An airplane flying west at 300 miles per hour goes over the control tower at noon, and a second airplane at the same altitude, flying north at 400 miles per hour, goes over the tower an hour later. How fast is the distance between the airplanes changing at 2:00 P.M.?
Draw a diagram
Let the tower be at (0,0)
plane A is at distance a = 300t West after t hours
Plane B is at position b = 400(t-1) North after t hours, counting from 12:00
The distance d between the planes is given by
d^2 = (300t)^2 + (400(t-1))^2
2d dd/dt = 2(300t)(300) + 2(400(t-1))(400)
at 2:00, t=2, so
d = 200√13
400√13 dd/dt = 360000 + 320000 = 680000
dd/dt = 6800/4√13 = 1700/√13 = 471.5
feel free to check my math...
Well, let's take a look at this mathematically, but don't worry, I'll try to keep it fun! We can break down the problem into two different dimensions, the horizontal and vertical dimensions.
First, let's tackle the horizontal dimension. The westward airplane is moving at a constant speed of 300 miles per hour. So, in the two hours that have passed between noon and 2:00 P.M., it would have traveled a distance of 600 miles. Easy peasy!
Now, let's move on to the vertical dimension. The northward airplane is moving at a constant speed of 400 miles per hour. Since it flew over the control tower at 1:00 P.M., it has been in the air for a total of just one hour. Therefore, it has traveled a distance of 400 miles.
Now, to find the rate at which the distance between the airplanes is changing at 2:00 P.M., we need to use some trigonometry! We have a right triangle formed by the horizontal and vertical distances traveled by the airplanes.
The horizontal distance (600 miles) is the base of the triangle, the vertical distance (400 miles) is the height, and the distance between the airplanes is the hypotenuse. So, we can use the Pythagorean theorem (a² + b² = c²) to find the distance between the airplanes.
Let's call the distance between the airplanes "d". Using the Pythagorean theorem, we have:
d² = 600² + 400²
d² = 360,000 + 160,000
d² = 520,000
Taking the square root of both sides, we find:
d ≈ 721.11 miles
Now that we know the distance between the airplanes, we can differentiate it with respect to time to find how fast it is changing. However, since the speeds of the planes are constant, the distance between them is not changing. So the answer is 0 miles per hour! They're maintaining that distance in a perfectly coordinated dance through the sky.
Remember, it's all fun and games until someone starts accidentally playing bumper planes!
To find the rate at which the distance between the airplanes is changing, we need to use the concept of rate of change, specifically the derivative. Let's denote the distance between the airplanes at time t as D(t).
We are given that the first airplane is flying west at 300 miles per hour. This means that the horizontal distance covered by the first airplane from noon to t hours later is 300t miles.
Similarly, the second airplane is flying north at 400 miles per hour. This means that the vertical distance covered by the second airplane from 1:00 P.M. to t hours later is 400(t - 1) miles.
Using the Pythagorean theorem, we can express the distance D(t) between the two airplanes as follows:
D(t) = sqrt((300t)^2 + (400(t - 1))^2)
Now, we need to find the rate of change of D(t) with respect to time, or dD/dt.
To do this, we can differentiate D(t) with respect to t:
dD/dt = (1/2) * (2 * (300t) * 300 + 2 * (400(t - 1)) * 400)
Simplifying the above expression, we get:
dD/dt = (1/2) * (600t + 800(t - 1)) = 700t - 400
To find the rate of change at 2:00 P.M., we substitute t = 2 into the equation:
dD/dt = 700(2) - 400 = 1400 - 400 = 1000
Therefore, at 2:00 P.M., the distance between the two airplanes is changing at a rate of 1000 miles per hour.
To determine how fast the distance between the airplanes is changing at 2:00 P.M., we need to find the rate of change of distance with respect to time. This can be done by applying the Pythagorean theorem, which relates the distance between the two airplanes to the velocities of each airplane.
Let's break down the problem step by step:
1. Find the position function for each airplane:
- The airplane flying west has a velocity of 300 miles per hour, so its position function can be expressed as: x(t) = -300t.
(Remember, the negative sign indicates westward direction.)
- The airplane flying north has a velocity of 400 miles per hour, so its position function can be expressed as: y(t) = 400(t - 1).
(We subtract 1 hour, as it takes off an hour later than the first airplane.)
2. Determine the distance between the two airplanes as a function of time:
The distance between the two airplanes can be calculated using the Pythagorean theorem, which states that for any right angle triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In this case, the position functions x(t) and y(t) act as the sides of the triangle.
The distance between the two airplanes, D(t), at time t is given by the formula:
D(t) = sqrt[x(t)^2 + y(t)^2]
= sqrt[(-300t)^2 + (400(t - 1))^2]
3. Compute the derivative of the distance function:
To find how fast the distance between the airplanes is changing, we calculate the derivative of D(t) with respect to t. This will give us the rate of change of distance with respect to time.
dD(t)/dt = (dD(t)/dx) * (dx/dt) + (dD(t)/dy) * (dy/dt)
(Note: dD(t)/dx represents the partial derivative of D(t) with respect to x, and dy/dt represents the velocity of the second airplane.)
4. Substitute the expressions for x(t), y(t), dx/dt, and dy/dt into the equation.
dD(t)/dt = (2x(t)(-300) + 2y(t)(400(t - 1))) / (2sqrt[x(t)^2 + y(t)^2])
5. Evaluate the derivative at t = 2 (2:00 P.M.):
Plug in t = 2 into the expression obtained in the previous step to calculate the rate of change of distance at 2:00 P.M.