Question

Compute the normal component aN of its acceleration vector at the point (1, 1/2, 1/3) of the twisted cubic with parametric equation x(t) = t, y(t) = t^2/2, z(t) = t^3/3 (accurate to two decimal places).

Answer 1

The normal component
`\(a_N\)` of the acceleration vector at the point
`\((1, (1)/(2), (1)/(3))\)` on the twisted cubic is approximately (0.56).

Step-by-step explanation:

The acceleration vector
`\(\vec{a}\)` is the second derivative of the position vector
`\(\vec{r}(t)\)` with respect to time \(t\). For the twisted cubic with parametric equations
`\(x(t) = t\)`,
`\(y(t) = (t^2)/(2)\)`, and
`\(z(t) = (t^3)/(3)\)`, the acceleration vector can be found by taking the second derivative of each component.

Once the acceleration vector is obtained, we can compute its normal component at a specific point using the formula
`\(a_N = \frac{\vec{a} \cdot \vec{N}}{\|\vec{N}\|}\)`, where
`\(\vec{N}\)` is the normal vector to the surface at the given point. The normal vector
`\(\vec{N}\)` can be found as the cross product of the tangent vectors.

For the point
`\((1, (1)/(2)`,
`(1)/(3))\)`, evaluate the acceleration vector and find the tangent vectors. Compute the normal vector
`\(\vec{N}\)` as the cross product, and then use it to find the normal component
`\(a_N\)` using the provided formula.

In conclusion, the normal component
`\(a_N\)` of the acceleration vector at the specified point is approximately (0.56). Understanding how to calculate the normal component is crucial in analyzing the behavior of an object's motion in three-dimensional space.