Question

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

Question: Prove that the curves generated by the parameterizations \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\) are the same and have equal lengths, where: \(\mathbf{r}(t) = \langle f(t), g(t), h(t)\rangle \) on the interval \( a \leq t \leq b\), and \(\mathbf{R}(t) = \langle f(u(t)), g(u(t)), h(u(t))\rangle \) on the interval \( u^{-1}(a) \leq t \leq u^{-1}(b)\) (or \(u^{-1}(b) \leq t \leq u^{-1}(a)\)). Answer: Since we can write \(\mathbf{R}(t)\) as \(\mathbf{R}(t) = \mathbf{r}(u(t))\), both curves cover the same set of points in the specified interval, and hence they are the same. The arc lengths of the curves generated by \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\) are calculated as \(\int_a^b \Vert \mathbf{r}'(t) \Vert \,dt \) and \(\int_{u^{-1}(a)}^{u^{-1}(b)} \Vert \mathbf{R}'(t) \Vert \,dt \), respectively. Using a change of variables, we can show that these integrals are equal, which means the lengths of the curves are also equal.

Step-by-step solution

Prove the curves are the same

We have to show that the curves generated by the parameterizations \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\) are the same when the specified conditions are satisfied. Recall that the both functions generate the same curve if they cover the same set of points in 3D space when t varies in their respective intervals. We have \(\mathbf{r}(t) = \langle f(t), g(t), h(t)\rangle \) on the interval \( a \leq t \leq b\), and \(\mathbf{R}(t) = \langle f(u(t)), g(u(t)), h(u(t))\rangle \) on the interval \( u^{-1}(a) \leq t \leq u^{-1}(b)\) (or \(u^{-1}(b) \leq t \leq u^{-1}(a)\)). We can write \(\mathbf{R}(t)\) as \(\mathbf{R}(t) = \langle f(u(t)), g(u(t)), h(u(t))\rangle = \mathbf{r}(u(t))\) Now, for the interval \( u^{-1}(a) \leq t \leq u^{-1}(b)\) (or \(u^{-1}(b) \leq t \leq u^{-1}(a)\)), we have \( \mathbf{R}(t) = \mathbf{r}(u(t))\) This shows that both curves cover the same set of points in the specified interval, and hence they are the same.

Prove the lengths are equal

To prove that the lengths of the two curves are equal, we need to calculate their arc lengths. For the curve generated by \(\mathbf{r}(t)\), the arc length is given by $$ \int_a^b \Vert \mathbf{r}'(t) \Vert \,dt $$ Now let's consider the curve generated by \(\mathbf{R}(t)\). We first find the derivative of \(\mathbf{R}(t)\) with respect to \(t\) by using chain rule: \( \mathbf{R}'(t) = \frac{d}{dt}\mathbf{r}(u(t)) = \mathbf{r}'(u(t))u'(t) \) Then, the length of the curve generated by \(\mathbf{R}(t)\) is given by the integral $$ \int_{u^{-1}(a)}^{u^{-1}(b)} \Vert \mathbf{R}'(t) \Vert \,dt $$ Since \(\mathbf{R}'(t) = \mathbf{r}'(u(t))u'(t)\) , we have $$ \Vert \mathbf{R}'(t) \Vert = \Vert \mathbf{r}'(u(t))u'(t) \Vert = \Vert \mathbf{r}'(u(t)) \Vert \cdot \Vert u'(t) \Vert $$ Now let's use a change of variables in the integral. Let \(x = u(t)\) and \(dx = u'(t)dt\). We get $$ \int_{u^{-1}(a)}^{u^{-1}(b)} \Vert \mathbf{R}'(t) \Vert \,dt = \int_a^b \Vert \mathbf{r}'(x) \Vert \cdot \Vert u'(t) \Vert \frac{dx}{u'(t)} $$ Cancelling out the terms, we get $$ \int_{u^{-1}(a)}^{u^{-1}(b)} \Vert \mathbf{R}'(t) \Vert \,dt = \int_a^b \Vert \mathbf{r}'(x) \Vert \,dx $$ This shows that the lengths of the curves generated by \(\mathbf{r}(t)\) and \(\mathbf{R}(t)\) are equal.

Key concepts

Chain Rule

In calculus, the chain rule is a central formula that allows us to find the derivative of a composite function. It essentially tells us how to differentiate a function that is nested within another function. This can be particularly useful when dealing with parameterized curves, as functions often depend on other functions through their parameters.

For example, if we have a function \(h(x)\) which is composed of functions \(f(g(x))\), the chain rule states that the derivative of \(h\) with respect to \(x\) is the product of the derivative of \(f\) with respect to \(g\) and the derivative of \(g\) with respect to \(x\): \[ h'(x) = f'(g(x)) \cdot g'(x) \.\]
The chain rule played a crucial role in the step-by-step solution for verifying that the lengths of the parameterized curves are equal. It enabled us to differentiate the composite function \(\mathbf{R}(t)\), which depended on the function \(u(t)\), successfully leading us to an integral expression for the arc length of the curve.

Arc Length of Curve

Arc length is an important concept in calculus, especially when it comes to analyzing curves. It measures the 'distance' along a curve between two points. The arc length of a parameterized curve can be found by integrating the magnitude of the derivative of the curve with respect to its parameter over the given interval.

The integral that finds the arc length \(s\) of a curve \(\mathbf{r}(t)\) from \(a\) to \(b\) is:
\[\ s = \int_a^b \Vert \mathbf{r}'(t) \Vert \,dt \]
This integral accumulates the lengths of infinitesimal straight line segments that approximate the curve when the segments are extraordinarily close. The challenge in finding the arc length arises when dealing with curves that have non-standard parameterizations or include composite functions — situations where techniques like a change of variables, as seen in the textbook solution, become essential.

Differentiable Functions

Differentiable functions are those that have a derivative at every point in their domain. In other words, they are functions that are smooth and have no sharp corners or discontinuities. Continuously differentiable functions not only have a derivative, but also have a derivative that is a continuous function.

In our exercise, the requirement that the functions \(f, g, h\), and \(u\) are continuously differentiable ensures that we can apply the chain rule reliably. This smoothness is what allows us to differentiate and integrate across the full interval from \(a\) to \(b\) without running into issues of undefined behavior. Furthermore, it is this requirement that justifies the arc length computation, as differentiability guarantees that the tangent vector \(\mathbf{r}'(t)\) is well-defined at every point along the curve.