Parameterizing a Piecewise Path

There are times when it is necessary to parameterize a path made up of pieces of different curves. This piecewise path may be open or form the boundary of a closed region as does the example shown in Figure $\PageIndex{4}$. In addition to determining a vector-valued function to trace out each piece separately, with the indicated orientation, we also need to determine a suitable range of values for the parameter $t$.

Note that there are many ways to parameterize any one piece, so there are many correct ways to parameterize a path in this way.

Example $\PageIndex{6}$: Parameterizing a piecewise path

Determine a piecewise parameterization of the path shown in Figure $\PageIndex{4}$, starting with $t=0$ and continuing on through each piece.

Solution

Our first task is to identify the three pieces in this piecewise path.

Note how we labeled these sequentially as $\vecs r_1$, $\vecs r_2$, and $\vecs r_3$. Now we need to identify the function for each and write the corresponding vector-valued function with the correct orientation (left-to-right or right-to-left).

Determining $\vecs r_1$: The equation of the linear function in this piece is $y = x$.

Since it is oriented from left-to-right between $t = 1$ and $t = 4$, we can write:

$$\vecs r_{1a}(t) = t\,\hat{\mathbf{i}}+ t \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 4 \nonumber$$

If we wish to begin this piece at $t = 0$, we just need to shift the value of $t$ one unit to the left. One way to do this is to write $\vecs r_{1a}$ in terms of $t_1$ instead of $t$ to make the translation easier to see.

Thus, we have $\vecs r_{1a}(t_1) = t_1\,\hat{\mathbf{i}}+ t_1 \,\hat{\mathbf{j}}$ for $1\le t_1\le 4$.

Figure $\PageIndex{4}$: A closed piecewise path

Subtracting $1$ from each part of this range of parameter values, we have: $0 \le t_1 - 1 \le 3$.

Now we let $t = t_1 - 1$. Solving for $t_1$, we obtain: $t_1 = t + 1$.

Replacing $t_1$ with the expression $t + 1$ will effectively shift the range of parameter values one unit to the left.

So, starting with $t = 0$, we have: $$\vecs r_1(t) = (t+1)\,\hat{\mathbf{i}}+ (t+1) \,\hat{\mathbf{j}} \quad\text{for}\quad 0 \le t \le 3 \nonumber$$

Double-check that this vector-valued function will trace out this segment in the correct direction before going on to $r_2$.

Determining $\vecs r_2$: This piece has a label showing the function whose graph it traces along. If it were oriented from left-to-right, we would have:

$$\text{Left-to-right:}\quad\vecs r_{2a}(t) = t\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-t}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 4 \nonumber$$

But since we need it to be oriented from right-to-left, we need to replace $t$ with $-t$ in the function and we need to divide through the range inequality by -1 to obtain the corresponding range. Thus we obtain:

$$\vecs r_{2b}(t) = -t\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(-t)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad -4 \le t \le -1 \nonumber$$

Check that it works!

Now we wish to have this piece start at $t = 3$ just after the first one finishes. Again let's make this easier to see by writing $r_{2b}$ in terms on $t_2$.

$$\vecs r_{2b}(t_2) = -t_2\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(-t_2)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad -4 \le t_2 \le -1 \nonumber$$

To force $r_2$ to start with $t = 3$ instead of $t = -4$, we need to add $7$ to each part of the inequality. This yields: $3 \le t_2 + 7 \le 6$.

Let $t = t_2 + 7$. Then solving for $-t_2$ (since this is what we need to replace in $r_{2b}$), we have: $-t_2 = 7-t$.

