A spectator watches a rowing race from the edge of a river bank. The lead boat is moving in a straight line that is 330 ft from the river bank. If the boat is moving at a constant speed of 50 ft/sec, how fast is the boat moving away from the spectator when it is 560 ft past her?
A) 18.63 ft/sec
B) 43.08 ft/sec
C) 61.95 ft/sec
D) 73.01 ft/sec

Question

A spectator watches a rowing race from the edge of a river bank. The lead boat is moving in a straight line that is 330 ft from the river bank. If the boat is moving at a constant speed of 50 ft/sec, how fast is the boat moving away from the spectator when it is 560 ft past her? ​
A) 18.63 ft/sec
B) 43.08 ft/sec
C) 61.95 ft/sec
D) 73.01 ft/sec

Quizplus · Accepted Answer

To find how fast the boat is moving away from the spectator, we can use the Pythagorean theorem to model the situation. Let $x$ be the distance of the boat from a point directly opposite the spectator on the other side of the river, $y$ be the distance from the spectator to the boat, and $330$ ft be the distance from the spectator to the point directly opposite on the other side of the river. We have $y^2 = 330^2 + x^2$. Differentiating both sides with respect to time $t$, we get $2y \frac{dy}{dt} = 2x \frac{dx}{dt}$. Given that $\frac{dx}{dt} = 50$ ft/sec (the speed of the boat) and $x = 560$ ft (the distance past the spectator), we can solve for $\frac{dy}{dt}$ (the speed at which the distance from the spectator to the boat is increasing).First, find $y$ when $x = 560$ ft: $y = \sqrt{330^2 + 560^2} = \sqrt{108900 + 313600} = \sqrt{422500} = 650$ ft.Then, solve for $\frac{dy}{dt}$: $\frac{dy}{dt} = \frac{x}{y} \cdot \frac{dx}{dt} = \frac{560}{650} \cdot 50 = \frac{56}{65} \cdot 50 = \frac{2800}{65} \approx 43.08$ ft/sec.

A Spectator Watches a Rowing Race from the Edge of a River