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 problem, we can use the concept of related rates. Related rates problems typically involve finding the rate at which one quantity changes with respect to another quantity. In this case, we want to find how fast the distance between the two ships is changing.
Let's start by assigning some variables and breaking down the given information:
Let:
- d(t) be the distance between the two ships at time t.
- t be the time elapsed in hours since noon. So, at 4:00 pm, t = 4.
Now, let's focus on finding an equation relating the variables.
We know that ship A is sailing south at a constant speed of 35 km/h. Since ship A is 100 km west of ship B at noon, we can express its horizontal position as x(t) = -100 km - (35 km/h) * t.
Similarly, ship B is sailing north at a constant speed of 25 km/h. Since ship B is stationary at noon, we can express its vertical position as y(t) = (25 km/h) * t.
Next, we need to find an equation for the distance between the two ships. We can apply the Pythagorean theorem to their positions:
d(t) = √[x(t)^2 + y(t)^2]
Substituting the equations for x(t) and y(t):
d(t) = √[(-100 km - (35 km/h) * t)^2 + ((25 km/h) * t)^2]
Now, to find how fast the distance between the ships is changing at 4:00 pm (t = 4), we can differentiate the equation with respect to time (t) and solve for d'(t).
d'(t) = [d(t)]'
= [√[(-100 km - (35 km/h) * t)^2 + ((25 km/h) * t)^2]]'
= ... (simplification required)
At this point, we need to simplify the equation and differentiate it with respect to t to find d'(t). Once we have d'(t), we can substitute t = 4 to find the rate of change of the distance between the ships at 4:00 pm.