At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

First draw a right triangle

Let side y be the distance by B in the time after noon this will be 15*7
Let side x be the distance traveled in total by A this will be 30+ (16*7)
so you have y=105 and x=142

Ship B is sailing at 15 knots, this is the rate it is changing, so it is also written as dy/dt=15

Ship A is sailing at 16 knots so this is written as dx/dt=16

you need the distance formula, write is as
h^2=x^2 + y^2
h is the hypotenuse of your triangle, or the distance between the two points

At this point there are two different approaches

APPROACH 1

h^2=x^2 + y^2 is the same as
h=(x^2+ y^2)^(1/2)

Find dh/dt

dh/dt=(1/2)(x^2+ y^2)^(-1/2) (2x dx/dt + 2y dy/dt)
dh/dt=(2x dx/dt + 2y dy/dt)/ [2(x^2+ y^2)^(1/2)]

plug in:
y=105
x=142
dx/dt=16
dy/dt=15

dh/dt=___

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APPROACH 2

h^2=x^2 + y^2

plug in your x and y and solve for h
h=(x^2+ y^2)^(1/2)
h=[(142)^2 + (105)^2]^(1/2)
h=(31189)^(1/2)

using h^2=x^2 + y^2 find dh/dt
2h dh/dt= 2x dx/dt + 2y dy/dt

solve for dh/dt by plugging in:
h=(31189)^(1/2)
y=105
x=142
dx/dt=16
dy/dt=15

Answer 2

33.95230566

Answer 3

The area of a circular sinkhole increases at a rate of 420 square yards per day. How fast is the radius of the sinkhole growing when its radius is 50 yards?

Answer 4

Helicopters.