Replacing $-t_2$ with $\left(7-t\right)$ in $\vecs r_{2b}$, we obtain:

$$\vecs r_{2}(t) = (7-t)\,\hat{\mathbf{i}}+ \left(2\sqrt{\frac{4-(7-t)}{3}}+4\right) \,\hat{\mathbf{j}} \quad\text{for}\quad 3 \le t \le 6 \nonumber$$

This can be combined with our earlier result for $r_1$ to write a piecewise-defined vector-valued function that traces out the first two pieces, starting at $t = 0$:

$$\vecs r(t) = \begin{cases} (t+1)\,\hat{\mathbf{i}} + (t+1) \,\hat{\mathbf{j}}, & 0 \le t \le 3 \\ (7-t)\,\hat{\mathbf{i}} + \left(2\sqrt{\frac{t - 3}{3}}+4\right) \,\hat{\mathbf{j}}, & 3 \lt t\le 6 \end{cases} \nonumber$$

Note that one small modification was made to the second range so that when $t = 3$, there is no confusion about which piece to evaluate.

Determining $\vecs r_3$: To determine this last piece we need to think a little differently. This is because it is a vertical segment, which cannot be represented with a function of the form, $y = f(x)$. Note that it could be represented by a function of the form $x = f(y)$. Letting $y = t$, we can write $x = f(t)$ and writing a parameterization in increasing $y$ values (bottom-to-top), we'd get: $\vecs r(t) = f(t) \,\hat{\mathbf{i}} + t \,\hat{\mathbf{j}}$.

The equation of this line is $x = 1$. Thus, if we wished to parameterize this segment with upward orientation (increasing values of $y$), we have:

$$\vecs r_{3a}(t) = 1\,\hat{\mathbf{i}}+ t \,\hat{\mathbf{j}} \quad\text{for}\quad 1 \le t \le 6 \nonumber$$

But since we wish to use a downward orientation (decreasing values of $y$), we need to use a decreasing function of $t$ for $y$. As before, the simplest case is to use $y = -t$. Then, in the general case, we'd trace a function $x = f(y)$ in a downwards orientation with $\vecs r(t) = f(-t) \,\hat{\mathbf{i}} - t \,\hat{\mathbf{j}}$.

In the case of $r_3$, this gives us:

$$\vecs r_{3b}(t) = 1\,\hat{\mathbf{i}}- t \,\hat{\mathbf{j}} \quad\text{for}\quad -6 \le t \le -1 \nonumber$$

Note that since $x = 1, \, f(-t) = 1$, that is, it did not change the first component since it was constant and not a variable function of the parameter $t$.

Also note that since we negated $t$, we also had to negate the range, dividing it through by $-1$.

As above, to facilitate the translation, we'll replace $t$ with $t_3$, giving us:

$$\vecs r_{3b}(t_3) = 1\,\hat{\mathbf{i}}- t_3\,\hat{\mathbf{j}} \quad\text{for}\quad -6 \le t_3 \le -1 \nonumber$$

Now, we wish this final piece to start at $t = 6$ where the second piece we formed above leaves off. We see that we need to add $12$ to the range of paramater $t$ to accomplish this, giving us a new range of $6 \le t_3 + 12 \le 11$.

Let $t = t_3 + 12$. Then solving for $-t_3$ (since this is what we need to replace in $r_{3b}$), we have: $-t_3 = 12-t$.

Replacing $-t_3$ with $\left(12-t\right)$ in $\vecs r_{3b}$, we obtain:

$$\vecs r_{3}(t) = 1\,\hat{\mathbf{i}} + (12 - t)\,\hat{\mathbf{j}} \quad\text{for}\quad 6 \le t \le 11 \nonumber$$

Check that this still traces out this vertical segment from top-to-bottom.

We can now state the final answer as a single piecewise-defined vector-valued function that traces out this entire path, starting when $t = 0$.

$$\vecs r(t) = \begin{cases} (t+1)\,\hat{\mathbf{i}} + (t+1) \,\hat{\mathbf{j}}, & 0 \le t \le 3 \\ (7-t)\,\hat{\mathbf{i}} + \left(2\sqrt{\frac{t - 3}{3}}+4\right) \,\hat{\mathbf{j}}, & 3 \lt t\le 6 \\ 1\,\hat{\mathbf{i}} + (12 - t)\,\hat{\mathbf{j}} & 6 \lt t \le 11 \end{cases} \nonumber$$

Be sure to verify that this single vector-valued function does indeed trace out the entire path!

