The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Related Rates: Hair on an Inflating Balloon

Date: 10/16/2003 at 22:53:46
From: Darcy
Subject: Derivative problems

A hair 2pi cm long lies as straight as possible on the surface of a 
spherical balloon while it is being inflated.  The balloon remains 
spherical at all times, and the hair, which doesn’t stretch or shrink, 
remains as straight as possible on its surface.

How is the radius of the balloon changing when it is 4 cm, if the ends
of the hair are moving apart at 1 cm/s at that instant?

At the same instant, how quickly is the midpoint of the hair 
approaching the line between the two ends?

Date: 10/18/2003 at 09:32:30
From: Doctor Luis
Subject: Re: Derivative problems

Hi Darcy,

The hair sits on the surface of the spherical balloon.  This means 
that it forms an arc for a circle of radius R(t), where R(t) is the 
radius of the sphere at time t.


You can imagine that as the balloon grows larger and larger, the 
curved hair will get straighter and straighter, and therefore the ends 
will grow further apart since they were closer together when they were 


The important thing to realize is that the hair is essentially an arc 
of length L on a circle of radius R.  This arc subtends an angle T on 
this circle, where L = R*T (formula for length of an arc).


You can see from that last diagram that the distance between the ends 
is e = 2R*sin(T/2) and the distance from the midpoint of the hair to 
the line between the ends is m = R - R*cos(T/2). 

The first question asks us to find dR/dt from de/dt = +1 cm/sec, and
R=4 cm.  The second question asks about dm/dt.

To solve the first question, we use the two equations L=R*T, 
e = 2R*sin(T/2), keeping in mind that L is constant and that both R 
and T change with time (you'll end up with dR/dt and dT/dt terms).

We can solve for dT/dt using the first equation by taking the 
derivative with respect to t:

  T = L/R  -> dT/dt = -(L/R^2)dR/dt   (rate of change of angle)

With the second equation, we can find the relationship between de/dt 
and dR/dt by taking that derivative:

      e = 2R*sin(T/2)
  de/dt = 2(dR/dt)sin(T/2) + 2R*cos(T/2) * (1/2)dT/dt
        = 2(dR/dt)sin(T/2) +  R*cos(T/2)*(-(L/R^2)dR/dt)
        = (dR/dt)*(2sin(T/2) - cos(T/2)*L/R)

Solving for dR/dt, we get

  dR/dt = (de/dt) / (2sin(T/2) - cos(T/2)*L/R)

we evaluate this at R=4 cm.  Remember that L=2pi cm, the angle T = L/R 
= (2pi cm)/(4cm) = pi/2 rad, and de/dt = +1 cm/sec.

See if you can find dm/dt by yourself.  Start by differentiating the
equation m = R - R*cos(T/2) with respect to time.

Let us know if you have any more questions.

- Doctor Luis, The Math Forum 

Date: 10/25/2003 at 00:08:25
From: Darcy
Subject: Derivative problems

Wow, thanks a lot for such an insightful answer.  The pictures really
helped me to see how to set it up.
Associated Topics:
College Calculus
High School Calculus

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.