Question

Sketch the space curve and find its length over the given interval. $$ \mathbf{r}(t)=\mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} $$ $$ [0,2] $$

Short answer

The sketched curve would start from the point (1,0,0) and move away from the z-axis, accelerating in the positive y direction. The length of the curve over the interval [0, 2] is \( 2\sqrt{13} \).

Step-by-step solution

Sketch the Curve

To acquire an intuitive understanding of the function and its behavior, it's often helpful to sketch the curve. Since it's a 3-dimensional curve, it's more complex than sketching a 2D curve. However, we can plot the curve by viewing it from different angles or sketching projections of it onto two dimensions. By inspecting the function \( r(t) = \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \), we can see that as \( t \) increases, the curve is going to go in a positive direction along the x-axis, as the x-coordinate (i-component) is always 1. In y-z plane, the curve accelerates away from the z-axis as \( t \) increases, due to the \( t^{2} \) and \( t^{3} \) terms. This gives us a rough idea of how the curve will look.

Calculate the Derivative

The length of a curve given by a vector function \( r(t) \) from \( a \) to \( b \) is given by \( \int_a^b |r'(t)| dt \). Therefore, we first need to calculate the derivative \( r'(t) \). Differentiating the function \( r(t) = \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \) with respect to \( t \), we have \( r'(t) = 2t \mathbf{j} + (3t^{2})\mathbf{k} \).

Find the Magnitude

Next, find the magnitude of the derivative \( r'(t) = 2t \mathbf{j} + (3t^{2})\mathbf{k} \). The magnitude of a vector \( v = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by \( |v| = \sqrt{a^2 + b^2 + c^2} \). Here, a=0, b=2t, and c=3t^2. So, \( |r'(t)| = \sqrt{(2t)^{2} + (3t^{2})^{2}} = \sqrt{13}t \).

Compute the Length

Using the formula for the length of a curve \( \int_a^b |r'(t)| dt \), we substitute the limits of interval [0, 2] and our value of \( |r'(t)| = \sqrt{13}t \), we find the length of the curve over the interval [0, 2]. It’ll be \( \int_0^2 \sqrt{13}t dt = \left. \frac{\sqrt{13}t^{2}}{2} \right|_0^2 = 2\sqrt{13} \).

Key concepts

Vector Function Calculus

Understanding the calculus of vector functions is essential for tasks like sketching space curves and determining their length. A vector function, such as \( \mathbf{r}(t) = \mathbf{i} + t^{2} \mathbf{j} + t^{3} \mathbf{k} \), describes a curve in three dimensions by expressing the x, y, and z coordinates as functions of a parameter, typically \( t \).

When working with vector functions, calculus involves finding derivatives and integrals, similar to single-variable calculus but applied to each of the component functions. While calculating lengths, areas, or volumes, we use integrals. The length of a space curve describes how \