Hope:  God, no.

Answer:  Why yes, yes it does.

In fact, here's a flux problem involving calculus and related rates.  Let's try it, shall we?

Problem:

A circular space station with area 4,200 square meters containing many high-ranking alien military personnel is propelling itself past Jupiter at 42,000 m/s.  They are traveling on a line tangent to an orbit 250,000 meters from Jupiter and the ring of their space station is parallel to that tangent (see highly artistic diagram below).  At the instant when the station is a million meters from Jupiter, at what rate is the flux through the station changing?

(Assume Jupiter's magnetic field to be a constant 420 Teslas.  Although they radiate outward from Jupiter, assume the magnetic field lines to be essentially parallel by the time they get to the station.  Also, disregard any effects that have no bearing on the problem.)
 
Diagram:

Solution:

Let the above angle F be the angle made by the station and the magnetic field lines.
 
Taking the derivative of Flux = BA cos F, we find that
d(flux)/dt = (-BA sin F)(dF/dt)

We need only to find dF/dt.

Using the Pythogorean Theorem, x^2 + y^2 = r^2
     Therefore, x^2 = (1,000,000)^2 - (250,000)^2 >>> x = 968,245 m

Taking the derivative of the Pythogorean Theorem,
     2x dx/dt + 2y dy/dt = 2r dr/dt >>> 2(968,245)(42,000) + 2(250,000)(0) = 2(1,000,000) dr/dt, solving which gives dr/dt = 40666 m/s (effectively negative)

We know that cos F = x/r by definition of cosine, so we can take the derivative and get: 

Because d(flux)/dt = BA(-sin F df/dt), we can substitute and get: 

We can then substitute directly and get: 

Which simplifies to:

d(flux)/dt = 463,111 Wb / S

That's a lot.  That's some current.  Let's hope the aliens thought ahead.

Phew.  Head to my final sort of wrap-up page, the musings
or go back to the contents