A question was posed about the following problem:

As always, we first must understand the question.

What are we asked to find?

We are asked the maximum height of the body. Since the height of the body, in meters, is \(s\), we must find the maximum value of \(s\).

What are the data?

We are given two pieces of data:

1. \(s\) is governed by the equation \(s=\frac{1}{2}gt^{2}+v_0t+s_0\).
2. In that equation, \(g\) is a non-zero positive number (that is, \(g > 0\)).

Do we understand all of the definitions?

The maximum of a function is its largest value over an interval.

Next, we must devise and execute a plan.

The intuitive approach

Students who have taken an introductory physics course will recognize \(s\) as the kinematic equation for displacement and acceleration, given an initial displacement. The body in our problem begins at some initial height \(s_0\), moving at some initial velocity \(v_0\). For each unit of time \(t\), the body moves a distance equal to its initial velocity plus the change in its velocity due to the rate of acceleration. We can make a data table to illustrate this.

Time (\(t\)) (seconds)	Height (\(s\)) (meters)
\(0\)	\((\frac{1}{2}g\times0^2)+(v_0\times 0)+s_0 = s_0\)
\(1\)	\((\frac{1}{2}g\times1^2)+(v_0\times 1)+s_0 = \frac{1}{2}g+v_0+s_0\)
\(2\)	\((\frac{1}{2}g\times2^2)+(v_0\times 2)+s_0 = 2g+2v_0+s_0\)
\(5\)	\((\frac{1}{2}g\times5^2)+(v_0\times 5)+s_0 = 12\frac{1}{2}g+5v_0+s_0\)

Notice that, as time passes (as \(t\) becomes larger), the first term (\(\frac{1}{2}gt^2\)) increases more than the second term (\(v_0t\)). (The third term, \(s_0\), does not change at all.)

If the mathematical expressions in the previous table make the motion difficult to visualize, we can assume that the body starts at some reference displacement, such as on the ground, where \(s_0 = 0 \mathrm{m}\). We can also assume that it starts from a dead stop, so its initial velocity \(v_0=0 \mathrm{m/s}\), and it is accelerating at \(g=1\mathrm{m/s^2}\) (in other words, the body's velocity increases every second by 1 m/s). With those assumptions, we can make a table that may be easier to visualize.

Time (\(t\)) (seconds)	Height (\(s\)) (meters)	Change in height since \(t-1\) (meters)
\(0\)	\(0\)	—
\(1\)	\(\frac{1}{2}\)	\(\frac{1}{2}\)
\(2\)	\(2\)	\(1\frac{1}{2}\)
\(3\)	\(4\frac{1}{2}\)	\(2\frac{1}{2}\)
\(4\)	\(8\)	\(3\frac{1}{2}\)
\(5\)	\(12\frac{1}{2}\)	\(4\frac{1}{2}\)

We can graph these data points.

Assuming no initial velocity (\(v_0 = 0\mathrm{m/s}\)) and a starting point on the ground (\(s_0 = 0\mathrm{m}\)).

Our intuition should be that the body is moving faster and faster and faster as time goes on. There seems to be no reason to think that the obect will slow down; there may not be a maximum height. But can we prove it?

The Algebraic Approach

Let's leave the physics to the side, and think about the math. We’ll use algebra to formalize our intuition. We are given an equation, \[s=\frac{1}{2}gt^{2}+v_0t+s_0\] and we want to find its maximum value. Our first observation should be that this is the equation of a quadratic function, so its graph is a parabola.  A parabola rises on both sides as the independent variable approaches negative and positive infinity.

A parabola.

In the intuitive approach, we discussed what happens as time goes on. Let's put that in algebraic terms.

