Calculus!!...Related Rates!!....Please I entail alot of HELP!?

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 specifically 1 m higher after the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how vigorous is the boat approaching the dock when it is 8 m from the dock?

---

Answers:    1 m.s no acceleration!

---

The rope, the dock, and the boat form a right triangle. When the boat is 8m from the dock, the rope is 8/(squareroot(65)) m long. So the ratio is:

8/(squareroot(65)) = x/1
x = 8/(squareroot(65))

---

OK, so when the boat is x m away from the dock the length of the rope is √(x^2 + 1) m. So we own
y = √(x^2 + 1)
dy/dt = -1
dx/dt = ?

dy/dt = dy/dx . dx/dt
So -1 = (1/2) (x^2 + 1)^(-1/2) . (2x) dx/dt
so dx/dt = -1 / [x . (x^2+1)^(-1/2)]
= -√(x^2+1) / x
So when x is 8 we have dx/dt = -√(65)/8 ≈ -1.008 m/s. So the boat is approaching the dock at approximately 1.008 m/s.

---

The closer it return with, the faster it get here.

---

More Auto Questions and Answers...

What is the entitle of biggest logistic ship surrounded by the world?

Boat engine keep dying, why?

1989 20Ft Bayliner wont shift into reverse lacking shutting sour the engine.?

Need to repair outboard that wont tilt up?

Problems next to a Force 40 HP?

What is the best type of bowrider that have more than 190hp for underneath close to 15 dignified?

This article contents is post by this website user, AutoQnA.com doesn't promise its accuracy.