Question

Sketch the curve represented by the vectorvalued function and give the orientation of the curve. \(\mathbf{r}(t)=(1-t) \mathbf{i}+\sqrt{t} \mathbf{j}\)

Short answer

The curve starts from the point (1,0) and moves along the y-axis upwards as \(x(t)\) decreases. Thus, the orientation of the curve is upwards.

Step-by-step solution

Analyze the individual components

The given function is \(\mathbf{r}(t) = (1-t) \mathbf{i} + \sqrt{t} \mathbf{j}\). This consists of two parts: \(x(t) = 1-t\) and \(y(t) = \sqrt{t}\). \(x(t)\) is a linear function and decreases with increasing \(t\). \(y(t)\) is a square root function which is defined only for non-negative values of \(t\) and increases with increasing \(t\).

Find the range of parametric function

As \(y(t) = \sqrt{t}\) is only defined for \(t \geq 0\), this is the lower bound for \(t\). There is no upper bound given in the task. Thus, the range of the parameter \(t\) is \([0, \infty)\).

Plot the function

Using the obtained values, plot the function. Keep in mind the orientation: as \(t\) increases, \(x(t)\) decreases starting from 1 and \(y(t)\) increases starting from 0 as \(t\) is non-negative. This will result in a curve in the 2D plane starting from the point (1,0) and moving along the y-axis upwards as \(x(t)\) decreases.