For calculus, I am asked to find instantaneous velocity. Here is the given data and question:
The table shows the position of a cyclist.
$$\begin{array}{c|c|c|c|c|c|c} \hline t\text{ (seconds)}&0 & 1&2&3&4&5\\ \hline s\text{ (meters)} & 0&1.4&5.1&10.7&17.7&25.8\end{array}$$
(a) Find the average velocity for each time period:
$\qquad$ (i) $[1,3]\qquad$ (ii) $[2,3]\qquad$ (iii) $[3,5]\qquad$ (iv) $[3,4]$(b) Use the graph of $s$ as a function of $t$ to estimate the instantaneous velocity when $t=3$.
I think that I should create the equation of tangent line and then put $t=3$ in the equation; for example let's take two arbitrary points, $t=1$ and $t=2$. We have two pairs, $(1,1.4)$ and $(2,5.1)$, and slope $m=\frac{5.1-1.4}{2-1}$ or $m=3.7$, so write equation in slope-form:
$$y-1=3.7(x-1.4)$$
$$y=3.7\cdot x-5.18+1 $$
$$y=3.7\cdot x-4.18$$
First of all, what I wanted to ask was: we have an approximation equation which expresses linear relationship between distance and time, so now I have two choices: directly put 3 into equation, or take the derivative, but derivative couldn't be taken because it is linear so by taking the derivative I will have a constant function, so that means I should put $t=3$, yes?
Please help me.
Best Answer
To answer you directly, you just want the slope of your line: 3.7.
But consider, please:
Below is an accurate scatter plot of your data. Despite what the instructions suggest, you do not know what the graph of $s$ looks like. However, you can imagine a curve that models the data points. This curve is the purple curve shown in the diagram.
Now, the instantaneous velocity at $t=3$ is approximately the slope of the tangent line shown above (approximate because the tangent line shown is tangent to the blue curve and the blue curve approximates the graph of $s$).
How can you estimate this slope using the tabular data? Well, it's essentially what you did: estimate the slope of the tangent line, and hence the instantaneous velocity at $t=3$, with the slope of a secant line between two of the given data points. (Note, please, you only need to estimate the slope of the line; you do not need to find the equation of the tangent line.)
But, you cannot select those two points randomly, this may give a bad estimate. In particular, you want $(3,10.7)$ to be "between" or one of the two points that you choose. Looking at the picture, it should be clear that the best points to choose are $(2,5.1)$ and $(4,17.7)$.
So, we will estimate the instantaneous velocity with the average velocity over $[2,4]$ (the average velocity over $[2,4]$ is the slope of the line connecting the points $[2,5.1]$ and $[4,17.7]$).
$${{\text {inst. vel.}}\atop {\text{ at }} t=3}\approx {17.7-5.1\over 4-2}={12.6\over2}=6.3. $$