Question

Change of variables Consider the parameterized curves \(\mathbf{r}(t)=\) \(\langle f(t), g(t), h(t)\rangle\) and \(\mathbf{R}(t)=\langle f(u(t)), g(u(t)), h(u(t))\rangle,\) where \(f, g, h,\) and \(u\) are continuously differentiable functions and \(u\) has an inverse on \([a, b]\) a. Show that the curve generated by \(\mathbf{r}\) on the interval \(a \leq t \leq b\) is the same as the curve generated by \(\mathbf{R}\) on \(\left.u^{-1}(a) \leq t \leq u^{-1}(b) \text { (or } u^{-1}(b) \leq t \leq u^{-1}(a)\right)\) b. Show that the lengths of the two curves are equal. (Hint: Use the Chain Rule and a change of variables in the arc length integral for the curve generated by \(\mathbf{R}\).)

Short answer

In summary, we showed that the two parameterized curves \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\) generate the same curve by plugging in the bounds of the intervals and comparing the generated points. We also proved that the lengths of the two curves are equal by using the chain rule and a change of variables in the integrals used to compute the arc lengths. Therefore, we conclude that the two parameterized curves generate the same curve, and have the same lengths.

Step-by-step solution

Analyze the relationship between the curves \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\)

We have two parameterized curves as follows: \(\mathbf{r}(t) = \langle f(t), g(t), h(t)\rangle \) and \(\mathbf{R}(t) = \langle f(u(t)), g(u(t)), h(u(t))\rangle \) Notice that \(\mathbf{R}(t)\) is obtained by substituting \(u(t)\) into \(\mathbf{r}(t)\) where u is an invertible function over the interval \([a, b]\).

Prove that the curves generated by the two functions are the same on the given intervals

To show that the two curves are the same, we will plug in the bounds of the interval for the curve generated by \(\mathbf{R}(t)\) and show it generates the same points as the curve of \(\mathbf{r}(t)\). At \(t = u^{-1}(a)\), the curve generated by \(\mathbf{R}(t)\) is: \(\mathbf{R}(u^{-1}(a)) = \langle f(u(u^{-1}(a))), g(u(u^{-1}(a))), h(u(u^{-1}(a)))\rangle = \langle f(a), g(a), h(a)\rangle \) which is the starting point for the curve generated by \(\mathbf{r}(t)\). Similarly, at \(t = u^{-1}(b)\), the curve generated by \(\mathbf{R}(t)\) is: \(\mathbf{R}(u^{-1}(b)) = \langle f(u(u^{-1}(b))), g(u(u^{-1}(b))), h(u(u^{-1}(b)))\rangle = \langle f(b), g(b), h(b)\rangle\) which is the endpoint for the curve generated by \(\mathbf{r}(t)\). The two curves generate the same points on the given intervals, therefore they are the same curves.

Show that the lengths of the two curves are equal using the Chain Rule and a change of variables

Let's denote the arc length of the curve generated by \(\mathbf{r}(t)\) by \(L_r\) and the arc length of the curve generated by \(\mathbf{R}(t)\) by \(L_R\). We have: \(L_r = \int_a^b ||\mathbf{r}'(t)||dt\) Since \(\mathbf{R}(t) = \mathbf{r}(u(t))\), using the chain rule we have: \(\mathbf{R}'(t) = \mathbf{r}'(u(t)) u'(t)\) By squaring both sides and taking the square root, thus we have: \(||\mathbf{R}'(t)|| = ||\mathbf{r}'(u(t)) u'(t)|| = ||\mathbf{r}'(u(t))|| |u'(t)|\) Now, let's find the arc length for the curve generated by \(\mathbf{R}(t)\): \(L_R = \int_{u^{-1}(a)}^{u^{-1}(b)} ||\mathbf{R}'(t)|| dt = \int_{u^{-1}(a)}^{u^{-1}(b)} ||\mathbf{r}'(u(t))|| |u'(t)| dt\) Now, we will replace \(t\) with its equivalent value in terms of \(u(t)\): 1. Let \(v = u(t)\) 2. Then \(t = u^{-1}(v)\) 3. And \(\frac{dt}{dv} = \frac{1}{u'(u^{-1}(v))}\) 4. Also, when \(t=u^{-1}(a)\), \(v=a\), and when \(t=u^{-1}(b)\), \(v=b\) 5. So we can rewrite the integral as: \(L_R = \int_a^b ||\mathbf{r}'(v)|| \cancel{|u'(u^{-1}(v))|} \frac{1}{\cancel{u'(u^{-1}(v))}} dv = \int_a^b ||\mathbf{r}'(v)|| dv = \int_a^b ||\mathbf{r}'(t)|| dt\) Finally, we have: \(L_r = L_R\) Which shows that the lengths of the two curves are equal.

Key concepts

Parameterized Curves

A parameterized curve is a way to describe the position of a point in space as a function of a parameter, usually denoted as \( t \). This parameter often represents time in physical contexts, allowing us to track how the point changes its position over time. In mathematics, a parameterized curve \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \) describes a path in three-dimensional space. It is defined by three continuously differentiable functions \( f(t) \), \( g(t) \), and \( h(t) \). This allows us to break down complex curves into more manageable parts.

In the given exercise, we have two curves:\( \mathbf{r}(t) \) and \( \mathbf{R}(t) = \langle f(u(t)), g(u(t)), h(u(t)) \rangle \). The curve \( \mathbf{R}(t) \) is simply a transformation of \( \mathbf{r}(t) \) using a function \( u(t) \). This transformation shifts the parameter \( t \) by applying \( u(t) \), essentially creating the same curve but reshaping the way we navigate over it within a given interval. Understanding this dynamic reshaping through parameterization is vital for analyzing different aspects like arc length and transformations.

Arc Length

The arc length of a curve is the distance along the path the curve traces out. It provides us with a way to measure how 'long' a curve is, analogous to how we measure straight-line distance, but for curves. For a parameterized curve \( \mathbf{r}(t) \), the arc length from \( a \) to \( b \) can be computed using the integral:

\[ L_r = \int_a^b ||\mathbf{r}'(t)|| dt \]
where \( ||\mathbf{r}'(t)|| \) represents the magnitude of the derivative of \( \mathbf{r}(t) \).

In the context of the exercise, even though \( \mathbf{r}(t) \) and \( \mathbf{R}(t) \) parameterize the same curve, their representation might look different due to the change of variables. However, the arc length, as shown in the solution, remains the same for both curves. This makes arc length a crucial parameterization-invariant property, meaning it doesn't change when we reparametrize the curve.

Chain Rule

The Chain Rule is a fundamental theorem in calculus used to differentiate compositions of functions. It states that if you have two functions \( f \) and \( g \), the derivative of their composition \( f(g(x)) \) is \( f'(g(x))g'(x) \). This rule finds wide application, particularly when dealing with parameterized curves and transformations like those from \( \mathbf{r}(t) \) to \( \mathbf{R}(t) \).

In the exercise, to determine how \( \mathbf{R}(t) \) behaves, the Chain Rule is used. Here, \( \mathbf{R}'(t) \) is expressed as \( \mathbf{r}'(u(t))u'(t) \). This describes how quickly \( \mathbf{R}(t) \) changes by accounting for the sensitivity of both the curve \( \mathbf{r} \) with \( u(t) \) and the function \( u \) itself. Using the Chain Rule simplifies the task of finding the derivative of such compositions, making computations much more manageable.

Inverse Functions

Inverse functions are essential when trying to reverse the effect of a function, i.e., going from the output back to the input. In the context of parameterized curves, an inverse function \( u^{-1} \) helps adjust the parameter interval when transforming curves. If a function \( u \) maps \( a \) to \( b \) such that the outputs are transformed, then the inverse \( u^{-1} \) undoes this transformation.

In the exercise, the inverse function \( u^{-1} \) is used to translate the interval \( [a, b] \) for \( \mathbf{r}(t) \) into a corresponding interval for \( \mathbf{R}(t) \). This is crucial because when we change variables, we want to ensure that both parameterizations cover the same segment of the curve. Practically, this ensures that the start and end points align perfectly across different parameterizations, verifying that both curves are indeed the same by tracing identical paths.