For any time (in seconds) \(t\), we know that the position of the body is \[s_1=\frac{1}{2}gt^{2}+v_0t+s_0\]. What happens in the next second, when the time is \(t+1\)? \[s_2=\frac{1}{2}g(t+1)^2+v_0(t+1)+s_0\]
\(t+1\) must be greater than \(t\), because \[1>0\] We can use the additive property of inequalities to add \(t\) to both sides while preserving the inequality, as \[t+1>t+0\] Simplifying, we have \[\begin{equation}t+1>t\end{equation}\]
If we restrict ourselves to cases where \(t\geq 0\) (an artificial restriction, but one that will suffice for our proof), then \((t+1)^2>t^2\). Observe that \[(t+1)^2 = t^2+2t+1\] Again, \(1>0\), and adding \(t^2\) to both sides, we have \[\begin{equation}t^2+1>t^2\end{equation}\]
We know, from our restriction, that \(t>0\). If we add that inequality to itself, we have \[\begin{align}t+t& >0+0\notag\\ 2t& >0 \tag{3} \end{align}\]
Adding inequality (2) to inequality (3), we have \[t^2+2t+1>t^2\]
And substituting \((t+1)^2\) for \(t^2+2t+1>t^2\), we have the inequality we were aiming to prove. \[(t+1)^2>t^2\tag{4}\]
We can use the multiplication property of inequalities to multiply both sides of inequality (4) by \(\frac{1}{2}g\) and \(v_0\), and we have \[\frac{1}{2}g(t+1)^2>\frac{1}{2}gt^2\tag{5}\] and \[v_0(t+1)^2>v_0t^2\tag{6}\]
We can add inequalities (5) and (6) to construct \[\frac{1}{2}g(t+1)^2+v_0(t+1)^2>\frac{1}{2}gt^2+v_0t^2\tag{7}\]
Finally, by the addition property of inequalities, we add \(s_0\) to both sides and have \[\frac{1}{2}g(t+1)^2+v_0(t+1)^2+s_0>\frac{1}{2}gt^2+v_0t^2+s_0\tag{8}\]
From inequality (8), we know that for any positive value of \(t\), the value of \(s\) is greater at \(t+1\). Therefore, \(s\) increases to infinity as \(t\) does, and there can be no maximum value.

The First Derivative Test

We have seen that algebra is sufficient to prove that \(s\) has no maximum. We can use calculus, however, for an alternative proof of our hypothesis.

Differentiating \(s\) with respect to \(t\), its first derivative is \[\frac{ds}{dt}=gt+v_0\]
Any maximum value of \(s\) would have to occur at a critical point, or root, of the first derivative. Therefore, we set the derivative equal to zero and solve for \(t\).
\[\begin{align*}gt+v_0& =0\\gt& = -v_0\\t& = -\frac{v_0}{g}\end{align*}\tag{9}\]
where equation (9) represents the only critical point of \(s\). Without knowing the value of v_0, we cannot determine whether \(\frac{ds}{dt}\) is greater or less than zero to the left and the right of \(-\frac{v_0}{g}\), so we must turn to the Second Derivative Test.
Further differentiating \(s\), we have \[\frac{d^2s}{dt^2}=g\]
Since we know that \(g\) is positive from our data, \(\frac{d^2s}{dt^2}\) is positive at all values of \(t\), and the graph of \(s\) is always concave up. That requires a conclusion that our critical point is a global minimum, i.e., \(s\) has no maximum.

A question was posed about the following problem (modified slightly from the original):

Edison Electric wants to build a wind farm to provide electricity for three towns. Residents are concerned about noise, so Edison Electric wants to build the wind farm equally distant from all three towns.  How can Edison Electric find such a place?

A question was posed about the following problem:

A line goes through the point \((-2, -4)\) and is parallel to the graph of the equation $y = -x + 5$. Find the equation of the line in either point-slope form or slope-intercept form.

First, we must understand the problem.
What are we asked to find?
The problem asks us to find an equation for a line, in either point-slope form or point-intercept form.
What are the conditions for a correct solution?

1. Our new line must pass through the point $(-2, -4)$.
2. Our new line must be parallel to the graph of the equation $y=-x+5$.