3.A The Derivative

The Derivative

The derivative is the instantaneous rate of change.

The derivative of $f(x)$ is denoted by $f'(x)$ and by $\displaystyle{\frac{df}{dx}}$.
We approximate this the derivative $f'(x)$ at the point $x=a$ with the formula \[ f'(a) \approx \frac{f(a+h) - f(a)}{h} \] where $h$ is a very, very small number.
The derivative measures the sensitivity of outputs to a small change in input. \[ \text{the derivative} \, \approx \, \frac{\text{change in output}}{\text{change in input}} \, = \, \frac{\Delta y}{\Delta x} \]
The derivative of a function $f$ at a point $a$ is the slope of the tangent line at $x=a$.
Economics. Given a cost function $C(x)$, the marginal cost at $x=a$ is the cost of producing one additional unit of a good. The formula is for the marginal cost at $x=a$ is $C(a+1)-C(a)$.
Physics. Given a displacement function $s(t)$, the velocity is $v(t) = s'(t)$, and the acceleration is $a(t)=v'(t)$.

Suppose that spending 3 extra hours on studying increases our percent grade on an exam by 14 percent.

Calculate the derivative, give the units, and give an interpretation of the number you get.
Is this rate of change constant? Explain.

Economists use the term “marginal cost” for the derivative of the cost of production. Here is the definition:

The marginal cost is the cost to produce 1 additional good.

Suppose that it costs $250,000 to produce 1000 iPhones.

If it costs $250,440 to produce 1002 iPhones, what’s the marginal cost of 1 iPhone?
Using the marginal cost from part 1, estimate how much it will cost to produce 1030 iPhones.

The total acreage of farmland has been shrinking in the United States since 1980.

Year	1980	1985	1990	1995	2000
Farmland (millions of acres)	1039	1012	987	963	945

Let $F(t)$ denote the amount of farmland at a time $t$. Use the data to estimate $F'(1995)$, the derivative at $t=1995$.
Did you use 1990 or 2000 as your estmation point? Did it matter? What does that mean?
You got a negative number for $F'(1995)$. Does that make sense with the situation? Why?
Give an example of a year that, if you had the data for it, would better help us estimate $F'(1995)$. Why?

A particle’s position over time is given by the formula \[ s(t)=2+3\sin(4t)+1.2t^2 \]

Find the velocity at $t=1$ and $t=2$ numerically (using small subintervals). Remember that the velocity is the derivative of the position.
Use desmos to graph the position curve $s(t)$ and the tangent lines at both $t=1$ and $t=2$.

Suppose that spending 3 extra hours on studying increases our percent grade on an exam by 14 percent.

The derivative is $\frac{14}{3} = 4.67$ percent per hour. The interpretation: for every hour that you spend studying, your grade increases by $4.67\%$.
This rate of change is not constant. After studying for a while, you reach “burn out”, or perhaps you run out of new material to master. In addition, your grade tops out at 100%, so we cannot increase forever.

The marginal cost is \[ \frac{250,440 - 250,000}{102-100}=\frac{440}{2} =220 \mbox{ dollars per phone} \]
I estimate that the cost of making 1030 phones is \[ 250,000 + 200 \cdot 30 = 256,600. \]

When estimating the derivative, we need to “stay local” in order to get the most reliable estimate. There are three equally good choices here: \[ \begin{array}{rcl} f'(1995) &\approx& \frac{f(2000)-f(1995)}{2000-1995}=\frac{945-963}{5} = - \frac{18}{5} = -3.6 \\ f'(1995) &\approx& \frac{f(1995)-f(1990)}{1995-1990}=\frac{963-987}{5} = - \frac{24}{5} = -4.8 \\ f'(1995) &\approx& \frac{f(2000)-f(1990)}{2000-1990}=\frac{945-987}{10} = - \frac{42}{10} = -4.2 \\ \end{array} \]
Any of the three estimates are good answers. Since we are estimating there isn’t a “right answer.” All three stay close to 1995, so each one is good.
We got a negative number because the acreage is decreasing.
If we had the data for 1994 or 1996, using these values would give us a better estimate. Using smaller and smaller intervals will give us a more accurate estimation. (But we cannot tell up front how “small” we need to be to get the accurracy we desire.)

Using desmos, we know that \[ s'(1) \approx \frac{s(1+h)-s(1)}{h} \approx -5.44 \] when $0 < h < 0.00001$. Similarly, we have \[ s'(1) \approx \frac{s(2+h)-s(2)}{h} \approx 5.03 \] when $0 < h < 0.00001$.
Here are the commands to plot the curve and the two tangent lines.

\[ \begin{array}{l} s\left(t\right)=2+3\sin(4t)+1.2t^{2} \\ \\ y = -5.44(t-1) + s(1) \\ \\ y = 5.03(t-2) + s(2) \\ \end{array} \]