Curve Representation of Vectorvalued Functions in 3D Space

Given the vector-valued function $r(t) = \langle 6\cos(t), 0, 6\sin(t) \rangle$, describe the curve in 3D space and explain the effect of a negative $y$-component like $-t$.

This problem involves understanding the representation of vector-valued functions in three-dimensional space. When dealing with such functions, it's crucial to visualize how each component affects the overall shape of the curve. In this case, the vector-valued function is composed of trigonometric components in the x and z directions, while the y component is zero. This means the curve will be a circle in the xz-plane when plotted over its domain.

When introducing a non-zero component to the y direction, especially a negative one like -t, the nature of the curve changes significantly. The curve no longer resides on a plane parallel to the xz-plane, but instead spirals downward as t increases. This is because the negative y-component adds a vertical component to the motion, pulling the curve in the negative y direction as the trigonometric components affect its circular path.

To tackle these kinds of problems, one should keep in mind the fundamental characteristics of vector functions. Analyze the individual components and their range of values, and understand how they contribute to the motion and shape of the curve in space. This is a common approach in various applications, including physics and engineering, where such vector functions describe paths of particles or objects. Conceptualizing these curves helps build intuition for more advanced topics in vector calculus, such as parametric surfaces and integrals of vector functions.

Related Problems