A man standing on a wharf is hauling in a rope attached to a boat, at the rate of 4 ft/sec. if his hands are 9 ft. above the point of attachment, what is the rate at which the boat is approaching the wharf when it is 12 ft away?

Question

Samantha Adkison · Accepted Answer

#sf(-5" ft/s")#

We are told:

#sf((dr)/dt=-4color(white)(x)"ft/s")#

We need to find #sf((d(d))/dt)#

Pythagoras gives us the value of r at that particular instant:

#sf(r^2=h^2+d^2)#

#sf(r^2=9^2+12^2)#

#sf(r^2=81+144)#

#sf(r=15 color(white)(x)ft)#

We know that:

#sf(r^2=81+d^2)#

Differentiating implicitely with respect to t:

#sf(2r.(dr)/dt=2d.(d(d))/dt)#

Putting in the numbers#rArr#

#sf(2xx15xx-4=2xx12.(d(d))/dt)#

#sf(-120/24=(d(d))/dt)#

#sf((d(d))/dt=-5color(white)(x)"ft/s")#

Elijah Abram · Answer

undefined To find the rate at which the boat is approaching the wharf, we can use related rates. Let $x$ be the distance between the boat and the wharf, and let $y$ be the length of rope between the boat and the point of attachment on the wharf.

Given: $\frac{dx}{dt} = 4 \text{ ft/sec}$, $y = 9$ ft

To find: $\frac{dy}{dt}$ when $x = 12$ ft

Using the Pythagorean theorem, $x^2 + y^2 = z^2$, where $z$ is the length of the rope.

Differentiating with respect to time:

$$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(z^2)$$
$$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$$

At the moment in question, $z = \sqrt{x^2 + y^2}$.

Substitute the given values and solve for $\frac{dy}{dt}$:

$$2(12)(4) + 2(9)\frac{dy}{dt} = 2\sqrt{12^2 + 9^2}\frac{dz}{dt}$$
$$96 + 18\frac{dy}{dt} = 2(15)\frac{dz}{dt}$$
$$18\frac{dy}{dt} = 30\frac{dz}{dt}$$
$$\frac{dy}{dt} = \frac{30}{18}\frac{dz}{dt}$$
$$\frac{dy}{dt} = \frac{5}{3}\frac{dz}{dt}$$

Given that $\frac{dx}{dt} = 4 \text{ ft/sec}$, we can solve for $\frac{dz}{dt}$ when $x = 12$ ft:
$$4^2 + 9^2 = z^2$$
$$16 + 81 = z^2$$
$$97 = z^2$$
$$z = \sqrt{97}$$

Differentiating with respect to time:
$$2z\frac{dz}{dt} = 0$$
$$2\sqrt{97}\frac{dz}{dt} = 0$$
$$\frac{dz}{dt} = 0$$

Now substitute $\frac{dz}{dt} = 0$ into the equation for $\frac{dy}{dt}$:
$$\frac{dy}{dt} = \frac{5}{3}(0)$$
$$\frac{dy}{dt} = 0$$

Therefore, when the boat is 12 ft away from the wharf, the rate at which the boat is approaching the wharf is 0 ft/sec.