how do you find the curvature k(t) of the curve r(t)=(-2sint)i+(-2sint)j+(-4cost)k?
To find the curvature k(t) of a curve r(t) in three-dimensional space, you need to follow these steps:
1. Calculate the first derivative, r'(t), of the given vector-valued function r(t). The derivative of a vector-valued function is obtained by differentiating its components separately. In this case, we have:
r'(t) = (-2cos(t))i + (-2cos(t))j + (4sin(t))k
2. Compute the second derivative, r''(t), by differentiating each component of r'(t):
r''(t) = (2sin(t))i + (2sin(t))j + (4cos(t))k
3. Determine the magnitude ||r'(t)|| (also known as the speed) of the first derivative. This can be done by taking the square root of the sum of the squares of its components:
||r'(t)|| = sqrt[(-2cos(t))^2 + (-2cos(t))^2 + (4sin(t))^2]
4. Calculate the magnitude ||r''(t)|| of the second derivative using the same approach:
||r''(t)|| = sqrt[(2sin(t))^2 + (2sin(t))^2 + (4cos(t))^2]
5. Finally, find the curvature k(t) by dividing ||r''(t)|| by ||r'(t)|| squared:
k(t) = ||r''(t)|| / ||r'(t)||^2
Substituting the formulas for ||r'(t)|| and ||r''(t)|| obtained in steps 3 and 4, the curvature of the given curve r(t) = (-2sin(t))i + (-2sin(t))j + (-4cos(t))k can be expressed as:
k(t) = sqrt[(2sin(t))^2 + (2sin(t))^2 + (4cos(t))^2] / [(-2cos(t))^2 + (-2cos(t))^2 + (4sin(t))^2]^2
Simplifying this expression will yield the curvature of the curve at any given point t.