by **Ahmed** on Mon Dec 20, 2010 12:03 am

Draw a right triangle. The base of the triangle should be along the ground from where the plane is to where the observer is. The hypotenuse should extend from the observer to the plane and the other side is just vertical and it has a length of 5 miles. Theta is the angle beside the observer formed by the hypotenuse and the base.

db/dt=-600 , b stands for base, t stands for time so the base is changing at a rate of 600 mi/h or in other words the derivative of the base with respect to time is 600. Note that I wrote db/dt=-600 because the length of the base is decreasing as the plane gets closer to the observer so the rate is negative.

We are asked to find dθ/dt and we are given db/dt so we need some equation that relates θ and b.

If you look at your triangle you'll notice tan(θ)=5/b

by implicit differentiation(with respect to time) we get, sec²(θ)dθ/dt=-5b^-2db/dt

Now you simply plug in the numbers given and solve

sec²(30°)dθ/dt=-5(5√3)^-2*-600

dθ/dt=30 degrees per hour or if you use π/6 instead of 30° you get the answer 30 radians per hour

If you're wondering where the 5√3 came from remember that tan(θ)=5/b so b=5/tan(θ)=5/tan(30°)=5/(1/√3)=5√3