Question

Sketch the plane curve and find its length over the given interval. $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{k}, \quad[0,4] $$

Short answer

The plane curve is rising more steeply as \(t\) increases, suggesting a form of parabolic trajectory in the xz-plane. The calculation of the length of this curve involves finding the derivative of the curve equation, computing its magnitude and then using these to calculate the definite integral which yields the length of the curve over the given time interval.

Step-by-step solution

Sketch The Curve

This parametric equation describes a curve that starts at the origin (when \(t=0\)) and moves along the xz-plane. As \(t\) increases, both x and z increase, but z increases at a greater rate (since \(z=t^{2}\), while \(x = t\)). Therefore, the curve begins rising more steeply as \(t\) increases. You can draw this curve using graphing software or by plotting points manually.

Find The Derivative of \(\mathbf{r}(t)\)

The length of a parametric curve requires computing the magnitude of the derivative of \(\mathbf{r}(t)\). Therefore, find \(\mathbf{r}'(t)\), which is the derivative of \(\mathbf{r}(t)\). Differentiating \(\mathbf{r}(t)\) with respect to \(t\) gives us \(\mathbf{r}'(t) = \mathbf{i} + 2t\mathbf{k}\).

Calculate The Magnitude of The Derivative

The magnitude of \(\mathbf{r}'(t)\) is given by \(||\mathbf{r}'(t)|| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1+4t^{2}}\).

Use The Formula For Length of a Parametric Curve

Substitute the value of \(||\mathbf{r}'(t)||\) in the formula \(L = \int_a^b ||\mathbf{r}'(t)|| dt\). This becomes \(L = \int_0^4 \sqrt{1+4t^{2}} dt\). This is an integral that can be computed using techniques from calculus.

Calculate The Length of The Curve

Calculate the integral from Step 4. This will require techniques of integration and possibly a table of integrals or technological help, depending on the complexity of the integral. The result gives the length of the curve from \(t=0\) to \(t=4\).

Key concepts

Vector Functions

Vector functions are fundamental in understanding various aspects of physics and engineering, especially when dealing with motion and curves in space. A vector function like \( \mathbf{r}(t) \), which was provided in the exercise, defines a vector for each value of the parameter \( t \).

Think of it as instructions for moving a point in space as time progresses. In the exercise, the vector function describes a curve in the xz-plane, signaling that for each time \( t \) there's a defined position given by \( t \mathbf{i}+t^{2} \mathbf{k} \). The \( \mathbf{i} \) and \( \mathbf{k} \) represent unit vectors along the x and z axes, respectively.

Vector functions allow us to describe complex paths, like the flight of a drone or the orbit of a planet, and understanding them opens the door to fields like kinematics and electromagnetism.

Integral Calculus

Integral calculus is like assembling the pieces of a puzzle to see the whole picture. At its heart, it deals with accumulation and adds up an indefinite number of infinitesimally small pieces. When we calculate areas, volumes, and as in this case, lengths, we are using integral calculus.

In this exercise, we're not just adding up lengths; we're computing the overall length of a curve - which can't be done using simple geometry. The integral \( \int_0^4 \sqrt{1+4t^{2}} dt \) represents the aggregation of countless infinitesimal line segments that make up the curve described by the vector function. It requires an understanding of integration techniques - from substitution to parts - that are essential artillery in a mathematician's armory.

Derivative of Vector Function

Taking the derivative of a vector function is akin to adding a speedometer to your car; it tells you not only the direction you're moving in but also how fast you're getting there. In the solution, when we determined \( \mathbf{r}'(t) \) as \( \mathbf{i} + 2t\mathbf{k} \), we found the instantaneous rate of change of our vector function.

It's this derivative that reveals the curve's behavior at each point. Think of it as peering under the hood of the vector function at each instant: Are you accelerating? Slowing down? Changing direction? The derivative charts this dynamic journey. In the context of our curve, we're not only interested in its direction but also in how it's steepening as \( t \) increases since this affects the curve's length.

Arc Length of Parametric Equations

The arc length of a curve described by parametric equations is not just a straight line measured with a ruler - it's the distance you would travel if you were walking along the curves and hills of the graph. To find it, we tally up the lengths of an infinite number of tiny, straight line segments that approximate the curve. This is where calculus shines.

The formula \( L = \int_a^b ||\mathbf{r}'(t)|| dt \) sums the magnitudes of these infinitesimal vectors along the curve. It’s like marking every step of the path with a microscopic dot and then counting them all. When students encounter integrals like \( \int_0^4 \sqrt{1+4t^{2}} dt \) for curve length, they employ calculus to effectively measure the 'walk' from the start point \( t=0 \) to the endpoint \( t=4 \) along the path dictated by the vector function. It encapsulates the curve in a single, elegant expression of its overall span.