“The volume of a cube is increasing at a rate of 10 cm3 per minute. How fast is the surface area increasing  when the length of an edge is 30 cm?”  Two of my students solved this problem in two different ways.

 

The first student used the fact that V = s^3 and A = 6s^2 to write V^(2/3) = A/6.  Taking the derivative implicitly wrt time and using the given information, she determined that 2/3V^(1/3)(dV/dt) = (1/6)dA/dt, so dA/dt = 1200 cm^2/ min.

 

The second student took the derivative of each of the two equations wrt time, resulting in two new equations:   dV/dt = 3s^2(dS/dt) and dA/dt = 12s(dS/dt).  Solving that last equation for dS/dt and substituting back into the first, dV/dt = 3s^2[(dA/dt)/12s] = (s/4)dA/dt.  Thus dA/dt = 10(4/30) = 4/3  cm^3/ minute.

Here is an analysis using function notation.  We can think of three 
relevant sets of ordered pairs, all with the same first coordinates 
(measure of time in minutes).   The first set of ordered pairs 
(function), V,  is a volume function with pairs of the form (t, V(t)).  
The second function, A,  is a surface area function with pairs of the 
form (t, A(t)).  The third function is a edge length function, e with 
pairs of the form (t, e(t)).  Here is the question:  These is a special 
time, m, when e(m) = 30.  Give the numerical value of A'(m). 

What do we  have to work with?  Well, we are told that for all relevant 
values of t, V'(t) = 10 (cm3/min).  We also have the following  
relationships between values of the relevant functions over the same set 
of values of t.  V(t) = (e(t))3 and A(t) = 6*(e(t))2.  Lets use  
implicit differentiation on each of the latter equations.  We have, for 
all relevant values of t, V'(t) = 3(e(t))2*e'(t)  and A'(t)= 
12*(e(t))*e'(t).  Perhaps we can use the first of these equations to get 
a value for  e'(m) and use then use the second to get a value for 
A'(m).  I get 4/3 cm/min for the value of A'(m). 

On the other hand, we have the option of writing A(t) = 6*(V(t))^(2/3).  
If we differeniate this implicitly, we get A'(t) = 4*(V(t))^(-1/3)*V'(t).

If we then go that special time, m, we again get  A'(m) = 4/3 cm/min. 

Perhaps a more explicit use of function notation displays the merit of 
the path on which both of your students set out.  In the first case, the 
problem may have been not being clear about that special time and what 
is true only then. 

Sincerely,

Richard Sisley

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