QUESTION 3
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
2d dd/dt = 200(t-3)
at t=7, d=50
2*50 dd/dt = 200(4)
dd/dt = 8 mph
i'm still not understand for this conclution
*conclusion not conclution
where do you get 200(t-3)???
To find the rate at which the boat's distance from point P is increasing, we can use the concept of derivative.
Let's define the position of the boat as a function of time, where the x-axis represents the east-west direction (positive to the east) and the y-axis represents the north-south direction (positive to the south).
Let t be the time elapsed since the boat leaves point P, and let d(t) be the distance between the boat and point P at time t.
Step 1: Determine the boat's position as a function of time.
Since the boat sails 30 miles to the east from point P and then changes direction to sail south, we can represent the boat's position using two separate functions:
At the initial stage (0 <= t < 2), the boat is moving east.
x(t) = 30t (since the boat is moving at a constant speed of 10 miles/hr to the east, the x-coordinate increases at a rate of 10 miles/hr.)
y(t) = 0 (since the boat is not moving in the north-south direction during this time.)
After changing direction at t = 2, the boat starts moving south.
x(t) = 30*2 = 60 (The boat's x-coordinate remains constant after changing direction.)
y(t) = 0 - 10(t - 2) = 20 - 10t (The boat's y-coordinate decreases linearly at a rate of 10 miles/hr.)
Step 2: Find the distance between the boat and point P as a function of time.
The distance between the boat and point P at time t can be calculated using the distance formula:
d(t) = √((x(t) - x(P))^2 + (y(t) - y(P))^2)
At point P (initial position), x(P) = 0 and y(P) = 0.
So, the distance function d(t) becomes:
d(t) = √((x(t))^2 + (y(t))^2)
Step 3: Differentiate the distance function with respect to time t to find the rate of change of distance with respect to t.
d'(t) = (d(t))' = (d(t))^2
Step 4: Calculate the rate at which the distance from point P is increasing at the given times.
a) 2 hours after leaving point P (2 < t < 7):
Since the boat has already traveled 30 miles to the east and started moving south, the distance from point P is constantly decreasing at a rate of 10 miles/hr.
Therefore, the rate at which the distance from point P is increasing 2 hours after leaving point P is 0 miles/hr.
b) 7 hours after leaving point P (7 < t):
After 7 hours, the boat is still moving south at a constant rate of 10 miles/hr, parallel to the y-axis. The x-coordinate of the boat remains constant at 60 miles.
Therefore, the rate at which the distance from point P is increasing 7 hours after leaving point P is also 10 miles/hr.