Expert: Scotto - 3/8/2010

Question
QUESTION: A rocket rises vertically from a point on the ground that is 100 m from an observer at ground level. The observer notes that the angle of elevation is increasing at a rate of 12 degrees  per second when the angle of elevation is 60 degrees . Find the speed of the rocket at that instant. Hint: remember to use radians.

ANSWER: Convert to polar coordinates.

Use x² + y² = r² and y/x = tanΘ., where x is a constant of 100.

Note that dΘ/dt = 12°/s.  In calculus, work needs to be done in radians.  12° = π/15 radians.

Once the derivative has been taken, note that an angle of 60° makes the height 100√3
and the hypoteneuse 2*base = 100.

To find the speed of the rocket, note that it is dy.
Since x is a constant, dy is found by taking the derivative of y/x = tanΘ
and getting (1/x)dy = sec²Θ dΘ.

The values of x, Θ, and dΘ are known, so dy can be found.


---------- FOLLOW-UP ----------

QUESTION: im not sure if i understood properly can u plz futher solve the problem so i can get and idea of how to do it

Answer
The x measurement is 100.  The y measurement is unknown.
From both of these, it is known that r² = x² + y².
Putting in x=100, we get r² = 10,000 + y².
Taking the derivative gives us 2r dr = 2y dy.

It is known that the angle Θ has the property that tanΘ = y/x.
Taking the derivative gives sec²(Θ) dΘ = (1/x) dy since x is a constant { namely, 100 }.
This can be used to say that dy = 100*sec²(Θ)*dΘ.

It is known that dΘ = 12°/sec.  Multiply by π/180 to get radians.  Thus, dΘ = 12π/180 = π/15.

We are given that Θ = 60°, and that, converted to radians by *π/180 is Θ = π/3.
It is known that sec(60°) = 1/cos(60°) = 1/(1/2) = 2.

Since we have x = 100, sec(60°) = 2, and dΘ = π/15,
these can all be put in dy = x*sec²(60°)*dΘ to find dy.