An airplane is flying on a flight path that will take it directly over a radar tracking station. The airplane is flying at an altitude of 6miles and s miles from the station. If s is decreasing at a rate of 400miles/hour when s=10miles. What is the speed of the Plane
To solve this problem, we can use the concept of related rates. Let's let the function S represent the distance between the airplane and the radar tracking station.
We are given that ds/dt = -400 miles/hour, which means that the distance S is decreasing at a rate of 400 miles/hour. We also know that when s = 10 miles, the distance S is equal to 6 miles.
Using the Pythagorean theorem, we can relate the variables s and S as follows:
S^2 = s^2 + 6^2
S^2 = s^2 + 36
Differentiating both sides with respect to time t, we get:
2S(dS/dt) = 2s(ds/dt)
Since we are interested in finding the speed of the airplane, we can substitute the given values into the equation:
2(6)(dS/dt) = 2(10)(-400)
12(dS/dt) = -8000
dS/dt = -8000/12
dS/dt = -666.67 miles/hour
Therefore, the speed of the airplane is approximately 666.67 miles per hour.
To find the speed of the plane, we need to apply the concept of related rates.
Let's denote the speed of the airplane as v miles/hour.
Given that the altitude of the airplane is 6 miles, we can form a right triangle between the airplane, the radar tracking station, and a point on the ground directly underneath the airplane.
Let's denote the distance from the airplane to the radar tracking station as s miles.
Using the Pythagorean theorem, we can write the equation: s^2 + 6^2 = v^2.
Now, we need to find the rate at which s is decreasing. Given that s is decreasing at a rate of 400 miles/hour when s is 10 miles, we can write ds/dt = -400 miles/hour.
To find the rate at which v is changing, we need to differentiate the equation s^2 + 6^2 = v^2 with respect to time t.
We have 2s(ds/dt) = 2v(dv/dt), where ds/dt represents the rate at which s is changing, and dv/dt represents the rate at which v is changing.
Plugging in the given values, we have 2(10)(-400) = 2v(dv/dt).
Simplifying the equation, we get -8000 = 2v(dv/dt).
Dividing both sides by 2v, we have -4000/v = dv/dt.
Now we need to find v when s = 10 miles. Plugging in the values, we have s^2 + 6^2 = v^2, which becomes 10^2 + 6^2 = v^2, or 100 + 36 = v^2.
Simplifying, we have v^2 = 136.
Taking the square root of both sides, we have v ≈ √136 ≈ 11.66 miles/hour.
Therefore, the speed of the plane is approximately 11.66 miles/hour.