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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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baldymcgee6Best ResponseYou've already chosen the best response.1
dw:1350689236706:dw
 one year ago

eva12Best ResponseYou've already chosen the best response.0
how do you get the answer or solve it?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
You must find a way to relate rate of the pully rope to the rate of the boat using the right angle triangle
 one year ago

eva12Best ResponseYou've already chosen the best response.0
i do not understand this problem or how to approach it to solve it
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
dw:1350690450622:dw
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
you can express the triangle as 7^2 + x^2 =r^2
 one year ago

eva12Best ResponseYou've already chosen the best response.0
then what will x be and would r will stand for rate rigth
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
\[\frac{ dr }{ dt } = 10ft/\min\]\[\frac{ dx }{ dt }_{r=100}\]
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
I'm just calling r the length of ROPE
 one year ago

eva12Best ResponseYou've already chosen the best response.0
so the equation 7^2+x^2=r^2 will be dr/dt=2x+2r
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
when we differentiate 7^2 + x^2 =r^2, we will get: 49 + 2x* dx/dt = 2r* dr/dt
 one year ago

eva12Best ResponseYou've already chosen the best response.0
then i pug in 100 for r and 10 for x in the equation
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
no, dr/dt = 10 ft/min.. we don't know what x is
 one year ago

eva12Best ResponseYou've already chosen the best response.0
49 +2x=2(100)= will = x is 75.5
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
no, we don't solve for x as x is always changing. that is essentially what the question is asking.
 one year ago

eva12Best ResponseYou've already chosen the best response.0
so then how do i approach then
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
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?
 one year ago

eva12Best ResponseYou've already chosen the best response.0
do i solve for 1 equation then the 2 one
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
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
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
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^249) 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(100^249) = your answer
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
no, it shouldn't be that big
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
you must have typed it in wrong in your calculator
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.1
you're very welcome
 one year ago
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