A boat sails 30 miles to the east from a point P, then it changes direction and sails to the south. If this boat is sailing at a constant speed of 10 miles/hr, at what rate is its distance from the point P increasing
a) 2 hours after it leaves the point P
b) 7 hours after it leaves the point P
The distance from point P after t hours is
d = sqrt[30^2 + (10 t)^2]
= sqrt(900 + 100 t^2)
dd/dt = [ (1/2)/( sqrt(900 + 100 t^2)]*200 t
Plug in the t value and compute the corresponding rate of change of d
To solve this problem, we can use the concept of rates and trigonometry. Let's break down the problem step by step.
Step 1: Find the coordinates of the boat after it has sailed 30 miles to the east.
Since the boat started at point P, let's assume the coordinates of P are (0,0). After sailing 30 miles to the east, the boat will be at the point (30,0) on the coordinate plane.
Step 2: Determine the rate at which the boat is moving along the x-axis.
Since the boat is sailing at a constant speed of 10 miles/hr, the rate at which it is moving on the x-axis is also 10 miles/hr.
Step 3: Determine the rate at which the boat is moving along the y-axis.
To determine the rate at which the boat is moving along the y-axis, we need to calculate the time it takes for the boat to change direction and sail south. Let's answer each part of the question separately.
a) 2 hours after it leaves point P:
Since the boat has been sailing east for 2 hours, and the speed is 10 miles/hr, it has traveled a distance of 2 * 10 = 20 miles. Therefore, the coordinates of the boat would be (30, -20).
To calculate the rate at which the boat is moving along the y-axis, we need to find the rate of change of the y-coordinate with respect to time. In this case, the boat is moving toward the south, so the rate is negative. Therefore, the rate at which the boat is moving along the y-axis is -20 miles/hr.
b) 7 hours after it leaves point P:
Since the boat has been sailing east for 7 hours, it has traveled a distance of 7 * 10 = 70 miles. However, the boat can only sail 30 miles to the east, so after traveling 30 miles, it changes direction. The remaining distance is 70 - 30 = 40 miles.
Since the boat is now sailing toward the south, we can use the Pythagorean theorem to determine the rate at which the boat is moving away from point P.
Using the Pythagorean theorem, the distance of the boat from point P is given by:
distance^2 = x^2 + y^2
At this point, the x-coordinate is still 30, and the y-coordinate is -20. Plugging these values into the equation, we get:
distance^2 = 30^2 + (-20)^2
distance^2 = 900 + 400
distance^2 = 1300
distance ≈ 36.06 miles
To find the rate at which the distance is increasing, we differentiate both sides of the equation with respect to time.
2 * distance * (rate of change of distance) = 2x * (rate of change of x) + 2y * (rate of change of y)
Since the rate at which the distance is increasing is what we want to find, we set the rate of change of x to 10 miles/hr (as calculated earlier) and the rate of change of y to -20 miles/hr (since the boat is moving south at a constant speed).
2 * (rate of change of distance) = 2 * 10 + 2 * -20
2 * (rate of change of distance) = 20 - 40
2 * (rate of change of distance) = -20
(rate of change of distance) = -20/2
(rate of change of distance) = -10 miles/hr
Therefore, 7 hours after leaving point P, the rate at which the distance of the boat from point P is increasing is -10 miles/hr.