Exercises:

For questions 41 - 44, provide a parameterization for each piecewise path. Try to write a parameterization that starts with $t = 0$ and progresses on through values of $t$ as you move from one piece to another.

41)

Answer

a. $\vecs r_1(t)= t\,\hat{\mathbf{i}} + t^4 \,\hat{\mathbf{j}}$ for $0 \le t \le 1$
$\vecs r_2(t)= -t\,\hat{\mathbf{i}} + \sqrt[3]{-t} \,\hat{\mathbf{j}}$ for $-1 \le t \le 0$

So a piecewise parameterization of this path is:
$\vecs r(t) = \begin{cases} t\,\hat{\mathbf{i}} + t^4 \,\hat{\mathbf{j}}, & 0 \le t \le 1 \\ \left(2-t\right)\,\hat{\mathbf{i}} + \sqrt[3]{2-t} \,\hat{\mathbf{j}}, & 1 \lt t\le 2 \end{cases}$

b. $\vecs r_1(t)= t\,\hat{\mathbf{i}} + \sqrt[3]{t} \,\hat{\mathbf{j}}$ for $0 \le t \le 1$
$\vecs r_2(t)= -t\,\hat{\mathbf{i}} + (-t)^4 \,\hat{\mathbf{j}}$ for $-1 \le t \le 0$

So a piecewise parameterization of this path is:
$\vecs r(t) = \begin{cases} t\,\hat{\mathbf{i}} + \sqrt[3]{t} \,\hat{\mathbf{j}}, & 0 \le t \le 1 \\ \left(2-t\right)\,\hat{\mathbf{i}} + \left(2-t\right)^4 \,\hat{\mathbf{j}}, & 1 \lt t\le 2 \end{cases}$

42)

43)

Answer

a. $\vecs r_1(t)= t\,\hat{\mathbf{i}} +0 \,\hat{\mathbf{j}}$ for $0 \le t \le 2$
$\vecs r_2(t)= -t\,\hat{\mathbf{i}} + \left(2 + t\right) \,\hat{\mathbf{j}}$ for $-2 \le t \le -1$
$\vecs r_3(t)= -t\,\hat{\mathbf{i}} + \left(-t\right)^3 \,\hat{\mathbf{j}}$ for $-1 \le t \le 0$

So a piecewise parameterization of this path is:
$\vecs r(t) = \begin{cases} t\,\hat{\mathbf{i}}, & 0 \le t \le 2 \\ \left(4-t\right)\,\hat{\mathbf{i}} + \left(t-2\right) \,\hat{\mathbf{j}}, & 2 \lt t\le 3 \\ \left(4-t\right) \, \hat{\mathbf{i}} + \left(4-t\right)^3 \,\hat{\mathbf{j}}, & 3 \lt t\le 4 \end{cases}$

b. $\vecs r_1(t)= t\,\hat{\mathbf{i}} + t^3 \,\hat{\mathbf{j}}$ for $0 \le t \le 1$
$\vecs r_2(t)= t\,\hat{\mathbf{i}} + \left(2 - t\right) \,\hat{\mathbf{j}}$ for $1 \le t \le 2$
$\vecs r_3(t)= -t\,\hat{\mathbf{i}} + 0 \,\hat{\mathbf{j}}$ for $-2 \le t \le 0$

So a piecewise parameterization of this path is:
$\vecs r(t) = \begin{cases} t\,\hat{\mathbf{i}} + t^3 \,\hat{\mathbf{j}}, & 0 \le t \le 1 \\ t\,\hat{\mathbf{i}} + \left(2 - t\right) \,\hat{\mathbf{j}}, & 1 \lt t\le 2 \\ \left(4-t\right) \, \hat{\mathbf{i}}, & 2 \lt t\le 4 \end{cases}$

44)