Rates of change (1 Viewer)

[Pace_T]

Hey i am having problems with this particular question can somebody help me?

1. An observer sees a plane directly overhead, 2 km above the ground. The plan is moving horizontally at a constant speed of 500km/h. Find the rate at which the distance will be increasing between the observer and the plan when the distance between them is 5km.

thanks

 

[SaHbEeWaH]

draw up a right-angled triangle

.....___x___
....|...........-''
..2|......-'' y
....|.-'

my best triangle..
let the hypotenuse = y, horizontal = x, vertical = 2

you're trying to find the rate at which y increases when y = 5, i.e. dy/dt
we know that the plane is flying at 500km/h, hence dx/dt = 500
all we need to find now is dx/dy because we know that dx/dt = dx/dy * dy/dt (our desired answer)

to find dx/dy, we try find an equation in x with respect to y then we differentiate it
we know from pythagoras that
y² = 2² + x² (pythagoras's theorem)
therefore x = sqrt(y² - 4)
so dx/dy = 1/2 * (y² - 4)^-1/2 * 2y
= y/(sqrt[y² - 4])

sub in y = 5 into dx/dy:
dx/dy = 5/sqrt(21)

now dx/dy * dy/dt = dx/dt
so 5/sqrt(21) * da/dt = 500
therefore the answer is 100*sqrt(21) kmh^-1 or 458kmh^-1

 

[Pace_T]

oh
silly me, i made a wee little mistake lol
thanks for that
i guess i didnt do too bad for doing all of applications of calculus to the physical world for 2u in one day
LOL!