find the curvature of r(t) = <t^2, ln t, t(ln t)> at the point (1,0,0)
To find the curvature of a curve, we need to evaluate three quantities: the tangent vector, the normal vector, and the curvature scalar.
1. Tangent vector (T):
The tangent vector represents the direction of motion along the curve. We can find it by taking the derivative of the vector-valued function r(t). So, let's differentiate r(t) to find T(t):
r(t) = <t^2, ln t, t(ln t)>
Differentiating each component with respect to t, we get:
T(t) = r'(t) = <2t, 1/t, ln t + t(1/t)>
2. Normal vector (N):
The normal vector represents the acceleration orthogonal to the tangent vector. We can find it by differentiating the tangent vector T(t). Let's differentiate T(t) to find N(t):
N(t) = T'(t) = <2, -1/t^2, (1/t - t/t^2)>
3. Curvature scalar (k):
The curvature scalar represents the rate at which the curve deviates from being a straight line. It can be calculated using the formula:
k(t) = |N(t)| / |T(t)|
where |N(t)| represents the magnitude of the normal vector and |T(t)| represents the magnitude of the tangent vector.
To find k(t), we need to calculate the magnitudes of N(t) and T(t), and then divide them:
|N(t)| = √(2^2 + (-1/t^2)^2 + ((1/t - t/t^2))^2)
|T(t)| = √((2t)^2 + (1/t)^2 + (ln t + t(1/t))^2)
Finally, we evaluate k(t) at the point (1, 0, 0) by substituting t = 1 into the expressions for N(t) and T(t) and then dividing the resulting magnitudes.