An airplane is flying at an altitude of 6 miles and passes directly over a radar antenna. When the plane is 10 miles away, the radar detects that the distance is changing at a rate of 240 mph. what is the speed of the plane? This is an implicit differentiation problem
To solve this implicit differentiation problem, we need to use the relationship between the distance, speed, and rate of change of distance.
Let's assume that the speed of the airplane is represented by 'v' (in mph), and the distance of the airplane from the radar antenna is represented by 'x' (in miles).
From the problem, we know that when the plane is 10 miles away, the radar detects that the distance is changing at a rate of 240 mph. This rate of change is given by dx/dt = 240 mph.
Now, we need to find the relationship between x, v, and dx/dt. Since the radar is directly above the antenna, we can form a right-angled triangle with the vertical distance (altitude) of 6 miles, horizontal distance x, and the hypotenuse (distance from radar to plane) c. Using the Pythagorean theorem, we have:
x^2 + 6^2 = c^2
x^2 + 36 = c^2 ------ (1)
We also know that the rate of change of distance (dx/dt) is related to the rate of change of c, which is the same as the speed of the airplane (v). So, we can derive the relationship between x, c, and v. Differentiating equation (1) with respect to time t:
2x(dx/dt) = 2c(dc/dt)
2x(dx/dt) = 2c(v)
Substituting the given values, we have:
2(10)(240) = 2c(v)
4800 = 2cv
cv = 2400
Therefore, the speed of the plane (v) is 2400 mph.