Math Insight

Scalar linear equation problems

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Total points: 10
  1. Solve the following differential equation: \begin{align*} \diff{ w }{t} &= \frac{1}{ 5 } \left(- w - 6 \right)\\ w(0) &= -15 \end{align*}

    Solution: $w(t) = $

    (If you round your answer, include at least 5 significant digits.)

  2. Imagine that, in response to a light input of magnitude $I$ (arbitrary units), the activity level $A(t)$ (arbitrary units) of a retinal neuron evolves according to $$\diff{A}{t} = \frac{1}{ 25 } \left(-A + \frac{ 100 I^{ 6 } }{ 160^{ 6 } + I^{ 6 } } \right).$$ The response of the neuron is determined by the nonlinear function $g(I) = \frac{ 100 I^{ 6 } }{ 160^{ 6 } + I^{ 6 } }$, the graph of which is shown below.

    (For the following questions, if you round your answers, include at least 5 significant digits in your response.)

    1. If for a long time, the light level has been at $I=65$, what is the activity level of the neuron?
      $A = $
    2. Now, imagine that at time $t=0$, the light level immediately changed to $I=130$ and stayed there. For $t>0$, determine the activity level of the neuron.

      $A(t) = $

      After a long time, the activity of the neuron will be $A=$
      .

    3. If the light level instead changed to $I=210$ at $t=0$, determine the activity level of the neuron for $t>0$.

      $A(t) = $

      After a long time, the activity of the neuron will be $A=$
      .

    4. If the light level instead changed to $I=620$ at $t=0$, determine the activity level of the neuron for $t>0$.

      $A(t) = $

      After a long time, the activity of the neuron will be $A=$
      .

    5. To summarize your results, sketch the three solutions on the following graph. For the graph, round the initial conditions and long time values of each curve to the nearest $5$. The temporal scale isn't given here, but make sure that the qualitative shape of each curve is correct.
      Feedback from applet
      Final values of curves:
      Initial conditions of curves:
      Number of curves:
      Speed profiles of curves:

  3. In the following, we explore a piecewise-linear approximation the neuron's action potential. Two ion channels play an important role in action-potential generation: the sodium (Na+) channel and the potassium (K+) channel. The opening and closing of these channels draws the transmembrane voltage of the neuron up or down, giving rise to the dynamic properties of the neuron.

    Note: include at least 3 significant digits if you round your answers.

    1. Neuron with a single channel. If a neuron had just a single channel, we could model the evolution of its voltage $V$ as $$C\diff{V}{t} = - g (V - E)$$ where $C$ is the (positive) capacitance of the neuron's membrane, $g$ is the (positive) conductance of the channel (the reciprocal of its resistance) and $E$ is its reversal potential. (We'll measure time $t$ in ms, voltage $V$ in mV, capacitance $C$ in μF, and conductance $g$ in mS.)

      For what value of $V$, will the system be at a steady state (or equilibrium), i.e., the change in the voltage, $dV/dt$, will be zero? $V=$

      If $V$ is larger than this value, the voltage will be
      since the derivative is
      . If $V$ is smaller than this value, the voltage will be
      since the derivative is
      .

      Let the channel be a potassium channel with a reversal potential of $E=E_{K} = -95$ mV and a conductance of $g=g_{K} = 1.2$ mS. If we set the capacitance to $C=1$ μF, the equation for $V$ becomes $$\diff{V}{t} = -1.2 \left(V + 95\right).$$

      If when $t=7$ ms, the voltage is $V(7) = -63$ mV, calculate the voltage for all time.

      $V(t) = $

      Plot $V(t)$ for $7 < t < 20$ on the below graph.

      Feedback from applet
      Final points of curves:
      Initial points of curves:
      Number of curves:
      Speed profiles of curves:

      You should have found that the solution decays exponentially to the reversal potential, with a time constant determined by the conductance.

    2. Now, we will model both the potassium channel and the sodium channel in the neuron. We assume that the currents from the channels sum linearly so that the combined model for the evolution of the voltage is $$C\diff{V}{t} = -g_{Na}(V-E_{Na}) - g_{K}(V-E_{K}),$$ where $g_{Na}$ is the sodium channel conductance, $g_K$ is the potassium channel conductance, $E_{Na}$ is the sodium reversal potential, and $E_K$ is the potassium reversal potential. For simplicity, we'll continue to set the capacitance to $C=1$ μF.

      For what value of $V$ will the neuron be at a steady state?

      $V=$

    3. Imagine that, for this neuron, the sodium reversal potential is $E_{Na} = 60$ mV and the potassium reversal potential is $E_K = -95$ mV. Moreover, when the neuron is at rest, its sodium channel conductance is $g_{Na} = 0.08$ mS and its potassium channel conductance is $g_K = 0.32$ mS.

      Calculate the resting potential $E_r$ of this neuron, i.e., the value of the steady state voltage when the neuron is at rest.

      $E_r =$
      mV

    4. An action potential is caused by changes in the sodium and potassium conductances. The conductances changes are triggered by changes in the voltage due to input, but here we'll just manually change the conductances to produce an action potential.

      Imagine that the neuron as been at rest for a long time. Therefore, its voltage has reached its steady state, which is $V=$
      mV. We'll use that value for the initial condition $V(0)$.

      At $t=0$, the action potential is somehow triggered, sending the sodium conductance immediately way up to $g_{Na} = 16.0$ mS, but only for a brief moment, for 1 ms. If $g_K$ remains at $0.32$ mS and $g_{Na} = 16.0$ mS for $0 \le t \le 1$, calculate $V(t)$ for that interval.

      $V(t) = $
      mV

      What is the value of the voltage at $t=1$ ms? $V(1)=$
      mV

    5. At $t=1$, the sodium conductance decreases to zero and the potassium conductance increases to $g_K = 4.0$ mS. Here, let's assume that $V(1) = _$ mV, the answer you entered in the previous part.

      Using the value $V(1) = _$ mV for the initial condition at $t=1$, calculate $V(t)$ for $1 \le t \le 4$, assuming that the conductances stayed fixed at $g_{Na}=0$ mS and $g_K=4.0$ mS during this interval.

      $V(t) =$
      mV

      At $t=4$, the voltage is $V(4) = $
      mV.

    6. Finally, for $t \ge 4$, the sodium and potassium conductances return to their resting values $g_{Na}=0.08$ mS and $g_K = 0.32$ mS. Using the value you calculated above, $V(4) = _$ mV, for the initial condition at $t=4$, calculate $V(t)$ for $t \ge 4$.

      $V(t)=$
      mV.

    7. Sketch the solution for $0 \le t \le 20$.
      Feedback from applet
      Final points of curves:
      Initial points of curves:
      Number of curves:
      Speed profiles of curves: