Problem with Intuitive meaning of derivative

Question

The derivative of a function in some variable is the rate of change of that function to the change of that variable. Suppose there is a phenomena in nature such that amount of a parameter, say mass, of that phenomena is specified by time (in year) by the following function

$$f(t) = 2t^3 + t - 1$$

It means in the third year I expect to have $2\times 3^3+3-1 = 56$ units of that phenomena. If the derivative of function of mass in time shows the rate of change for each year, so the mass of that phenomena in the third year must be the sum of the changes of masses in the current moment and the first and the seconds years. So there must be

$$ f(3) = \sum_{i=0}^{2}f^{'}(i) = \sum_{i=0}^{2}6i^2+1 = 1 + 7 + 25 = 33 $$

and $33 \neq 56$ which confuses me. I would be appreciated if you explain what is wrong here.

Thanks in advance

Answer 1

I don't know where you found that sum formula from, but its definitely wrong. However, if we use another handy tool of calculus, the integral, we can make a more meaningful expression$$f(3)=f(0)+\int_0^3f'(t)dt$$Where you went wrong was assuming that derivative meant change in $f$ between unit time intervals. Rather, it is the instantaneous rate of change, the limit of the rate of change as you continue to reduce the size of the interval.

Answer 2

As you correctly mention the derivative gives you the change per time. In order to get the absolute change for a small time interval $\Delta t$ you have to multiply the derivative with this Intervall $f'(t)\Delta t$. Note that now also the units are correct. Now if you sum over this changes up to $t=56$, you would get $33$. The problem is that $f‘(t)$ is not constant. This means that $\Delta t$ has to be infinitesimally small. Taking the limit of the resulting sum we obtain the integral as in Rushabh Mehta‘s answer. Let $\Delta t=\frac t n$, then $$ f(t)=f(0)+\lim_{n\to\infty}\sum_{k=0}^n f'\bigg(k\cdot\frac t n\bigg)\cdot\frac t n=f(0)+\int_0^t f'(t)\,\text dt $$ The sum is called Riemann sum.

Answer 3

$\sum_\limits{i=0}^2 t'(i) = t'(0) + t'(1) + t'(2)$

Or the area of the green bars illustrated above.

The concept you are going after, that the area under the curve $f'(t)$ evaluated from from 0 to 3 should equal $f(3).$

But you still need to adjust for the constant term in $f(t)$ that is:

$\int_0^3 f'(t)\ dt = f(3) - f(0)\\ f(3) = f(0) + \int_0^3 f'(t)\ dt$

But the bigger piece that you are missing is that in evaluating at as you have, you have the lower sums, and you are missing the red areas.

Since $f'(t)$ is monotonic, You can say

$t'(0) + t'(1) + t'(2)\le f(3) - f(0) \le t'(1) + t'(2) + f'(3)$

And then you could make a increasingly fine partitions, until you squeeze the value in the middle.