An airplane flying 390 feet per second at an altitude of 5,000 feet flew directl

Question

An airplane flying 390 feet per second at an altitude of 5,000 feet flew directly over an observer. The figure shows the relationship of the airplane to the observer at a later time. (a) The equation is . (b) Find the value of x when y is 12,000. x = (Round to the nearest integer as needed.) (c) How fast is the distance from the observer to the airplane changing at the time when the airplane is 12,000 feet from the observer? That is, what is dy/dt at the time when dx/dt = 390 and y= 12,000? dy/dt = (Round to the nearest integer as needed.)

Explanation / Answer

dx/dt = 390 feet / second

Altitude = 5000

a) By pythagorean theorem :

x^2 + 5000^2 = y^2

y^2 = x^2 + 25000000

y = sqrt(x^2 + 25000000) ---> ANSWER

b)

x^2 + 5000^2 = y^2

x^2 + 5000^2 = 12000^2

x^2 = 12000^2 - 5000^2

x^2 = 119000000

x = 10909 ----> ANSWER

c)

We have x^2 + 5000^2 = y^2

Lets derive this implicitly with respect to t :

2x * dx/dt + 0 = 2y * dy/dt

2(10908.71211463571441) * 390 = 2*12000*dy/dt

dy/dt = 354.5331437256607183 --> exact value

dy/dt = 355 ----> ANSWER

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