How Do You Solve Related Rates Problems in Calculus?

[viper2308]

Homework Statement

A boat is pulled into a dock by means of a winch 12 ft. above the deck of a boat.
(A) the winch pulls in rope at a rate of 4 ft/sec. What is the speed of the boat when there is 13ft of rope out. What happens to the speed of the boat as it gets closer to the dock?
(B)The boat is traveling at 4ft/sec. Determine the speed at which the winch pulls in the rope when there is 13ft of rope out. What happens to the speed at which the winch pulls in rope as the boat gets closer?

Homework Equations

12^2+x^2=z^2 z is the hypotenuse

The Attempt at a Solution

Found Derivative: 2x(dx/dt)=2z(dz/dt)
dx/dt=z(dz/dt)/x I got the answer to A to be 10.4 ft/sec, is this correct?

For B I got 20/13 ft/sec, is this correct?

 

[mighty2000]

it is important to approach this problem using the appropriate equations and units. Let's break down each part of the problem and find the correct solutions.

A) To find the speed of the boat when there is 13 ft of rope out, we can use the Pythagorean theorem to relate the length of the rope (z) to the distance between the dock and the boat (x). This gives us the equation:

12^2 + x^2 = z^2

We can also use the chain rule to relate the rate at which the rope is being pulled in (dz/dt) to the rate at which the boat is moving (dx/dt). This gives us:

2x(dx/dt) = 2z(dz/dt)

Solving for dx/dt, we get:

dx/dt = z(dz/dt)/x

Now we can plug in the given values:

z = 13 ft
dz/dt = 4 ft/sec

To find x, we can use the Pythagorean theorem again:

12^2 + x^2 = 13^2
x = √(13^2 - 12^2) = √25 = 5 ft

Now we can plug in these values to find the speed of the boat (dx/dt):

dx/dt = (13 ft)(4 ft/sec)/5 ft = 10.4 ft/sec

So your answer of 10.4 ft/sec is correct.

As for the second part of the question, as the boat gets closer to the dock, the length of the rope (z) will decrease, which means the speed of the boat (dx/dt) will also decrease. This is because the boat is being pulled in at a constant rate (4 ft/sec) while the length of the rope is decreasing. So as the boat gets closer to the dock, its speed will decrease.

B) For this part, we are given the speed of the boat (dx/dt = 4 ft/sec) and we need to find the speed at which the rope is being pulled in (dz/dt) when there is 13 ft of rope out. Again, we can use the Pythagorean theorem to relate z and x:

12^2 + x^2 = z^2

And the chain rule to relate dz/dt and dx/dt:

2x(dx/dt) = 2