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# Parameterizing a Piecewise Path

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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 the 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)

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.

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)