Section 2.2: An Example


In the last section, we discussed how finding the "steepness" of a curve (i.e., slope of tangent line) at a point is an answer to the question: "how sensitive is the value of $y$ to small changes in $x$?"


In this section, we will take a look at a more applied problem.


With careful measurement we could determine that the height of a dropped nugget is given by $h(t)=h_0-kt^2$, where

  1. $t$ is time measured seconds
  2. $h_0$ is the initial height of the nugget
  3. $k$ is some constant

Natural question

"How fast is the nugget falling at time $t$?"

More concrete

Let $k=4.9$ and $h_0=500$ feet. In this case, $h(t)=500-4.9t^2$ and the graph of $h$ is as follows. Note: Click on the point to drag it around.

Let's explore what happens after 3 seconds into the nugget's fall. We see that when $t=3$,

\begin{align} h(3)=500-4.9(3)^2=455.9\text{ feet}. \end{align}

One second later, the height is

\begin{align} h(4)=500-4.9(4)^2=421.6\text{ feet}. \end{align}

From this, we can calculate the average velocity for the nugget between 3 seconds and 4 seconds. Let's do that now.

Note: The value we computed for the average velocity is negative, which indicated that the nugget is falling. The average speed is the absolute value of the velocity.

Recall that the value that we just computed is the slope of the chord over the interval $[3,4]$. How does this value compare to the actual speed at $t=3$?

We can improve our approximation of the velocity at $t=3$ by finding the average velocity over smaller and smaller intervals. For example, if we compute the average velocity over the interval $[3,3.01]$ the value will be much closer to the actual velocity at $t=3$ than what we computed above. Let's do that now.

We can do this for a really long time! What we really want to know is what happens to(3)
\begin{align} \frac{h(3+\Delta t)-h(3)}{\Delta t} \end{align}

as $\Delta t$ gets smaller and smaller. Let's do some algebra like we did in the last section and see if we can get our hands on something useful.

In the language of the previous section, we have just determined that the slope of the tangent line to $h(t)=500-4.9t^2$ at $t=3$ is $-29.4$. In other words, $h'(3)=-29.4$.


The moral of the story is that finding the instantaneous velocity of a position function at a fixed moment in time is the same problem as finding the slope of the tangent line to a function at a fixed point.