How to show a parametric equation is continuous?

In summary: The curve has a vertical tangent at that point, which corresponds to the discontinuity in the derivative. So in summary, a parametric equation, r(t), is considered smoothly parametrized if its derivative is continuous and does not equal zero for all t in the domain of r. For example, the tractrix, r(t)=(t-tanht)i+(secht)j, is smoothly parametrized except at t=0, where its derivative is discontinuous.
  • #1
WalkingInMud
5
0
A parametric equation, say r(t), is smoothly parametrized if:

1. its derivative is continuous, and
2. its derivative does not equal zero for all t in the domain of r.

Now that sounds simple enough. Now let's say we have the tractrix:

r(t) = (t-tanht)i + sechtj, ...

then r'(t) = [ 1/(1+x^2) ]i + [ tantsect ]j, right?


Now, without reverting to MatLab or Maple to view the graph, how do we mathematically explain that r'(t) is continuous (or not)?

Do I just state that is is/isn't -by inspection, or ...?
 
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  • #2
WalkingInMud said:
A parametric equation, say r(t), is smoothly parametrized if:

1. its derivative is continuous, and
2. its derivative does not equal zero for all t in the domain of r.

Now that sounds simple enough. Now let's say we have the tractrix:

r(t) = (t-tanht)i + sechtj, ...

then r'(t) = [ 1/(1+x^2) ]i + [ tantsect ]j, right?
No, not right. r'(t)= [1- sech t]i- tanh t sech t j/b]. I don't know why you have changed to "x" in the first component but you surely should not have tan and sec rather than tanh and sech in the second component.


Now, without reverting to MatLab or Maple to view the graph, how do we mathematically explain that r'(t) is continuous (or not)?

Do I just state that is is/isn't -by inspection, or ...?
A vector valued function is continuous if and only if its components are continuous. Where are sech t and tanh t continuous?
 
  • #3
How does this look?

r(t)=(t-tanht)i+(secht)j ...so
r'(t)=(1-secht)i+(-tanhtsecht)j

Now, secht is continuous for all real t, and also
tanht is continuous for all real t, BUT...

since tanh(0)=0 i.e. (-tanhtsecht)=0 for t=0, and furthermore sech(0)=1 i.e. (1-secht)=0 for t=0,

...r'(0)=0 so r(t) is smoothly parametrized for all t NOT EQUAL TO 0

Is this OK?
 
  • #4
WalkingInMud said:
r(t)=(t-tanht)i+(secht)j ...so
r'(t)=(1-secht)i+(-tanhtsecht)j

Now, secht is continuous for all real t, and also
tanht is continuous for all real t, BUT...

since tanh(0)=0 i.e. (-tanhtsecht)=0 for t=0, and furthermore sech(0)=1 i.e. (1-secht)=0 for t=0,

...r'(0)=0 so r(t) is smoothly parametrized for all t NOT EQUAL TO 0

Is this OK?
Yes, that is correct. And if you look at the graph of the tractrix, you can see what happens at t= 0.
 

1. How do you define continuity for a parametric equation?

Continuity for a parametric equation is defined as the property of a function where small changes in the input result in small changes in the output. In other words, as the input values get closer and closer, the output values also get closer and closer.

2. What is the importance of showing continuity in a parametric equation?

Showing continuity in a parametric equation is important because it ensures that the function is well-behaved and predictable. It also allows us to make accurate predictions and draw conclusions about the behavior of the function.

3. How do you show that a parametric equation is continuous at a specific point?

To show that a parametric equation is continuous at a specific point, we need to check three conditions: 1) the function is defined at that point, 2) the limit of the function exists at that point, and 3) the limit is equal to the value of the function at that point.

4. Can a parametric equation be continuous but not differentiable?

Yes, a parametric equation can be continuous but not differentiable. This can happen if the function has sharp turns or corners, which would make it impossible to find a derivative at those points.

5. How can we use the graph of a parametric equation to show continuity?

We can use the graph of a parametric equation to show continuity by looking for any breaks, jumps, or discontinuities in the graph. If there are none, then the function is continuous. Additionally, we can zoom in on a specific point on the graph and see if the function remains smooth and connected, which would also indicate continuity.

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