Question

Find the curvature \(K\) of the curve. $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k} $$

Short answer

The curvature \( K \) of the curve \( r(t) = ti + t^2j + \frac{t^2}{2}k \) is \( K = \frac{\sqrt{5 + 4t^2}}{(\sqrt{1 + 5t^2})^3} \)

Step-by-step solution

Compute the first and second derivative

Based on the function \( r(t)=ti + t^2j + \frac{t^2}{2}k \), for the first derivative \( r'(t) = i + 2tj + t*k \). Apply the differentiation rule again to obtain the second derivative which results in \( r''(t) = 2j + k \)

Calculate the cross product of r'(t) and r''(t)

Remember that the cross product of two vectors yields another vector. Here, the cross product is obtained as \( r'(t) \times r''(t) = (2i - j - 2tk) \)

Get the norm of the cross product and r'(t)

Magnitude of r'(t) will be \( \|r'(t)\| = \sqrt{1^2+(2t)^2+t^2} = \sqrt{1+5t ^2} \). Similarly, evaluate magnitude of the cross product as \( \|r'(t) \times r''(t)\| = \sqrt{(2)^2+(-1)^2+(-2t)^2} = \sqrt{5+4t^2} \)

Enter values in the curvature formula

Substitute obtained values in the formula to get \( K = \frac{\|r'(t) \times r''(t)\|}{\|r'(t)\|^3} = \frac{\sqrt{5 + 4t^2}}{(\sqrt{1 + 5t^2})^3} \).

Key concepts

Derivatives in Vector Functions

Understanding derivatives in vector functions is crucial when dealing with curves in three-dimensional space. A vector function, like \( \mathbf{r}(t) \) in the exercise, represents a curve by assigning a vector to each value of \( t \) that corresponds to a point in space. The derivative of this function, \( \mathbf{r}'(t) \) signifies the rate of change of the position vector as \( t \) changes, which corresponds to the tangent vector of the curve at each point.

When we compute \( \mathbf{r}'(t) \) for our given function, we differentiate each component of the vector function with respect to \( t \) independently, resulting in the components \( i + 2tj + tk \). Similarly, taking the second derivative, \( \mathbf{r}''(t) \) gives us the acceleration vector of the curve, which is \( 2j + k \) for our case. This second derivative is essential for finding the curvature as it helps determine how the direction of the tangent vector changes.

Cross Product of Vectors

The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space which results in a third vector that is perpendicular to the plane containing the first two vectors. It is denoted by a \( \times \) between the two vectors.

The rule for computing the cross product involves multiplying the magnitudes of the vectors with the sine of the angle between them and a direction given by the right-hand rule. But when in coordinate form, as in our exercise, we calculate it by finding the determinant of a matrix constructed from the two vectors. As in our example, \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) yields \( 2\mathbf{i} - \mathbf{j} - 2t\mathbf{k} \), which is crucial in finding curvature as it provides the numerator in the curvature formula, representing the rate of change of the tangent vector with respect to the path's length.

Norm of a Vector

The norm of a vector, often referred to as the magnitude or length of a vector, is a measure of its size. Mathematically, for a vector \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), the norm is found using the euclidean distance formula: \( \|\textbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).

In the context of curvature, the norms of \( \mathbf{r}'(t) \) and the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) are necessary to calculate the curvature \( K \). For the given function, the norm of \( \mathbf{r}'(t) \) is \( \sqrt{1 + 4t^2 + t^2} \) and the norm of the cross product is \( \sqrt{5 + 4t^2} \). These give us a measure of the 'strength' of the vectors which is important because curvature intuitively represents the amount by which a curve deviates from being a straight line, and so it is influenced by the magnitude of changes in the direction – measured by these norms.