91Ó°ÊÓ

The unit step function, \(u_c(t)\), is defined as: $$ u_{c}(t)= \begin{cases} 0, & \text{ if } t < c\\ 1, & \text{ if } t \geq c \end{cases} $$ In our case, we have \(c = \pi\). The function \(u_\pi(t)\) will be equal to 0 for \(t < \pi\) and 1 for \(t \geq \pi\).

The function \(f(t) = t^2\) is a simple quadratic function. To sketch the graph, we can start by plotting a few points: \((0, 0)\), \((1, 1)\), \((-1, 1)\), \((2, 4)\), \((-2, 4)\), etc. Using these points, we can draw a curve representing the function \(f(t) = t^2\). For \(t \geq 0\), this is a parabola opening upwards with its vertex at the origin.

Now that we have the graph of \(f(t) = t^2\), we can apply the unit step function \(u_\pi(t)\). This means that the graph of the function \(f(t-\pi) u_{\pi}(t)\) will be the same as the graph of \(f(t)\) for \(t \geq \pi\) and 0 for \(t < \pi\). First, find the function \(f(t-\pi)\), which is obtained by replacing \(t\) with \(t-\pi\) in the original function: $$ f(t-\pi) = (t-\pi)^2 = t^2 - 2\pi t + \pi^2 $$ The graph of this function will be the same as the graph of \(f(t)\) but shifted to the right by \(\pi\). Thus, for \(t\geq\pi\), the graph of the function \(f(t-\pi)\) will start at \((\pi, 0)\) and open upwards like a parabola. Now, applying the unit step function \(u_\pi(t)\) means that the graph of the function \(f(t-\pi) u_{\pi}(t)\) will be the same as the graph of \(f(t-\pi)\) for \(t \geq \pi\), and will be 0 for \(t < \pi\). So we have a parabola starting at \((\pi, 0)\) and opening upwards for \(t \geq \pi\), and the graph will lie on the \(t\)-axis for \(t < \pi\).

With all the information gathered in the previous steps, we can now sketch the final graph of the given function, \(f(t-\pi) u_{\pi}(t)\). - For \(t < \pi\): The graph will lie on the \(t\)-axis. - For \(t \geq \pi\): The graph is a parabola starting at \((\pi, 0)\) and opening upwards. By combining both parts, we have the complete sketch of the given function.