Calculus Related Rate Question Boat

A boat is pulled in to a dock by means of a rope with one end attached to the bow of the boat, the other end passing though a ring attached to the dock at a point 4 feet higher than the bow of the boat. If the rope is pulled in at the rate of 2ft/sec how fast is the boat approaching the dock when 10 feet of rope are out?
Please show work
thanks

Let x = distance from boat to dock
Let r = length of rope from boat to ring

By Pythagorean theorem:
x² + 4² = r²
x² = r² − 16

When r = 10
x² = 100 − 16 = 84
x = 2 √21

x² = r² − 16

Differentiate both sides with respect to t (time)
2x dx/dt = 2r dr/dt + 0
dx/dt = (r dr/dt) / x

Now plug in known vales
Rope is pulled in at rate of 2ft/sec -----> dr/dt = −2
10 feet of rope are out -----> r = 10, x = 2 √21

dx/dt = 10 * −2 / (2 √21) = −10/√21 = −2.18

dx/dt < 0 -----> boat is approaching dock at a rate of 2.18 ft/sec

Mαthmφm

A right triangle is formed by the height = vertical distance from bow to ring, the base is the distance from bow to dock, and hypotenuse = length of rope from bow to ring.

Let x = base distance
Let L = rope length

L^2 = 4^2 + x^2

x = sqrt(L^2 - 16)

dx/dt = dx/dL * dL/dt

. . . dx/dL = L / sqrt(L^2 - 16)

dx/dt = L / sqrt(L^2 - 16) * dL/dt

. . . given L = 10 ; dL/dt = - 2

dx/dt = 10 / sqrt(10^2 - 16) * (- 2)

dx/dt = - 20 / sqrt(84) = - 10 / sqrt(21)

distance x is decreasing at about 2.182 ft/sec