x x' = 2(8)(-1)
rr' = -8
So, r(8) = √65
r'√65 = -8
r' = 0.992 m/s
Makes sense; since the boat is far away as compared to the height, the rope length is essentially the distance away, so dr/dt is about the same as dx/dt.
The closer the boat gets, the slower the rope length decreases; when the boat is very near the dock, the rope length essentially stays at 1 meter.
a) A boat is pulled into a dock by a rope attached to the bow ("front") of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?
b)A particle is moving along the curve y=√x. As the particle passes through the point (4,2), its x-coordinate increases at a rate of 3cm/s. How fast is the distance from the particle to the origin changing at this instant?
c)The price (in dollars) p and the quantity demanded q are related by the equation: p2+2q2=1100.
If R is revenue, dR/dt can be expressed by the following equation: dR/dt=A(dp/dt),
where A is a function of just q.
i)Find A.
ii)Find dR/dt when q=20 and dp/dt=4.
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