I found the equation which was the pythagoras theorem and then took the derivative of the equation but not sure what to plug in where. Oh and I found C(t) to be 743.30
An airplane passes over an airport at noon traveling 500 mi/hr due
West. At 1:00 pm, another airplane passes over the same airport at the same
elevation traveling due North at 550 mi/hr. Assuming both planes maintain their
(equal) elevations, how fast is the distance between them changing at 2:30 pm?
To solve this problem, we can use the Pythagorean theorem to find the distance between the two airplanes at any given time. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Let's denote the distance between the airplanes as "d" at time t. According to the problem, the first airplane travels west at a constant speed of 500 mi/hr, and the second airplane travels north at a constant speed of 550 mi/hr.
At time t, the first airplane would have traveled for t hours. Therefore, the distance it covers in the west direction is 500t miles. Similarly, the second airplane would have traveled for (t - 1) hours since it passed over the airport at 1:00 pm. Hence, the distance it covers in the north direction is 550(t - 1) miles.
Using the Pythagorean theorem, we get:
d^2 = (500t)^2 + (550(t - 1))^2
Now, we can use implicit differentiation to find how the distance, d, changes with respect to time, t. Differentiating both sides of the equation with respect to t gives:
2d * dd/dt = 2(500t) * 500 + 2(550(t - 1)) * 550 * (dt/dt)
The term dt/dt is equal to 1, so simplifying further, we have:
2d * dd/dt = 1000t * 500 + 1100(t - 1) * 550
Substituting t = 2.5 (since we need to find the rate of change at 2:30 pm), we can plug in the values and solve for dd/dt:
2(743.30) * dd/dt = 1000(2.5) * 500 + 1100(2.5 - 1) * 550
Simplifying the equation, we can solve for dd/dt, which represents the rate at which the distance between the planes is changing at 2:30 pm.