Differential Calculus Problem, A bombardier is sighting on a street....

Question

A bombarider is sighting on a street on the ground directly ahead. If the bomber is flying 2 miles above the ground at 240 mi/hr, how fast must the sighting instrument be turning when the angle between the path of the bomber and the line sight is 30 degrees?

Ans. 3/2pi degrees/sec

Answer 1

Hi Ren,

If we picture the bomber in the upper left hand corner of our picture, and the ground target in the lower right hand corner, it forms a right triangle. Then we can define:

x = the number of horizontal miles that the bomber is from the target

y = the height of the bomber (2 miles)

θ = the angle of the sighting instrument (initially 30 degrees)

t = time in seconds

As the bomber flies to the right, x increases and we can picture the sighting instrument rotating downward in a clockwise direction.

From basic geometry we know that tan(θ) = y / x = 2 / x.

At the moment, since θ = 30 degrees, then tan(30) = 2 / x so x = 2 / tan(30) = 2 * sqrt(3) miles.

Taking the derivative of each side with respect to t and using implicit differentiation gives us

sec²(θ) * dθ/dt = - 2 / x² * dx/dt

We can plug in what we have for θ and x.

We can also plug in dx/dt = 240 mi/hr, noting that we have to convert to mi/sec, and that the result should be negative since the distance between the bomber and the target is shrinking:

dx/dt = -240 mi/hr * 1 hr/ 3600 sec = -1/15 mi/sec

So we have sec²(30) * dθ/dt = - 2 / (2 * sqrt(3))² * -1/15

(2 / sqrt(3))² * dθ/dt = - 2 / (2 * sqrt(3))² * -1/15

4/3 * dθ/dt = - 2/12 * -1/15

dθ/dt = 1/120 degrees/sec

Although this doesn’t match the answer provided, the answer provided also seems unusual in that the units are stated in degrees, but involve pi as if it were radians.

Please let me know if I can be more helpful, thanks!

Joe