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

Question

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? (Round your answer to two decimal places.)

Answer 1

You need to picture a right triangle formed vertically by the dock, the base is the water and the hypotenuse is the rope.

 

The algebraic relationship between these three sides is the pythagorean theorem.  a²+b²=c².

 

Now implicitly differentiate that relationship 2a(da/dt)+2b(db/dt)=2c(dc/dt).

 

We know the following:

a=1(the dock is a meter above the bow)

da/dt = 0 (that height is not changing)

b=7(the dock is 7 feet away from the boat)

db/dt = ? (that is the speed we are looking for

c = well...they didn't give it to use but we can find it using the pythagorean theorem because we know the dock is 1 meter and the boat is 7 m away so sqrt(1²+7²)=sqrt(50)=5sqrt(2)

dc/dt = 1 (the rope is being pulled in at 1 m/s)

 

We can know plug into our differentiated equation

2(1)(0)+2(7)(db/dt)=2(5sqrt(2))(1)

 

and solve

 

db/dt = 5sqrt(2)/7