calculus
Find the average velocity of an object whose position is given by:
$$s(t)=\frac{13}{t+2}$$
Using the following intervals:
i.[11,12]
ii. [11, 11.1]
iii. [11, 11.001]
iv. [11, 1.00001]
Then guess the value of the instantaneous velocity at t=11
I've look at a couple of examples and I'm still not completely understanding how to solve it. Do I find the average between all of the intervals answers?
Please Help!!!
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 $[4,17.7]$).
$${{\text {inst. vel.}}\atop {\text{ at }} t=3}\approx {17.7-5.1\over 4-2}={12.6\over2}=6.3. $$
I think you need some physical reasoning here. Once you fire your bullet vertically, it's only slowed by gravity until it's velocity is zero and then it starts to fall back. So, in absolute value, velocity is maximum in the moment of firing and when it returns to the surface. If you think about the vertical direction as positive, then the maximum occurs precisely in the firing moment. The reason why the derivative test doesn't work here is the following. Mathematically, it's stated as the proposition:
"If $f$ has a relative maximum in an open interval, then $f'$ is zero in this point"
The problem here is that velocity is a decreasing linear function of time, say $v(t) = v_{0} - gt$, where $v_{0}$ is the firing velocity and $g$ is gravity. Also, you're working on a closed interval $[0, t_{max} ]$, where $t_{max}$ is the instant corresponding to maximum height, calculated as above by you. Notice that $v(t)$ has no relative minimum in $(0, t_{max} )$, and that's why the test doesn't work (indeed, it's derivative is a constant). On the other hand, Weierstrass Theorem assures you that the continuous function $v(t)$ attains a maximum in the compact interval $[0, t_{max} ]$. Since you know it's not in the interior, it must be one end-point, and it's clear that it's actually $0$.
Best Answer
Yes, you're supposed to find the average velocity over each interval. In this case, we have $$ s(t)=\frac{13}{t+2} $$ (if you mean something else, please say so). To find the average velocity over an interval $[t_1,t_2]$, the formula is $$ v=\frac{s(t_2)-s(t_1)}{t_2-t_1} $$ So, for i) we get $$ v = \frac{\frac{13}{12+2}-\frac{13}{11+2}}{12-11}\approx-0.0714 $$ Can you figure out the rest?