Question

An object moves along a path given by $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ for \(0 \leq t \leq 2 \pi\) a. Show that the curve described by \(\mathbf{r}\) lies in a plane. b. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the curve described by \(\mathbf{r}\) is a circle?

Short answer

A short answer could be as follows: Given the parametric representation of a curve by the vector function \(\mathbf{r}(t)\), we find that the curve lies in a plane when the scalar triple product \([\mathbf{r}(t),\mathbf{r}'(t),\mathbf{r}''(t)] = 0\). For the curve to represent a circle, it must have a constant magnitude of the first derivative (speed) and a constant radius. The required condition for the curve to be a circle is \(ab + cd + ef = 0\).

Step-by-step solution

Find the tangent vector

To find the tangent vector, we need to calculate the derivative of the vector function \(\mathbf{r}(t)\) with respect to \(t\). Using the rules of differentiation, we have: $$\mathbf{r}'(t)=\langle -a \sin t + b \cos t, -c \sin t + d \cos t, -e \sin t + f \cos t\rangle$$

Find the normal vector to the curve

We can find the normal vector to the curve by calculating the second derivative of the vector function \(\mathbf{r}(t)\) with respect to \(t\). Using the rules of differentiation, we have: $$\mathbf{r}''(t)=\langle -a \cos t - b \sin t, -c \cos t - d \sin t, -e \cos t - f \sin t\rangle$$

Calculate the scalar triple product of \(\mathbf{r}(t)\), \(\mathbf{r}'(t)\), and \(\mathbf{r}''(t)\)

Dot product the first derivative \(\mathbf{r}'(t)\) with the cross product of \(\mathbf{r}(t)\) and \(\mathbf{r}''(t)\) to check if they lie in the same plane. If the scalar triple product is equal to 0, then the curve lies in a plane. $$[\mathbf{r}(t),\mathbf{r}'(t),\mathbf{r}''(t)] = \mathbf{r}'(t) \cdot (\mathbf{r}(t) \times \mathbf{r}''(t))$$ By taking the cross product of \(\mathbf{r}(t)\) and \(\mathbf{r}''(t)\) and then the dot product with \(\mathbf{r}'(t)\), we find that the scalar triple product is indeed 0. Thus, the curve lies in a plane.

Analyze conditions for the curve to be a circle

For the curve to be a circle, it must satisfy the basic properties of a circle. In particular, the distance from the center of the circle to any point on the curve should be constant. To satisfy this condition, the following must hold true: 1. The magnitude of the first derivative vector (i.e. the speed value) must be a constant. 2. The radius of the circle should be the same for all \(t\) values. Let's analyze these conditions.

Check if the magnitude of the first derivative is constant

The magnitude of the first derivative (the tangent vector) is given by: $$|\mathbf{r}'(t)| = \sqrt{(-a \sin t + b \cos t)^2 + (-c \sin t + d \cos t)^2 + (-e \sin t + f \cos t)^2}$$ Simplifying the expression, we have: $$|\mathbf{r}'(t)|^2 = a^2 + b^2 + c^2 + d^2 + e^2 + f^2 - 2(ab + cd + ef)\cos t\sin t$$ To make the magnitude of the tangent vector constant, the term with \(\cos t\sin t\) should be zero. Thus, the condition for this to hold true is: $$ab + cd + ef = 0$$

Check if the radius is constant

The distance from a point on the curve to the center of the circle (the radius) can be written as the magnitude of the position vector \(\mathbf{r}(t)\). Let's calculate \(|\mathbf{r}(t)|^2\) and understand the conditions under which it would remain constant for all \(t\) values. $$ |\mathbf{r}(t)|^2 =|a \cos t + b \sin t|^2 + |c \cos t + d \sin t|^2 + |e \cos t + f \sin t|^2$$ Simplifying this gives: $$ |\mathbf{r}(t)|^2 = a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + 2(ab + cd + ef) \cos^2 t$$ We have already found the condition that makes \(2(ab + cd + ef)\) equal to zero. When this condition is satisfied, the radius will be constant, and the curve will be a circle.

Conclusion

To summarize, the curve described by \(\mathbf{r}\) lies in a plane. For it to be a circle, the following condition must be satisfied: $$ab + cd + ef = 0$$

Key concepts

Tangent Vector

The term tangent vector is critical in understanding the motion of an object along a path. It is, simply put, the vector that is tangent to the path at any given point. To compute it, we differentiate the vector function that represents the path. In the context of our exercise, the tangent vector \( \mathbf{r}'(t) \) is calculated by differentiating \( \mathbf{r}(t) \) with respect to \( t \).

This differentiation process, as seen in the solution steps, provides us with the velocity of the object at each point \( t \) along the curve. The magnitude of this velocity can reveal a lot about the motion; for instance, if it remains constant, we may infer an important attribute of the path, such as it being circular. Understanding the concept of the tangent vector is fundamental in recognizing how objects move within a space.

Normal Vector

Complementary to the tangent vector, the normal vector serves as an orthogonal (perpendicular) vector to the curve at a given point. To find it, one might compute the second derivative of the vector function, finding \( \mathbf{r}''(t) \).

This vector points towards the center of curvature of the path and thus describes how the path is curving at that point. If you have a curve with a constant radius, like a circle, the normal vector at any point will always point towards the center of that circle. For our exercise, calculating the normal vector provides information on the curvature of the path an object takes. By examining the second derivative, we gain further insight into the motion and the forces that may be acting upon the object.

Scalar Triple Product

The concept of the scalar triple product gives us a method to determine if three vectors lie within the same plane. This is found by performing a dot product on one vector with the cross product of the other two vectors, often denoted with the notation \( [\mathbf{r}(t), \mathbf{r}'(t), \mathbf{r}''(t)] \).

If the result of this operation is zero, then the three vectors are coplanar, meaning they describe a geometry confined to a single plane. This relevance to our exercise is pivotal in establishing that the path described by \( \mathbf{r}(t) \) does indeed lie within a plane, as is the case with circular motions or plane curves.

Vector Differentiation

The process of vector differentiation is at the heart of understanding motion along a path in vector functions. It involves finding the rate at which the vector function changes with respect to an independent variable, often time. In this exercise, differentiating the position vector function \( \mathbf{r}(t) \) with respect to \( t \) not only yields the tangent vector but also allows further investigation into the characteristics of motion.

Through vector differentiation and subsequent analysis, such as checking the magnitude's consistency, we can deduce whether the motion describes a circle by ensuring the derivative’s magnitude remains constant. Thus, comprehending vector differentiation is instrumental in unraveling the behaviors of complex motions in multidimensional space.