1. Nov 3, 2007

### jen333

A and B are walking on straight paths that meet at right angles. A approaches at 2m/sec; B moves away from the intersection at 1m/sec. At what rate is the angle $$\vartheta$$ changing when A is 10m from the intersection and B is 20m from the intersection. Ans in degrees per second.

attempted solving the question using tan ratio where:
tan$$\vartheta$$ (t)= A/B= (10-2t)/(20+t)

I know i have to take the derivative of this equation in order to get d$$\vartheta$$
/dt, but how?

2. Nov 3, 2007

### Dick

Just do it. Differentiate tan(theta(t)) and don't forget the chain rule. The chain rule gives you the dtheta(t)/dt part.

3. Nov 3, 2007

### jen333

thx for you response. ok, so

sec^2(theta) (d(theta)/dt)= (B(dA/dt)-A(db/dt))/B^2

where A=10-2t and B=20+t

if i'm doing this right, then how would i solve for t?

4. Nov 3, 2007

### Dick

The way you've set it up, t=0, yes? All you need is theta and A and B.

5. Nov 4, 2007

### HallsofIvy

Staff Emeritus
t is "when A is 10m from the intersection and B is 20m from the intersection." As Dick sad, since you cleverly used 10-2t and 20+t as the lengths of the sides, A= 10 and B= 20 when t= 0.

6. Nov 6, 2007

### luznyr

hey, thought i would do this question randomly for some exam study, heres how i did it (not sure if its right, hopefully is though, lol).

EDIT: sry stuffed up my working, here new working

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Last edited: Nov 6, 2007