Expert: Paul Klarreich - 6/14/2008

Question
a sphere is being filled at a constant rate of 4cm^3/sec. At what rate is the surface area changing when the volume is 30 cm^3.

Answer
Questioner:   inthujah
Category:  Calculus
Private:  No
 
Subject:  calculus related rates... grade 12 calculus
Question:  a sphere is being filled at a constant rate of 4cm^3/sec. At what rate is the surface area changing when the volume is 30 cm^3.
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Hi, Inthujah,

Is this your first attempt at R-R problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.

3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is the key step.

4. Now differentiate implicitly, then substitute the known quantities and rates, and solve for the unknown rate.

AND, Please check the archives for other Related Rates examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.
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Variables:

r = radius of sphere
A = area of sphere.
V = volume of sphere.

Rates:

dr/dt not known.
dV/dt = 4 cm^3/sec
dA/dt TO BE FOUND.

Relations:

A = 4 pi r^2
V = 4 pi r^3/3

Since you have dV/dt and need dA/dt, you should express A in terms of V: (i.e. eliminate r)

A = 4 pi r^2

V = 4 pi r^3/3

3V/4pi = r^3

r = (3V/4pi)^1/3

A = 4 pi (3V/4pi)^2/3

A = 4 pi (3/4pi)^2/3 V^2/3

   4 pi 3^2/3 V^2/3
A = -----------------
      (4 pi)^2/3

A = (4 pi)^1/3  3^2/3 V^2/3

A = (4 pi)^1/3  9^1/3 V^2/3

A = (36 pi)^1/3  V^2/3

Whew!

Now Differentiate:

dA/dt = (36 pi)^1/3 (2/3) V^-1/3 dV/dt

        (36 pi)^1/3 (2/3) dV/dt
dA/dt = -------------------------
                V^1/3

Use V = 30, dV/dt = 4

        (36 pi)^1/3 (2/3) (4)
dA/dt = -------------------------
                (30)^1/3


Simplify a bit:

        (6 pi)^1/3 (8/3)
dA/dt = ------------------
            (5)^1/3

I think that's about it. [I hope I didn't blow too many exponents.]