This seems like it should be simple....

[passingmtl]

OK, here's the question, straight from my Calculus test:

"A quantity Q is increasing by 3300 per year at the present time. This means that _______. (Provide a statement that is always true, involving either Q'(0), Q(0), Q(3300) or Q'(3300)."

OK, so I realize the the derivative is 3300 for t=the present year, and that therefore the function is increasing at t=the present year. However, I have no way of knowing whether the present year is the initial year of the function (i.e. t=0) or if the present year is t=3300 (which I reason to be possible if the initial year of the function was 3300 years ago), or if the present year is some value of t between 0 and 3300 or even after 3300. So, how the heck am I supposed to come up with a statement that is "always true" for any of those Q or Q' options??

I would really appreciate any thoughts you have on this.

Thanks!

 

[BigGlenntheHeavy]

\(\displaystyle \frac{dQ}{dt} \ = \ 3300, \ (Rate \ of \ change \ per \ time, \ t \ in \ years).\)

\(\displaystyle Hence, \ \int dQ \ = \ \int 3300dt \ \implies \ Q \ = \ 3300t+C\)

\(\displaystyle Now, \ since \ Q \ is \ a \ function \ of \ t, \ \implies \ Q(t) \ = \ 3300t+C.\)

\(\displaystyle Ergo, \ Q(0) \ = \ 3300(0)+C, \ therefore \ Q(0) \ = \ C\)

\(\displaystyle This \ means \ that \ at \ the \ beginning \ (t=0), \ Q(0) \ equals \ some \ constant \ C, \ always \ true.\)