I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain how to solve this problem.
A boy is standing on a dock watching a boat moving north away from him at a speed of $5000$ ft/min. A girl is standing $1000$ ft east of the boy and is watching the same boat. How fast is the boat moving away from the girl when it is $12500$ ft away from the boy? (Calculator allowed).
My thought process was as such:
We're looking for the derivative of the distance between the girl and the boat. The distance is the hypotenuse of the right-angled triangle formed between the boat, boy, and girl - which is $\sqrt {(5000t)^2 + 1000^2}$.
However, when I wrote out the derivative and plugged in the value for t
(when the boat is $12500$ ft away from the boy, $t=\frac {12500}{5000}=2.5$), I didn't get the correct answer.
Any help (with steps shown) would be really amazing.
No comments:
Post a Comment