Answer: 438.3 mph
Explanation:
Let z be the distance from the plane to the station. You should draw a right triangle
for the diagram with z on the hypotenuse, 1 on the vertical side, and x on the horizontal
side. The 1 never changes, but x changes with time.
Find the attached file for the solution.
Answer:
The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 424 mi/h
Explanation:
Given;
vertical position of the plane, h = 1 mi
when plane is 2 mi away from the station, this position and vertical position forms a right - angled triangle.
Let the vertical position = y = 1 mi
Let the 2 mi position = hypotenuse = p
Let the remaining side of the triangle, which is horizontal = x
x² = p² - y²
[tex]x = \sqrt{p^2 -y^2} = \sqrt{2^2 -1^2} = \sqrt{3}[/tex]
Again;
p² = x² + y²
the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station, can be determined by differentiating P with respect to time t.
[tex]p^2.\frac{dp}{dt} = x^2.\frac{dx}{dt} +y^2.\frac{dy}{dt} \\\\2p.\frac{dp}{dt} = 2x.\frac{dx}{dt} + 2y.\frac{dy}{dt}\\\\2p.\frac{dp}{dt} = 2x.\frac{dx}{dt} + 0\\\\\frac{dp}{dt} = \frac{2x}{2p} .\frac{dx}{dt} \\\\\frac{dp}{dt} = \frac{x}{p} .\frac{dx}{dt} = \frac{\sqrt{3}}{2} . 490(\frac{mi}{h}) = 424 \ mi/h[/tex]
Therefore, the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 424 mi/h
