Related Rates
I am having trouble finding the equation to use to fit this all together.
At noon, ship A is 100km west of ship B. Ship A is sailing south at 35 km/s and ship B is sailing north at 25 km/s. How fast is the distance between the ships changing at 4:00pm.
To solve this related rates problem, we need to understand the relationship between the variables and their rates of change. Let's break down the problem and determine the relevant variables:
Given information:
- Ship A is sailing south at 35 km/s.
- Ship B is sailing north at 25 km/s.
- At noon, Ship A is 100 km west of Ship B.
Let's assign the following variables:
- Let x represent the distance Ship A has traveled in kilometers since noon.
- Let y represent the distance Ship B has traveled in kilometers since noon.
- Let d represent the distance between the two ships at any given time.
We need to find d'(t), the rate at which the distance between the two ships is changing at 4:00 pm. To do this, we can differentiate the equation involving the variables, x, y, and d, with respect to time (t).
From the given information, we can find the relationship between x, y, and d:
d = 100 + x + y
Now, let's differentiate the equation with respect to time (t):
d/dt(d) = d/dt(100 + x + y)
The derivative of d with respect to t is d'(t). The derivatives of x and y with respect to t are dx/dt and dy/dt, respectively.
d'(t) = d/dt(100) + d/dt(x) + d/dt(y)
Since 100 is a constant, its derivative is zero.
d'(t) = 0 + dx/dt + dy/dt
Now, we need to determine dx/dt and dy/dt. Given that Ship A is sailing south at 35 km/s and Ship B is sailing north at 25 km/s:
dx/dt = -35 km/s (negative because Ship A is moving in the south direction)
dy/dt = 25 km/s
Plugging these values into the related rates equation:
d'(t) = 0 + (-35) + 25
d'(t) = -10 km/s
Therefore, the distance between the ships is changing at a rate of -10 km/s (negative because the ships are moving closer together) at 4:00 pm.