Rating Between Shark Helmets Related Rates Calculus: Offshore?
Related Rates Calculus: Offshore? - rating between shark helmets
Hello everyone,
I need help with this problem on the related rates in Calculus AB. I understand better what gives me the problem, but I need pictures to describe the sea
A shark in search of food, is to swim parallel to the beach and 90 meters wide. The shark swam at a constant speed of 30 meters per second. At time t = 0, the shark is directly opposite the lifeboat. What is the speed of a shark near the lifeguard station, where the distance between them is 150 meters?
Thank you.
|
1 comments:
Let d be the distance between them.
Then,
d ^ 2 = 90 ^ 2 + (30T) ^ 2 = 8100 + 900T ^ 2
The differentiation with respect to time gives
2DD '= 1800T
d '= 1800T / 2D
We are told that D = 150 feet. We know from the Pythagorean theorem that 150 ^ 2 = 90 ^ 2 + x (^ 2, which is the horizontal distance of travel of sharks). Solving for x yields 120 feet. The shark would take 4 seconds to go so far. So now we have everything we need to calculate d '.
d '= 1800 (4) / 2 (150) = 24 meters per second.
Post a Comment