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
at 10 mph, it is still going east at t=2
so, distance is increasing at 10 mph
at t=3, it turns south. After that, the distance d from P is
d^2 = 30^2 + (10(t-3))^2
since it has only been sailing south for (t-3) hours.
2d dd/dt = 200(t-3)
at t=7, d=50
2*50 dd/dt = 200(4)
dd/dt = 8 mph
To find the rate at which the boat's distance from point P is increasing, we can use the concept of related rates.
Let's break down the problem into two parts - the boat's distance east and its distance south from point P.
a) 2 hours after it leaves point P:
In 2 hours, the boat will have traveled 2 * 10 = 20 miles to the east. At this point, it changes direction and starts moving south.
Since the boat is moving directly south, its distance east from point P remains constant at 20 miles.
The distance the boat travels south depends on how long it has been moving south. After 2 hours, it will have traveled 2 * 10 = 20 miles to the south.
To find the boat's distance from point P, we can use the Pythagorean theorem:
Distance from P = √(distance east)^2 + (distance south)^2
At this point, the boat's distance from point P is √(20^2 + 20^2) = √(400 + 400) = √800 ≈ 28.28 miles.
To find the rate at which the boat's distance from point P is increasing:
We differentiate the distance equation with respect to time:
d(distance from P)/dt = (1/2)*(2*(distance east)*(rate of change of distance east) + 2*(distance south)*(rate of change of distance south))
Since the boat's distance east is constant at 20 miles, the rate of change of distance east is 0.
Therefore, d(distance from P)/dt = (1/2)*(2*(20)*(0) + 2*(20)*(rate of change of distance south))
= (1/2)*(0 + 2*(20)*(rate of change of distance south))
= 20*(rate of change of distance south)
The rate at which the boat's distance from point P is increasing 2 hours after it leaves point P is directly proportional to the rate of change of its southward distance. Since the boat is traveling south at a constant speed of 10 miles/hr, the rate of change of its southward distance is also 10 miles/hr.
Therefore, d(distance from P)/dt = 20*(10) = 200 miles/hr.
So, the rate at which the boat's distance from point P is increasing 2 hours after it leaves point P is 200 miles/hr.
b) 7 hours after it leaves point P:
In 7 hours, the boat will have traveled 7 * 10 = 70 miles to the east. At this point, it changes direction and starts moving south.
Since the boat is moving directly south, its distance east from point P remains constant at 70 miles.
The distance the boat travels south depends on how long it has been moving south. After 7 hours, it will have traveled 7 * 10 = 70 miles to the south.
Using the Pythagorean theorem, the boat's distance from point P is √(70^2 + 70^2) = √(4900 + 4900) = √9800 ≈ 98.99 miles.
To find the rate at which the boat's distance from point P is increasing:
We differentiate the distance equation with respect to time:
d(distance from P)/dt = (1/2)*(2*(distance east)*(rate of change of distance east) + 2*(distance south)*(rate of change of distance south))
Since the boat's distance east is constant at 70 miles, the rate of change of distance east is 0.
Therefore, d(distance from P)/dt = (1/2)*(2*(70)*(0) + 2*(70)*(rate of change of distance south))
= (1/2)*(0 + 2*(70)*(rate of change of distance south))
= 70*(rate of change of distance south)
The rate at which the boat's distance from point P is increasing 7 hours after it leaves point P is directly proportional to the rate of change of its southward distanc