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.)
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
33.95230566
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?
Helicopters.
To find the rate at which the distance between the ships is changing at 7 PM, we can use the concept of related rates from calculus.
Let's break down the problem and establish some notation:
- The position of ship A at any given time t will be denoted as (x, y), where x is the distance traveled west and y is the distance traveled north.
- The position of ship B at any given time t will be denoted as (0, y'), where y' is the distance traveled north.
- The distance between the ships at any given time t will be denoted as D.
We know that at noon (t = 0), ship A is 30 nautical miles due west of ship B. Therefore, we have:
D^2 = (x - 0)^2 + (y - y')^2 [Equation 1]
Now, let's differentiate Equation 1 with respect to time t:
2D * (dD/dt) = 2(x - 0)*(dx/dt) + 2(y - y')*(dy/dt)
Simplifying:
dD/dt = [(x - 0)*(dx/dt) + (y - y')*(dy/dt)] / D
We need to find dD/dt at 7 PM (t = 7 hours).
Given that ship A is sailing west at 16 knots (1 knot = 1 nautical mile per hour), dx/dt = -16 knots.
Also, ship B is sailing north at 15 knots, dy/dt = 15 knots.
To find x and y at 7 PM, we need to determine their values at noon (t = 0) and consider the time elapsed.
Since ship A is sailing west at a constant rate of 16 knots, x = 16t.
Since ship B is sailing north at a constant rate of 15 knots, y' = 15t.
At noon (t = 0), ship A is 30 nautical miles due west of ship B. So when t = 0, x = -30 nautical miles.
Therefore, at 7 PM (t = 7 hours), we have:
x = 16t = 16 * 7 = 112 nautical miles
y' = 15t = 15 * 7 = 105 nautical miles
Now, substitute these values along with dx/dt and dy/dt into the expression for dD/dt:
dD/dt = [(x - 0)*(dx/dt) + (y - y')*(dy/dt)] / D
= [(112 - 0)*(-16) + (y - 105)*(15)] / D
Now, we need to find D at 7 PM. Using Pythagoras' theorem and the distances traveled by each ship, we have:
D^2 = (x - 0)^2 + (y - y')^2
= (112)^2 + (y - 105)^2
Finally, calculate the value of dD/dt at 7 PM:
dD/dt = [(112 - 0)*(-16) + (y - 105)*(15)] / D
= [(-16)*(112) + (y - 105)*(15)] / sqrt((112)^2 + (y - 105)^2)
This expression will give you the rate at which the distance between the ships is changing at 7 PM. Simplifying the arithmetic will provide the final answer in knots.