Math Insight

Forward Euler and linear approximations

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  1. For the following differential equation \begin{align*} \diff{ v }{ t } &= - 0.1 v + 6\\ v(0) & = 7.9 \end{align*} calculate a Forward Euler approximation to $v(21.0)$ using a time step of $\Delta t = 7.0$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ v }{ t }(t) &= - 0.1 v{\left (t \right )} + 6. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 7.0$, yielding an estimate of $v(7.0)$.

      What is the initial condition? $v(0) = $

      Based on that result and the differential equation, what is the slope of $v(t)$ at time $t=0$?
      $\diff{ v }{t}(0) = - 0.1 v{\left (0 \right )} + 6 =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = v(0) + \diff{ v }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=\Delta t = 7.0$.
      $v(7.0) \approx L(7.0) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 7.0$, yielding an estimate of $v(14.0)$.

      Given the value of $v(7.0)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=\Delta t = 7.0$?
      $\diff{ v }{t}(7.0) = - 0.1 v{\left (7.0 \right )} + 6 =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 7.0$.
      $L(t) = v(7.0) + \diff{ v }{t}(7.0) (t-7.0) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=2\Delta t = 14.0$.
      $v(14.0) \approx L(14.0) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 7.0$, yielding an estimate of $v(21.0)$.

      Given the value of $v(14.0)$ estimated above and the differential equation, what is the slope of $v(t)$ at time $t=2\Delta t = 14.0$?
      $\diff{ v }{t}(14.0) = - 0.1 v{\left (14.0 \right )} + 6 =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 14.0$.
      $L(t) = v(14.0) + \diff{ v }{t}(14.0) (t-14.0) =$

      We use this linear approximation to approximate the value of $v(t)$ at $t=3\Delta t = 21.0$.
      $v(21.0) \approx L(21.0) =$

  2. For the following logistic differential equation \begin{align*} \diff{ s }{ t } &= 1.3 s \left(- \frac{s}{890} + 1\right)\\ s(0) & = 623 \end{align*} calculate a Forward Euler approximation to $s(2.4)$ using a time step of $\Delta t = 0.8$. In this problem, we guide your through the steps of the calculation. It may easier to understand the steps by writing the differential equation using an explicit argument of $t$ \begin{align*} \diff{ s }{ t }(t) &= 1.3 \left(- \frac{1}{890} s{\left (t \right )} + 1\right) s{\left (t \right )}. \end{align*}

    1. The following steps will guide you in calculating the Forward Euler approximation after one step of size $\Delta t = 0.8$, yielding an estimate of $s(0.8)$.

      What is the initial condition? $s(0) = $

      Based on that result and the differential equation, what is the slope of $s(t)$ at time $t=0$?
      $\diff{ s }{t}(0) = 1.3 \left(- \frac{1}{890} s{\left (0 \right )} + 1\right) s{\left (0 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=0$.
      $L(t) = s(0) + \diff{ s }{t}(0) (t-0) =$

      We use this linear approximation to approximate the value of $s(t)$ at $t=\Delta t = 0.8$.
      $s(0.8) \approx L(0.8) =$

    2. The following steps will guide you in calculating the Forward Euler approximation after the second step of size $\Delta t = 0.8$, yielding an estimate of $s(1.6)$.

      Given the value of $s(0.8)$ estimated above and the differential equation, what is the slope of $s(t)$ at time $t=\Delta t = 0.8$?
      $\diff{ s }{t}(0.8) = 1.3 \left(- \frac{1}{890} s{\left (0.8 \right )} + 1\right) s{\left (0.8 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=\Delta t = 0.8$.
      $L(t) = s(0.8) + \diff{ s }{t}(0.8) (t-0.8) =$

      We use this linear approximation to approximate the value of $s(t)$ at $t=2\Delta t = 1.6$.
      $s(1.6) \approx L(1.6) =$

    3. The following steps will guide you in calculating the Forward Euler approximation after the third step of size $\Delta t = 0.8$, yielding an estimate of $s(2.4)$.

      Given the value of $s(1.6)$ estimated above and the differential equation, what is the slope of $s(t)$ at time $t=2\Delta t = 1.6$?
      $\diff{ s }{t}(1.6) = 1.3 \left(- \frac{1}{890} s{\left (1.6 \right )} + 1\right) s{\left (1.6 \right )} =$

      Those two numbers are enough information to make a linear approximation at $t=2\Delta t = 1.6$.
      $L(t) = s(1.6) + \diff{ s }{t}(1.6) (t-1.6) =$

      We use this linear approximation to approximate the value of $s(t)$ at $t=3\Delta t = 2.4$.
      $s(2.4) \approx L(2.4) =$

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Math 1241, Fall 2020