Question

A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 5 ft higher than the front of the boat. The rope is being pulled through the ring at a rate of 0.6 ft/sec. How fast is the boast approaching the dock when 13 ft of rope are out?

Answer 1

Refer to attached picture.

Answer 2

We want to know how quickly the boat is coming in, which is dx/dt, right?

Answer 3

right.

Answer 4

So if the rope length is h, how can we write an equation relating x to h?

Answer 5

yes.

Answer 6

h=sqrt(x^2+3^2)

Answer 7

5^2 + x^2 = h^2. Solve for x instead of h.

Answer 8

oh ok.

Answer 9

And the height of the ring is 5, not 3.

Answer 10

sorry typo

Answer 11

x=sqrt(h^2-5^2)

Answer 12

Yes. That's the right equation. Now take the derivative with respect to h.

Answer 13

dx/dh?

Answer 14

dx/dh=h/sqrt(h^2-5^2)

Answer 15

Yes.

Answer 16

And what is dx/dt?

Answer 17

0.6?

Answer 18

That's dh/dt.

Answer 19

oh right.

Answer 20

so I take the derivative again, but with respect to x?

Answer 21

Not quite. We have dx/dh and we have dh/dt, but we want dx/dt.

Answer 22

Sound familiar?

Answer 23

oh ok.

Answer 24

so dx/dt=.6h/sqrt(h^2-5^2)

Answer 25

Yes. Very good.

Answer 26

then just plug in h?

Answer 27

Indeed.

Answer 28

wonderful! thankyou!

Answer 29

My pleasure =)