Suppose that two boats leave a dock at different times. One heads due north, the other due east. Find the rate at which the distance between the boats is changing when the first boat is 56 miles from the dock traveling at a speed of 35 miles per hour and the second boat is 20 miles from the dock traveling at a speed of 32 miles per hour.
Ahoy there, matey! Looks like we have a classic case of boats in motion. Now, to find the rate at which the distance between the boats is changing, we need to put on our math pirate hats.
Let's call the distance between the first boat and the dock x, and the distance between the second boat and the dock y. To find the rate at which the distance between the boats is changing, we need to find the derivative of the distance between the boats with respect to time.
Using the Pythagorean theorem, we know that the distance between the two boats is given by the equation d^2 = x^2 + y^2.
Now, we can find the derivative of both sides with respect to time (let's say, t) using the chain rule:
2d * dd/dt = 2x * dx/dt + 2y * dy/dt.
In this case, dx/dt represents the rate the first boat is moving, dy/dt represents the rate the second boat is moving, and dd/dt represents the rate the distance between the boats is changing.
Substituting the given values, we have:
2 * 56 * dd/dt = 2 * 35 * dx/dt + 2 * 32 * dy/dt.
Simplifying the equation, we get:
112 * dd/dt = 70 * dx/dt + 64 * dy/dt.
Dividing both sides by 112, we have:
dd/dt = (70/112) * (dx/dt) + (64/112) * (dy/dt).
Now, plugging in the values, we have:
dd/dt = (70/112) * (35) + (64/112) * (32).
Calculating this, we get:
dd/dt ≈ 21.25 + 18.29.
So, the rate at which the distance between the boats is changing is approximately 39.54 miles per hour. Arrr, that's how the boats be movin'!
To find the rate at which the distance between the boats is changing, we can use the concept of related rates. Let's consider the two boats as point A and point B.
Let x be the distance (in miles) that the first boat has traveled north, and let y be the distance (in miles) that the second boat has traveled east.
Using the Pythagorean theorem, we have:
x^2 + y^2 = d^2
Where d is the distance (in miles) between the two boats.
Differentiating both sides of the equation with respect to time t, we obtain:
2x(dx/dt) + 2y(dy/dt) = 2dd/dt
Now, we need to find dx/dt, dy/dt, and dd/dt.
Given that the first boat is traveling north at a speed of 35 miles per hour, we have:
dx/dt = 35 mph
Similarly, the second boat is traveling east at a speed of 32 miles per hour, giving us:
dy/dt = 32 mph
At the given instant, the first boat is 56 miles from the dock (meaning x = 56) and the second boat is 20 miles from the dock (meaning y = 20). Now, we need to find dd/dt.
Substituting the given values into the equation x^2 + y^2 = d^2, we have:
56^2 + 20^2 = d^2
3136 + 400 = d^2
d^2 = 3536
Taking the square root of both sides, we find:
d = √3536 ≈ 59.43 miles
Now, we can substitute dx/dt = 35 mph, dy/dt = 32 mph, and d = 59.43 miles into the equation 2x(dx/dt) + 2y(dy/dt) = 2dd/dt to solve for dd/dt:
2(56)(35) + 2(20)(32) = 2(59.43)(dd/dt)
3920 + 1280 = 118.86(dd/dt)
5200 = 118.86(dd/dt)
Dividing both sides of the equation by 118.86, we find:
dd/dt = 5200 / 118.86 ≈ 43.73 mph
Therefore, the distance between the boats is changing at a rate of approximately 43.73 miles per hour.
To find the rate at which the distance between the boats is changing, we can use the concept of related rates.
Let's assume that the first boat has been traveling for time t1 (in hours), and the second boat has been traveling for time t2 (in hours).
Since the first boat is traveling due north, its position can be represented as x1 = 35t1 (since it is 35 miles per hour), and the second boat, traveling due east, can be represented as x2 = 32t2 (since it is 32 miles per hour).
The distance between the two boats can be calculated using the Pythagorean theorem:
d^2 = x1^2 + x2^2
Taking the derivative of both sides with respect to time (d/dt):
2d * dd/dt = 2x1 * dx1/dt + 2x2 * dx2/dt
Since we are given the values for x1, x2, dx1/dt, and dx2/dt at a particular moment, we can substitute them into the equation and solve for dd/dt, which is the rate at which the distance between the boats is changing.
At the given moment:
x1 = 56 miles, x2 = 20 miles, dx1/dt = 35 miles per hour, dx2/dt = 32 miles per hour
Plugging in these values:
2d * dd/dt = 2(56) * (35) + 2(20) * (32)
Simplifying:
2d * dd/dt = 3920 + 1280
2d * dd/dt = 5200
Dividing both sides by 2d:
dd/dt = 5200 / (2d)
Now, substitute the value of d (the distance between the boats) at the given moment:
d = sqrt(x1^2 + x2^2) = sqrt((56)^2 + (20)^2) = sqrt(3136 + 400) = sqrt(3536) =~ 59.47 miles
dd/dt = 5200 / (2 * 59.47)
dd/dt ≈ 87.17 miles per hour
Therefore, the rate at which the distance between the boats is changing is approximately 87.17 miles per hour.
The distance traveled after t hours is:
If boat N is at (0,y)
boat E is at (x,0),
then the distance z from N to E is
z^2 = x^2+y^2
At the given moment, z=√(56^2+20^2) = √3536 = 4√221
2z dz/dt = 2x dx/dt + 2y dy/dt
Now plug in the numbers (after canceling out all those useless 2's):
4√221 dz/dt = 20*32 + 56*35
dz/dt = 2600 / 4√221 = 43.72 mi/hr