Question

Find the curvature \(K\) of the plane curve at the given value of the parameter. $$ \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}, \quad t=1 $$

Short answer

The curvature \( K \) of the given plane curve at \( t=1 \) is \( K = \frac{2}{5 \sqrt{5}} \).

Step-by-step solution

Compute the first derivative of the given vector

The first derivative of \( r(t) \) is \( r'(t) = \frac{dr}{dt} = \frac{d(t)}{dt} \mathbf{i} + \frac{d(t^{2})}{dt} \mathbf{j} = 1 \mathbf{i} + 2t \mathbf{j} \).

Compute the second derivative of the given vector

The second derivative of \( r(t) \) is \( r''(t) = \frac{d^2r}{dt^2} = \frac{d(1)}{dt} \mathbf{i} + \frac{d(2t)}{dt} \mathbf{j} = 0 \mathbf{i} + 2 \mathbf{j} \).

Compute the cross product of the first and second derivatives

The cross product of \( r'(t) \) and \( r''(t) \) is \( r'(t) \times r''(t) = (1 \mathbf{i} + 2t \mathbf{j}) \times (0 \mathbf{i} + 2 \mathbf{j}) = -2 \mathbf{k} \). The magnitude of the cross product, \( ||r'(t) \times r''(t)|| \), is then \( |-2| = 2 \).

Compute the magnitude of the first derivative

The magnitude of \( r'(t) \) is \( ||r'(t)|| = \sqrt{(1)^{2} + (2t)^{2}} \). Substituting \( t=1 \) gives \( ||r'(1)|| = \sqrt{1+4} = \sqrt{5} \). The cube of this magnitude is \( ||r'(1)||^{3} = (\sqrt{5})^{3} = 5 \sqrt{5} \).

Compute the curvature

Finally, the curvature \( K \) is given by \( K = \frac{||r'(t) \times r''(t)||}{||r'(t)||^3} \). Substituting the calculated values gives \( K = \frac{2}{5 \sqrt{5}} \).