Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 4 mi away from the station.

Question

A plane flying horizontally at an altitude of 3 mi and a speed of 460 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 4 mi away from the station. (Round your answer to the nearest whole number.)

Answer 1

p'=b'= 460 mph at 3 miles high and 4 miles from the station

construct a right triangle with hypotenuse =h= 4, height = a = 3

3rd side = base = sqr(16-9)=sqr7 = b

use the Pythagorean Theorem

h^2= a^2 +b^2

take the derivative with respect to time,

hh' = aa' +bb', a'=0

hh' = bb'

solve for h' = dh/dt = rate of change of distance to the station

plug in the values for b,b' and h

h'= bb'/h = 460sqr7/4

h' = about 304.2614008 mph

h' = about 304 mph

=

rate of change in distance from the station

Answer 2

Distance at moment t is D = √[3² + (460t)²] = √(9 + 211600t²)

Rate at moment t: dD/dt = 211600t/D; D = 4; 9 + 211600t² = 16; t = √7/211600 = √7/460

dD/dt = 460√7/4 mi/h = 304.26 mi/h