a boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the…

Explanation:

Step1: Establish a relationship

Let the horizontal distance of the boat from the dock be $x$ (in meters) and the length of the rope be $l$ (in meters). By the Pythagorean - theorem, $x^{2}+1^{2}=l^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+1 = l^{2}$ with respect to $t$ gives $2x\frac{dx}{dt}=2l\frac{dl}{dt}$. Then $\frac{dx}{dt}=\frac{l}{x}\cdot\frac{dl}{dt}$.

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

When $x = 9$, using $x^{2}+1^{2}=l^{2}$, we have $l=\sqrt{9^{2}+1^{2}}=\sqrt{81 + 1}=\sqrt{82}$.

Step4: Substitute the known values

We know that $\frac{dl}{dt}=-1$ (negative because the length of the rope $l$ is decreasing). Substituting $x = 9$, $l=\sqrt{82}$, and $\frac{dl}{dt}=-1$ into $\frac{dx}{dt}=\frac{l}{x}\cdot\frac{dl}{dt}$, we get $\frac{dx}{dt}=\frac{\sqrt{82}}{9}\times(-1)\approx - 1.01$.

Answer:

$1.01$