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Rate of Change of Angle

Date: 8/12/96 at 0:31:2
From: da bellm
Subject: Rate of Change of Angle

Dear Dr. Math,

I'm an engineering student at the University of Adelaide in South 
Australia.  This is a problem that I was given about 4 years ago,  
back in high school.  No one I've asked has been able to give me a 
satisfactory answer to it.

Two roads, BA and CA, meet at an angle of 60 degrees. A landmark 
situated at B, 500 metres from A, is visible to drivers approaching 
A along CA. A vehicle at X is moving along CA towards A at 
72 kilometres per hour.
           /  o
         /  60
      A                   X           C

a)  Find the rate at which BX is changing when the vehicle is 500m 
from A. (This bit is easy,  but leads on to part b.)

b)  Find the rate at which angle BXA is changing in radians per second 
when the distance BX is least,  and give a physical explanation of why 
it is whatever it is. (This is the hard bit!)

Have a nice day, 

Date: 8/24/96 at 19:30:46
From: Doctor Mike
Subject: Re: Rate of Change of Angle

Hi  Dave, 
Since you already understand part (a) I'll skip that one. 
Part (b) is a fantastic calculus problem! The straightforward way
is incredibly long and messy, but there's a quick way if you look
at it right.  The key is the Chain Rule for derivatives.
Drop a perpendicular from B to the road CA and let that be the
origin of a coordinate system with +x toward C and +y toward B.
This origin is 250 metres from A and 250*sqrt(3) metres from B.
The only *physical explanation* I see is the observation that
BX is least exactly when the vehicle location X is at the origin.
To keep consistent units, use 20 metres per second rather than 
72kph.  With respect to some arbitrary reference time t=0 let p(t)
be the x-coordinate of the vehicle location on road CA at time 
t seconds.  If the reference time is when the vehicle is M metres
from the origin, then p(t) = M-20t and p'(t) = -20 is constant.
Let d(t) be the distance BX at time t seconds, and a(t) be the
angle BXA at time t seconds.  Then cos(a(t)) = p(t)/d(t).  
Take derivatives of both sides to get : 
                             d(t)*p'(t) - p(t)*d'(t)  
       -sin(a(t)) * a'(t) = ------------------------- 
You want to solve for a'(t) when BX is least, which is when X is
at the origin, a(t) = pi/2, and d'(t) = 0 for a d(t) minimum.
Answer: (2/75)*sqrt(3) or about 0.046188 radians per second.
I hope this is what you were looking for. 

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
High School Calculus

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