Math Insight

Graphical solution of pure-time differential equations

Math 201, Spring 22
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Total points: 3
  1. Our goal is to understand what a pure-time differential equation can tell us about the graph of the unknown function.
    1. Recall using the slope of the tangent line to find the derivative of a function at a point. If a function is decreasing, the slope is negative, so the derivative is negative. Likewise, if the derivative is negative at a point, the function is
      there.
    2. If a function is increasing, the slope is positive, so the derivative is positive. Likewise, if the derivative is positive at a point, the function is
      there.
    3. If the function is flat, or has a horizontal tangent line, the slope and derivative are zero. Then if the derivative is zero at a point, the function is
      there. In other words, the graph is
      .

  2. Let's use our understanding of what the derivative tells us about the shape of a graph to sketch the solution to the following pure-time differential equation: \begin{align*} \diff{ w }{t} &= f( t)\\ w(0) &= -2 \end{align*} Here's a graph of $f$.

    The function $w(t)$ is the solution to the differential equation. We don't yet know what the function $w(t)$ is, but we do know that at each time point $t$, its derivative must be equal to the value of $f$ at time $t$, as graphed above.

    1. For what values of $t$ is the graph of $w$ horizontal? (If there are multiple, enter them in increasing order, separated by commas.)

      For what values of $t$ is $w$ increasing?

      For what values of $t$ is $w$ decreasing?

    2. Our next goal is to sketch what the solution $w(t)$ looks like.

      First, identify the initial condition (given at the beginning of this problem): $w($
      $)=$
      . Start your sketch of the solution by plotting this point.

      Feedback from applet
      initial condition:
      peak location:
      shape:
      (You'll finish graphing the solution on the above axes in part c.)
    3. Next, use your results from part a to determine whether $w$ starts off by increasing or by decreasing. Since when $t=0$, $w$ is
      , we need the graph the of $w(t)$ to be
      the initial condition for values of $t$ just above $0$.

      However, $w(t)$ does not keep moving in the same direction. Instead, it stops and reverses direction at $t=$
      . Make sure your sketch accurately represents what happens to $w(t)$ at this value of $t$.

  3. Let's use our understanding of what the derivative tells us about the shape of a graph to sketch the solution to the following pure-time differential equation, where $f$ is graphed below: \begin{align*} \diff{ v }{t} &= f( t)\\ v(0) &= 5 \end{align*}
    1. For what values of $t$ is the graph of $v$ horizontal? (If there are multiple, enter them in increasing order, separated by commas.)

      For what values of $t$ is $v$ increasing? It is increasing for
      .

      For what values of $t$ is $v$ decreasing? It is decreasing for
      .

    2. Now we will sketch the solution on the below axes.

      First, identify the initial condition: $v($
      $)=$
      . If you are sketching by hand, you'll want to start your sketch of the solution by plotting this point. In the below applet, you will have to adjust other points later to make sure the initial condition is satisfied.

      Next, use your results from part a to determine whether $v$ begins by increasing or decreasing. Since when $t=0$, $v(t)$ is
      , we need the graph to be
      the initial condition for values of $t$ just above $0$.

      From the above graph of the derivative of $v(t)$ (i.e, the graph of $f$), you can determine that $v(t)$ stops and changes directions at $t=$
      . (If there are multiple values, enter them in increasing order, separated by commas.) Make sure your sketch of the solution $v(t)$ changes direction at these values of $t$ to reflect this reality. At each point, determine whether the graph should have a local maximum, minimum, or neither. Enter maximum, minimum, or neither, separated by commas, in the same order as the corresponding values of $t$:

      For the sketch to be considered accurate, it must match the initial condition, be increasing and decreasing over the correct intervals, and have local maxima and minima at the correct values of $t$. Since we don't have a formula for $f$, we aren't concerned with the precise values of $v(t)$.

      Feedback from applet
      initial condition:
      peak locations:
      shape: