Question

Sketch the plane curve represented by the vector-valued function, and sketch the vectors \(\mathbf{r}\left(t_{0}\right)\) and \(\mathbf{r}^{\prime}\left(t_{0}\right)\) for the given value of \(t_{0} .\) Position the vectors such that the initial point of \(\mathbf{r}\left(t_{0}\right)\) is at the origin and the initial point of \(\mathbf{r}^{\prime}\left(t_{0}\right)\) is at the terminal point of \(\mathbf{r}\left(t_{0}\right) .\) What is the relationship between \(\mathbf{r}^{\prime}\left(t_{0}\right)\) and the curve? $$ \mathbf{r}(t)=e^{t} \mathbf{i}+e^{2 t} \mathbf{j}, \quad t_{0}=0 $$

Short answer

The vector \(\mathbf{r}(0) = \mathbf{i}+\mathbf{j}\) and vector \(\mathbf{r}'(0) = \mathbf{i} + 2\mathbf{j}\). The curve sketched out by \(\mathbf{r}(t)=e^{t} \mathbf{i}+e^{2t} \mathbf{j}\) is an exponential curve moving upward faster in the \(j\) direction than in the \(i\) direction. The vector \(\mathbf{r}'(t_{0})\) provides the direction and speed of curve change at the point corresponding to given \(t_{0}\).

Step-by-step solution

Evaluate the vector \(\mathbf{r}(t)\) at \(t_{0}\)

Using the given vector-valued function, we replace \(t\) with \(t_{0}\) to find \(\mathbf{r}(t_{0})\): \[\mathbf{r}(0)=e^{0} \mathbf{i}+e^{2*0} \mathbf{j}=\mathbf{i}+\mathbf{j}\]

Find the derivative \(\mathbf{r}'(t)\) and evaluate it at \(t_{0}\)

To differentiate our vector-valued function, we take the derivative of each component with respect to \(t\), and then evaluate this derivative at \(t_{0} = 0\):\[\mathbf{r}'(t) = e^t \mathbf{i} + 2e^{2t} \mathbf{j}\]So, \[\mathbf{r}'(0) = e^0 \mathbf{i} + 2e^{2*0} \mathbf{j} = \mathbf{i} + 2\mathbf{j}\]

Sketch the curve and the vectors

Now it's time to sketch the vector \(\mathbf{r}(t)=e^{t} \mathbf{i}+e^{2t} \mathbf{j}\), the vector \(\mathbf{r}(0)\), and the tangent vector \(\mathbf{r}'(0)\). As \(t\) changes, \(\mathbf{r}(t)\) traces out an exponentially curved path that moves upward faster in the \(j\) direction than in the \(i\) direction, because of the power of \(2\) in the exponent for the \(j\) component. The vector \(\mathbf{r}(0)\) is positioned at the origin and \(\mathbf{r}'(0)\) is a tangent to the curve at \(\mathbf{r}(0)\), pointing in the direction of the curve increase.

Interpret the physical meaning

The vector \(\mathbf{r}'(t_{0})\) is tangential to the curve at the point corresponding to \(t = t_{0}\). It gives the rate of change and direction in which the curve is moving at \(t = t_{0}\). Thus, its direction indicates the direction in which the curve is moving, and its length corresponds to the speed of the change.