A shark, looking for dinner, is swimming parallel to a straight beach and is 90 meters offshore. The shark is swimming at a constant speed of 30 meter per second. At time t = 0, the shark is directly opposite a lifeguard station. How fast is the shark moving away from the lifeguard station when the distance between them is 150 meters?
To find the rate at which the shark is moving away from the lifeguard station when the distance between them is 150 meters, we can use the concept of related rates.
Let's assign variables:
- Let x be the distance between the lifeguard station and the shark at time t.
- Let y be the distance the shark has traveled along the beach at time t.
We are given that the shark is swimming parallel to the beach, 90 meters offshore, at a constant speed of 30 meters per second. This means the distance y the shark has traveled along the beach is equal to the distance x between the lifeguard station and the shark at any given time t.
We are looking to find the rate at which the shark is moving away from the lifeguard station, which is the rate of change of x with respect to time, dx/dt.
Using the Pythagorean theorem, we can write an equation relating x and y:
x^2 + y^2 = d^2
where d is the distance between the lifeguard station and the shark (150 meters).
Differentiating both sides of the equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since y = x (as the shark is swimming parallel to the beach), we can substitute y with x in the equation:
2x(dx/dt) + 2x(dy/dt) = 0
Now, let's substitute the known values into the equation:
2(150)(dx/dt) + 2(90)(30) = 0
Simplifying the equation, we have:
300(dx/dt) + 5400 = 0
Solving for dx/dt, we find:
300(dx/dt) = -5400
dx/dt = -5400/300
dx/dt = -18 meters per second
Therefore, the shark is moving away from the lifeguard station at a rate of 18 meters per second.
To find out how fast the shark is moving away from the lifeguard station when the distance between them is 150 meters, we can use the concept of related rates.
Let's define the following variables:
- x: represents the distance between the shark and the lifeguard station.
- t: represents time in seconds.
Given:
- The shark's initial distance from the lifeguard station is 90 meters.
- The shark is swimming at a constant speed of 30 meters per second.
We need to find dx/dt, the rate at which the distance x is changing with respect to time t when x = 150 meters.
To solve this, we'll differentiate both sides of the Pythagorean theorem equation that relates the distance between the shark and the lifeguard station:
x^2 + (90)^2 = (150)^2
Differentiating both sides with respect to time t using the chain rule, we get:
2x * dx/dt + 0 = 0
2x * dx/dt = 0
dx/dt = 0 / (2x)
dx/dt = 0
Therefore, the rate at which the distance between the shark and the lifeguard station is changing when x = 150 meters is 0. This means that the shark is not moving away from the lifeguard station at that specific moment.
make a diagram, you will have a right-angled triangle
let the horizontal distance covered by the shark since t=0 be x m , and
let the distance between the shark and the lifeguard be y m
x^ + 90^2 = y^2
2x(dx/dt) + 0 = 2y(dy/dt)
when y = 150
x^2 + 90^2 = 150^2
x = √14400 = 120
dy/dt = 2x(dx/dt)/(2y)
= 2(120)(30)/300
= 24 m/sec