# Parameterizing a Piecewise Path

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- Contributed by Paul Seeburger
- Professor (Mathematics) at Monroe Community College

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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:

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

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: \[\vecs r_1(t) = (t+1)\,\hat{\mathbf{i}}+ (t+1) \,\hat{\mathbf{j}} \quad\text{for}\quad 0 \le t \le 3\]

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:

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

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:

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

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 \(t_2\).

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

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:

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

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}\]

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:

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

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:

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

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:

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

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:

\[\vecs r_{3}(t) = 1\,\hat{\mathbf{i}} + (12 - t)\,\hat{\mathbf{j}} \quad\text{for}\quad 6 \le t_3 \le 11\]

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_3 \le 11

\end{cases}\]

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) a. b.

**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)