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A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 7 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 10 ft/min, at what rate will the boat be approaching the dock when 100 ft of rope is out?

Calculus1
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|dw:1350689236706:dw|
how do you get the answer or solve it?
You must find a way to relate rate of the pully rope to the rate of the boat using the right angle triangle

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Other answers:

i do not understand this problem or how to approach it to solve it
can someone help me
|dw:1350690450622:dw|
you can express the triangle as 7^2 + x^2 =r^2
then what will x be and would r will stand for rate rigth
\[\frac{ dr }{ dt } = -10ft/\min\]\[\frac{ dx }{ dt }|_{r=100}\]
I'm just calling r the length of ROPE
so the equation 7^2+x^2=r^2 will be dr/dt=2x+2r
when we differentiate 7^2 + x^2 =r^2, we will get: 49 + 2x* dx/dt = 2r* dr/dt
then i pug in 100 for r and 10 for x in the equation
no, dr/dt = -10 ft/min.. we don't know what x is
49 +2x=2(100)= will = x is 75.5
no, we don't solve for x as x is always changing. that is essentially what the question is asking.
so then how do i approach then
okay... so here's what we do, we have to equations that will help us solve this problem, namely: 7^2 + x^2 =r^2 49 + 2x* dx/dt = 2r* dr/dt are you with me so far?
do i solve for 1 equation then the 2 one
First lets deal with this equation: 49 + 2x* dx/dt = 2r* dr/dt what are we looking for, well the question asks, "at what rate will the boat be approaching the dock?" The rate at which the boat is approaching the dock is represented by dx/dt (the rate of change of x), so we need to re arrange this equation for dx/dt
dx/dt=2r*dr/dt-49/2x
good.. what do we know, we know what dr/dt is, and we know what that r=100 at the point of interest. but we don't know what x is at this given instance, so what we need to do is express x in terms of r in the other other equation: 7^2 + x^2 =r^2 x = sqrt(r^2-49) now substitute that into our change of rates equation, and we know that r =100, and dr/dt= -10: dx/dt = (2(100)*(-10)-49)/2(sqrt(100^2-49) = your answer
no, it shouldn't be that big
you must have typed it in wrong in your calculator
-10.27
thanks
you're very welcome

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