Question

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 that is 1 m higher than the bow of the boat. if the rope is pulled in at a rate of 1 m/s, how fast (in m/s) is the boat approaching the dock when it is 9 m from the dock? (round your answer to two decimal places.)

1. -/1 points

at noon, ship a is 100 km west of ship b. ship a is sailing south at 40 km/h and ship b is sailing north at 20 km/h. how fast (in km/h) is the distance between the ships changing at 4:00 p.m.? (round your answer to one decimal place.)

Explanation:

Step1: Establish the relationship

Let $x$ be the distance between the boat and the dock, and $y$ be the length of the rope. Using the Pythagorean theorem, we have $x^{2}+1^{2}=y^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+1 = y^{2}$ with respect to $t$, we get $2x\frac{dx}{dt}=2y\frac{dy}{dt}$, which simplifies to $x\frac{dx}{dt}=y\frac{dy}{dt}$.

Step3: Find the value of $y$ when $x = 9$

When $x = 9$, by the Pythagorean - theorem $y=\sqrt{x^{2}+1}=\sqrt{9^{2}+1}=\sqrt{81 + 1}=\sqrt{82}$.

Step4: Substitute the known values

We know that $\frac{dy}{dt}=-1$ (negative because the length of the rope $y$ is decreasing). Substitute $x = 9$, $y=\sqrt{82}$, and $\frac{dy}{dt}=-1$ into the equation $x\frac{dx}{dt}=y\frac{dy}{dt}$. We have $9\frac{dx}{dt}=\sqrt{82}\times(-1)$.

Step5: Solve for $\frac{dx}{dt}$

$\frac{dx}{dt}=-\frac{\sqrt{82}}{9}\approx - 1.01$. The negative sign indicates that the distance $x$ is decreasing.

Answer:

$1.01$