To determine
To find: The quantities in the given problem.
Answer
The quantities in the given problem is dxdt=500 mi/h.
Explanation
Given:
A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station.
Calculation:
If let t be time (in hours) and x be the horizontal distance traveled by the plane (in mi), then we are given that dxdt=500 mi/h.
To determine
To find: The unknown.
Explanation
The rate at which the distance from the plane to the station is increase when it is 2 mi from the station. Let y be the distance from the plane to the station then find dydt when y=2 mi.
To determine
To draw: The picture of the plane position at time t.
Explanation
Draw the picture of the plane position at time t is given below.
From Figure 1, it is observed that the right angle triangle.
To determine
To write: The equation that relates the quantities.
Answer
The Equation is dydt=xy⋅dxdt.
Explanation
By the Pythagorean Theorem, y2=x2+1.
Here, the horizontal distance x increases as time t increases.
Therefore, the distance y also increases as time t increases.
Hence x and y both are the function of the time t.
Obtain dydt when y=2 mi and the speed of the plane is dxdt=500 mi/h.
Calculate the value of x when y=2 mi using y2=x2+1.
Substitute y=2 in y2=x2+1.
22=x2+14=x2+1x2=4−1x=±3
Since distance cannot be negative, x=3.
Differentiate the function y2=x2+1 with respect to the time t.
ddt(y2)=ddt(x2+1)2ydydt=2xdxdtdydt=xy⋅dxdt
To determine
To find: The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
Answer
The required rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is dydt=2503 mi/h ≈433 mi/h.
Explanation
Substitute 2 for y and 3 for x and 500 for dxdt in dydt.
dydt=32(500)=2505 mi/h ≈433 mi/h
Therefore, the required rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is dydt=2503 mi/h ≈433 mi/h.