A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 7 m from the dock?
IM SO CONFUSED ON HOW TO DO THIS PROBLEM ANY HELP IS WELCOMED!! THANK YOU
IM SO CONFUSED ON HOW TO DO THIS PROBLEM ANY HELP IS WELCOMED!! THANK YOU
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DRAW the figure...your model has a right triangle with sides x , 1 and hypotenuse y
and you are given that dy / dt = - 1 { m/s } , find dx / dt when x = 7
work it !...{ ≈ - 1.01 m/s }
and you are given that dy / dt = - 1 { m/s } , find dx / dt when x = 7
work it !...{ ≈ - 1.01 m/s }
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The bow, pulley, and horizontal distance to dock forms a right triangle.
let x = the length of rope between the bow and the pulley.
let D = horizontal distance from bow to dock
x^2 = D^2 + 1^2
D^2 = x^2 - 1
D = sqrt(x^2 - 1)
dD/dt = dD/dx * dx/dt
. . . dD/dx = x / sqrt(x^2 - 1)
dD/dt = x / sqrt(x^2 - 1) * dx/dt
. . . given dx/dt = -1 ; x = 7
dD/dt = 7 / sqrt(7^2 - 1) * (-1)
dD/dt = - 7 sqrt(3) / 12 ... about - 1.01 m/s <== minus sign = distance decreasing.
boat approaching dock at 1.01 m/s
let x = the length of rope between the bow and the pulley.
let D = horizontal distance from bow to dock
x^2 = D^2 + 1^2
D^2 = x^2 - 1
D = sqrt(x^2 - 1)
dD/dt = dD/dx * dx/dt
. . . dD/dx = x / sqrt(x^2 - 1)
dD/dt = x / sqrt(x^2 - 1) * dx/dt
. . . given dx/dt = -1 ; x = 7
dD/dt = 7 / sqrt(7^2 - 1) * (-1)
dD/dt = - 7 sqrt(3) / 12 ... about - 1.01 m/s <== minus sign = distance decreasing.
boat approaching dock at 1.01 m/s