Question

A helicopter flies parallel to the ground at an altitude of 1/2 kilometer and at a speed of 2 kilometers per minute. If the helicopter passes directly over the White House, at what rate is the distance between the helicopter and the White House changing 3 minutes after the helicopter flies over the White House?

Answer 1

Answer: Hello there!

If we define the x-axis as the one parallel to the ground and y as the perpendicular ( or the height in this case) then you know that the y position of the helicopter is equal to 0.5 kilometers, and is constant.

And the x component is x = (2km/min)*t where t is in minutes, and we could define t = 0 when the helicopter is over the white house.

Also, in this case, we can define the position of the white house as y = 0 and x = 0

Then our position vector (x,y) for the helicopter is ( 2*t, 0.5) and the position of the white house is (0,0)

then the distance between the helicopter and the white house is:

$d(t) =I(2*t,0.5) - (0,0)I = I(2*t, 0.5)I= \sqrt{ ((2*t)^2 +0.5^2)}$

where the distance is calculated as the magnitude of the difference in the position between the two objects

Now we want to know the rate of change in the distance when t = 3.

This is equivalent to calculate the derivative of d(t) valuated in t = 3 minutes, this is d'(3)

then we need to derivate our equation:

$d'(t) = \frac{dd(t)}{dt} = \frac{d(\sqrt{ ((2*t)^2 +0.5^2)})}{dt} = 0.5*((2*t)^2 +0.5^2)^{-1/2} *4*t$

now we can replace t  by 3, and see the value of the rate of change 3 minutes after the helicopter was directli over the white house:

$d'(3) = 0.5*((2*3)^2 +0.5^2)^{-1/2} *4*3 = 0.99$

The rate of change of the distance is 0.99 kilometers per minute square.