Related Rates Calculus Question?

Question

Ship B sailing south at 16km/h is northwest (N45W) of a second ship A sailing east at 12 km/h. At what rate are the two ships approaching each other or seperating at that instant?

What i did was draw the diagram but after that i did not know what to do? should i flip the 12 to make it west?
dB/dt = 16km/h [S]
dA/dt = 12km/h [E]

noo this is exactly what it says on the work sheet provided...but im guessing since its 45 degrees...there might be some pythagroem therom involved

i just relalized that the diagonal space in between would be 20...because of the pythagreom triple....12, 16, 20

Answer 1

Draw the coordinate system: the first ship is on the y-axis (travels south) and the second ship on the x-axis (travels east). At the time of interest, the angle of the line drawn between the ships forms a 45º angle with the x and y axes. (Southbound ship is 45º west of north from the eastbound ship.)

You will see that the ship positions form a right triangle with the origin. Let the position of the ships by x and y; the distance between them is then

s = √[x^2 + y^2]

Take the derivative of this with respect to t:

ds/dt = 0.5/√[x^2 + y^2] * (2x*dx/dt + 2y*dy/dt); however, in a right triangle with corner angles = 45º, x = y so substitute x for y:

ds/dt = 0.5/√[2x^2] * 2x*(dx/dt + dy/dt)

ds/dt = 0.5/x√2 * 2x*(dx/dt + dy/dt)

ds/dt = 1/√2 * (dx/dt + dy/dt)

You are given that dy/dt = -16km/h and dx/dt = 12km/h (south is negative direction, east is positive). Therefore

ds/dt = 1/√2 (12 - 16) = -4/√2 km/h = -2√2 km/h

The negative sign means that the distance is reducing.

Note: regarding your added comment, the pythagorean relation applies to the distances of the ships from the origin, not the velocities. To get the velocity relation, you have to take the derivative as I did above.

Answer 2

You have an isosceles right triangle. Ships A and B are at opposite ends of the hypotenuse. Ship B is sailing toward the right angle and ship A is sailing away from it.

Let
h = distance between ships A and B
x = distance of ship A from right angle
x = distance of ship B from right angle (they will be the same)

Given

dA/dt = 12 km/h
dB/dt = -16 km/h (since distance is decreasing)

h² = x² + x² = 2x²
h = x√2

Differentiate implicitly.

h² = x² + x²

2h(dh/dt) = 2x(dA/dt) + 2x(dB/dt)
h(dh/dt) = x(dA/dt) + x(dB/dt)
(x√2)(dh/dt) = x(12) + x(-16) = -4x
√2(dh/dt) = -4
dh/dt = -4/√2 = -2√2 km/hour

Answer 3

Does it give you info on how far apart they are?
dB/dt = - 16 because it is approaching the other ship.
dA/dt = +12