Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

Parameterizing a Piecewise Path

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 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 6: Parameterizing a piecewise path

piecewise12-1.pngDetermine a piecewise parameterization of the path shown in Figure 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 r1, r2, and r3. 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 r1: 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:

r1a(t)=tˆi+tˆjfor1t4

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 r1a in terms of t1 instead of t to make the translation easier to see.

Thus, we have r1a(t1)=t1ˆi+t1ˆj for 1t14.

Figure 4: A closed piecewise path

Subtracting 1 from each part of this range of parameter values, we have: 0t113.

Now we let t=t11. Solving for t1, we obtain: t1=t+1.

Replacing t1 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: r1(t)=(t+1)ˆi+(t+1)ˆjfor0t3

Double-check that this vector-valued function will trace out this segment in the correct direction before going on to r2.

Determining r2: This piece has a label showing the function whose graph it traces along. If it were oriented from left-to-right, we would have:

Left-to-right:r2a(t)=tˆi+(24t3+4)ˆjfor1t4

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:

r2b(t)=tˆi+(24(t)3+4)ˆjfor4t1

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 r2b in terms on t2.

r2b(t2)=t2ˆi+(24(t2)3+4)ˆjfor4t21

To force r2 to start with t=3 instead of t=4, we need to add 7 to each part of the inequality. This yields: 3t2+76.

Let t=t2+7. Then solving for t2 (since this is what we need to replace in r2b), we have: t2=7t.

Replacing t2 with (7t) in r2b, we obtain:

r2(t)=(7t)ˆi+(24(7t)3+4)ˆjfor3t6

This can be combined with our earlier result for r1 to write a piecewise-defined vector-valued function that traces out the first two pieces, starting at t=0:

r(t)={(t+1)ˆi+(t+1)ˆj,0t3(7t)ˆi+(2t33+4)ˆj,3<t6

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 r3: 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: r(t)=f(t)ˆi+tˆj.

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:

r3a(t)=1ˆi+tˆjfor1t6

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 r(t)=f(t)ˆitˆj.

In the case of r3, this gives us:

r3b(t)=1ˆitˆjfor6t1

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 t3, giving us:

r3b(t3)=1ˆit3ˆjfor6t31

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 6t3+1211.

Let t=t3+12. Then solving for t3 (since this is what we need to replace in r3b), we have: t3=12t.

Replacing t3 with (12t) in r3b, we obtain:

r3(t)=1ˆi+(12t)ˆjfor6t11

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.

r(t)={(t+1)ˆi+(t+1)ˆj,0t3(7t)ˆi+(2t33+4)ˆj,3<t61ˆi+(12t)ˆj6<t11

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)

Counterclockwise-oriented boundary of a closed region formed by y = x^4 and y equals the cube root of x. Clockwise-oriented boundary of a closed region formed by y = x^4 and y equals the cube root of x.

Answer
a. r1(t)=tˆi+t4ˆj for 0t1
r2(t)=tˆi+3tˆj for 1t0

So a piecewise parameterization of this path is:
r(t)={tˆi+t4ˆj,0t1(2t)ˆi+32tˆj,1<t2

b. r1(t)=tˆi+3tˆj for 0t1
r2(t)=tˆi+(t)4ˆj for 1t0

So a piecewise parameterization of this path is:
r(t)={tˆi+3tˆj,0t1(2t)ˆi+(2t)4ˆj,1<t2

42)

Counterclockwise-oriented boundary of a closed region formed by y = x^3 and y = 4x. Clockwise-oriented boundary of a closed region formed by y = x^3 and y = 4x.

43)

Counterclockwise-oriented boundary of a closed region formed by y = x^3 and y = 2 - x and the x-axis. Clockwise-oriented boundary of a closed region formed by y = x^3 and y = 2 - x and the x-axis.

Answer
a. r1(t)=tˆi+0ˆj for 0t2
r2(t)=tˆi+(2+t)ˆj for 2t1
r3(t)=tˆi+(t)3ˆj for 1t0

So a piecewise parameterization of this path is:
r(t)={tˆi,0t2(4t)ˆi+(t2)ˆj,2<t3(4t)ˆi+(4t)3ˆj,3<t4

b. r1(t)=tˆi+t3ˆj for 0t1
r2(t)=tˆi+(2t)ˆj for 1t2
r3(t)=tˆi+0ˆj for 2t0

So a piecewise parameterization of this path is:
r(t)={tˆi+t3ˆj,0t1tˆi+(2t)ˆj,1<t2(4t)ˆi,2<t4

44)

Counterclockwise-oriented boundary of a closed region formed by y = 1-x/2 and y = 3x/2 - 3 and y = 1 plus the square root of x. Clockwise-oriented boundary of a closed region formed by y = 1-x/2 and y = 3x/2 - 3 and y = 1 plus the square root of x.


This page titled Parameterizing a Piecewise Path is shared under a not declared license and was authored, remixed, and/or curated by Paul Seeburger.

Support Center

How can we help?