# Related Rate -finding the Rate of Change of an Angle

1. Oct 6, 2005

### jaggtagg7

here is the problem i was trying to do:

A baseball player stands 2 feet from home plate and watches a pitch fly by. Find the rate D(theta)/dt at which his eyes must move to wach a fastball with dx/dt=-130 ft/s as it crosses homeplate at x=0.

now there is a nice diagram of a right trianlge with x labled as the distance from the ball from the plate and theta as the angle from the player's eyes to the ball.

where i'm confused is how exactly i relate these. not sure what trig function to use, or then how to solve it.

also what would be some general rules to follow when solving any related rate problem involving an angle?

PS: this is not a hw problem, but rather on i was trying to solve for fun, so i'm not really interested in the answer but more of how you solve it.

2. Oct 6, 2005

### amcavoy

Please scan in the diagram if you can.

I'd have to look at it, but generally what you do is this:

Set up θ in terms of x. Then you can differentiate and notice that dθ/dx = (dθ/dt) / (dx/dt), which you can solve for dθ/dt.

Alex

Last edited: Oct 6, 2005
3. Oct 6, 2005

4. Oct 6, 2005

### amcavoy

Well if you know geometry, you will notice that the angle can be given as:

$$\tan{\theta}=\frac{x}{2}\implies\theta=\arctan{\frac{x}{2}}$$

Use the info I posted above when taking this derivative (I assume you know the derivative of arctan(x)).

Alex

5. Oct 6, 2005

### jaggtagg7

ok, gotcha thus far. but now what i am unsure of is how you would implicity take the derivative for arctan(x/2). i can't recall how u would treat the (1/2)x in the derivative.

something like this?

dθ/dt = [1/(1+ x^2/4)] dx/dt

that cant be right, because when i substituted, i ened up with dθ/dt= 1/1 *-130
...... :/