Ship A is traveling west at 40kkm/h and ship B is traveling north at 30mk/h. Both are headed for the Los Angeles Harbor.
(a) At what rate are the boats approaching each other when ship A is 3km and ship B is 4km from the dock?
(b) At what rate would the distance between the boats be increasing if they started at the dock with the same speeds and after ship A has gone 5km and ship B has gone 12km?
Thanks in advance for the help!!
(a) at the moment specified, A and B are 5km apart
If ship A is a km from port and
ship B is b km from port,
the distance between them is given by
d^2 = a^2 + b^2
2d dd/dt = 2a da/dt + 2b db/dt
2(5) dd/dt = 2(3)(-40) + 2(4)(-30)
10 dd/dt = -240 + -240
dd/dt = -48
(b) Same formula, only now distances are increasing.
distance at time given is 13
2(13)dd/dt = 2(5)(40) + 2(12)(30)
26 dd/dt = 400 + 480
dd/dt = 33.8
To solve this problem, we can use the concept of related rates. Related rates is a method of finding how the rate of change of one quantity is related to the rate of change of another quantity.
(a) Let's start by labeling the distance between ship A and the dock as x and the distance between ship B and the dock as y. We are given that ship A is traveling west at 40 km/h, and ship B is traveling north at 30 km/h.
We want to find the rate at which the boats are approaching each other, which is the rate of change of the distance between them. Let's call this rate of change dz/dt.
Using the Pythagorean theorem, the relationship between x, y, and z is given by:
x^2 + y^2 = z^2
Differentiating both sides of this equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
We are given that x = 3 km, y = 4 km, dx/dt = -40 km/h (since the ship is traveling west), and dy/dt = 30 km/h. We need to find dz/dt, which is the rate at which the distance between the boats is changing when x = 3 km and y = 4 km.
Plugging in these values, we have:
2(3)(-40) + 2(4)(30) = 2(5)(dz/dt)
-240 - 240 = 10(dz/dt)
-480 = 10(dz/dt)
dz/dt = -48 km/h
Therefore, when ship A is 3 km and ship B is 4 km from the dock, the boats are approaching each other at a rate of 48 km/h.
(b) Similarly, we can solve for the rate at which the distance between the boats is increasing when ship A has gone 5 km and ship B has gone 12 km.
Let's label the distance traveled by ship A as x1 and the distance traveled by ship B as y1.
We know that dx1/dt = -40 km/h and dy1/dt = 30 km/h.
Using the Pythagorean theorem again, the relationship between x1, y1, and the distance z1 between the boats is given by:
x1^2 + y1^2 = z1^2
Differentiating both sides of this equation with respect to time t, we get:
2x1(dx1/dt) + 2y1(dy1/dt) = 2z1(dz1/dt)
We want to find the rate at which the distance between the boats is increasing, dz1/dt, when x1 = 5 km and y1 = 12 km.
Plugging in these values, we have:
2(5)(-40) + 2(12)(30) = 2(z1)(dz1/dt)
-400 + 720 = 2(z1)(dz1/dt)
320 = 2(z1)(dz1/dt)
dz1/dt = 320 / (2z1)
To find z1, we can use the Pythagorean theorem again:
z1^2 = x1^2 + y1^2
z1^2 = 5^2 + 12^2
z1^2 = 25 + 144
z1^2 = 169
z1 = √169
z1 = 13 km
Plugging in z1 = 13 km, we have:
dz1/dt = 320 / (2 * 13)
dz1/dt = 320 / 26
dz1/dt = 12.31 km/h
Therefore, when ship A has gone 5 km and ship B has gone 12 km, the distance between the boats is increasing at a rate of approximately 12.31 km/h.