Question

Suppose a curve is given by \(\mathbf{r}(t)=\langle f(t), g(t)\rangle,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous, for \(a \leq t \leq b .\) Assume the curve is traversed once, for \(a \leq t \leq b\) and the length of the curve between \((f(a), g(a))\) and \((f(b), g(b))\) is \(L\). Prove that for any nonzero constant \(c\), the length of the curve defined by \(\mathbf{r}(t)=\langle c f(t), c g(t)\rangle,\) for \(a \leq t \leq b,\) is \(|c| L\)

Short answer

Answer: Multiplying the components of a vector function defining a curve by a nonzero constant c scales the length of the curve by a factor of |c|.

Step-by-step solution

Length Formula for a Curve Defined by a Vector Function

The length of any curve can be given by the formula: $$L = \int_a^b \|\mathbf{r}'(t)\| dt$$ where \(\mathbf{r}'(t)\) is the derivative of the vector function and its norm represents the length of the tangent vector. This integral represents summing of all the infinitesimally small arc segments along the curve from \(t=a\) to \(t=b\).

Find the Derivative of \(\mathbf{r}(t)=\langle f(t), g(t)\rangle\)

We have \(\mathbf{r}(t)=\langle f(t), g(t)\rangle\). We need to find its derivative: $$\mathbf{r}'(t) = \langle f'(t), g'(t) \rangle$$

Compute the Length of the Curve for \(\mathbf{r}(t)\)

Using the length formula and the derivative we found in Step 2, we can compute the length of the curve given by \(\mathbf{r}(t)\): $$L = \int_a^b \|\mathbf{r}'(t)\| dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2} dt$$

Find the Derivative of \(\mathbf{r}(t)=\langle cf(t), cg(t)\rangle\)

We have \(\mathbf{r}(t) = \langle cf(t), cg(t) \rangle\). We also need to find its derivative: $$\mathbf{r}'(t) = \langle c\cdot f'(t), c\cdot g'(t) \rangle$$

Compute the Length of the Curve for \(\mathbf{r}(t)=\langle cf(t), cg(t)\rangle\)

Using the length formula and the derivative we found in Step 4, we can compute the length of the curve given by \(\mathbf{r}(t) = \langle cf(t), cg(t) \rangle\): $$L' = \int_a^b \|\mathbf{r}'(t)\| dt = \int_a^b \sqrt{(c\cdot f'(t))^2 + (c\cdot g'(t))^2} dt$$

Compare the Lengths of the Two Curves

We want to show that \(L' = |c|L\). Let's simplify the expression for \(L'\): $$L' = \int_a^b \sqrt{c^2(f'(t)^2 + g'(t)^2)} dt = |c| \int_a^b \sqrt{f'(t)^2 + g'(t)^2} dt = |c|L$$ From our simplification, it is clear that \(L' = |c|L\), proving that the length of the curve defined by \(\mathbf{r}(t) = \langle cf(t), cg(t) \rangle\) is \(|c|\) times the length of the original curve.

Key concepts

Vector Functions

Vector functions are a powerful tool in mathematics, particularly within the realm of multivariable calculus. A vector function, sometimes called a vector-valued function, assigns a vector to each variable in its domain. Most commonly, these functions are used in three-dimensional space and can be written as \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), where \( x(t) \) , \( y(t) \) , and \( z(t) \) are scalar functions that define the components of the vector function.

When it comes to a curve in space, a vector function is particularly useful because it defines the position of a point along the curve at any given time \( t \). The way the curve is traced as \( t \)'s value changes can help us understand the curve's behavior, its shape, and its geometric properties. Moreover, if we need to calculate the length of such a curve, we start with finding the vector function that defines its path.

Derivatives in Calculus

Derivatives are one of the fundamental concepts in calculus, often thought of as the rate of change of a function. If you have a scalar function, say \( f(t) \) representing some quantity with respect to time, the derivative \( f'(t) \) gives you its rate of change at any instant \( t \).

In the context of vector functions, the derivative extends this idea to higher dimensions. For the vector function \( \mathbf{r}(t) = \langle f(t), g(t) \rangle \), we look for the rate of change of each component function. Therefore, the derivative of this vector function would be \( \mathbf{r}'(t) = \langle f'(t), g'(t) \rangle \). This gives us the tangent vector at any point \( t \) on the curve, which is central to understanding the curve's geometry.

For curve length calculation, the derivative plays a critical role because we need to know the length of the tangent at many points along the curve. The integral of the norm (magnitude) of these tangent vectors then gives us the total length of the curve.

Integral Calculus

Integral calculus, the counterpart to differential calculus, concerns itself with accumulation. If derivatives are about instant rates of change, integrals are about total accumulation over an interval. The most common type of integral in calculus is the definite integral, noted as \( \int_a^b f(x) dx \) which gives the total accumulation of \( f(x) \) from \( a \) to \( b \).

In the case of curve length calculation using vector functions, we're dealing not with the accumulation of a simple scalar function but rather the cumulative length of a vector function's tangent vectors over an interval. The definite integral that gives us curve length from \( t=a \) to \( t=b \) is written as \( L = \int_a^b \|\mathbf{r}'(t)\| dt \) where \( \|\mathbf{r}'(t)\| \) represents the magnitude of the tangent vector to the curve at \( t \) — essentially, how fast the point on the curve is moving.

This integral, then, sums up all the little bits of 'movement' along the curve, giving us the entire length — a concept that showcases the power and application of integral calculus in analyzing spatial paths.