Question

Consider the curve \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle,\) for \(t>0 .\) Find all points on the curve at which \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal.

Short answer

Based on the given curve \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle,\) find the points, if any, where the curve and its derivative, \(\mathbf{r}'(t)\), are orthogonal.

Step-by-step solution

Find the derivative of the curve

To find the derivative of the curve \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle,\) differentiate each component of the curve with respect to \(t\): \(\mathbf{r}'(t)=\langle\frac{d(\sqrt{t})}{dt},\frac{d(1)}{dt},\frac{d(t)}{dt}\rangle=\langle\frac{1}{2\sqrt{t}},0,1\rangle.\)

Compute the dot product of the curve and its derivative

Compute the dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}'(t)\): \(\mathbf{r}(t) \cdot \mathbf{r}'(t)=(\sqrt{t})(\frac{1}{2\sqrt{t}}) + (1)(0) + (t)(1) = \frac{1}{2} + t.\)

Equate the dot product to zero and solve for \(t\)

Set the dot product equal to zero and solve for \(t\): \(\frac{1}{2} + t =0 \implies t =- \frac{1}{2}\) Since \(t\) must be positive, there are no positive real solutions for \(t\).

Find the corresponding points on the curve

Since there are no positive real solutions for \(t\), there are no points on the curve at which \(\mathbf{r}\) and \(\mathbf{r}'\) are orthogonal.

Key concepts

Orthogonal Vectors

In vector calculus, orthogonal vectors play a vital role, particularly when studying curves. Two vectors are considered orthogonal if their dot product is zero. This means that the vectors are perpendicular to each other. In practical terms, if you imagine two vectors starting from the same point, they will form a right angle if they are orthogonal.

In the context of curves, determining the orthogonality between a curve vector \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}'(t)\) can reveal interesting properties about the curve. Specifically, points at which these vectors are orthogonal often highlight changes in the curve's direction or other significant features. This idea is useful for analyzing the geometric behavior of the curve.

Curve Derivative

The derivative of a curve in vector calculus represents how the curve changes with respect to its parameter, typically time \(t\). Finding the derivative \(\mathbf{r}'(t)\) involves differentiating each component of the curve's vector function.

For a curve given by \(\mathbf{r}(t) = \langle \sqrt{t}, 1, t \rangle\), the process involves:

• Taking the derivative of each component.
• Deriving \(\frac{1}{2\sqrt{t}}\) for the first component, \0\ for the constant term, and \1\ for the linear term \(t\).

This results in the derivative vector \(\mathbf{r}'(t) = \langle \frac{1}{2\sqrt{t}}, 0, 1 \rangle\).

By interpreting \(\mathbf{r}'(t)\), one can understand the rate and direction of the change of the curve at any point along the parameter \(t\).

Dot Product

The dot product is a fundamental concept in vector calculus used to determine the angle between two vectors. The dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as the sum of the products of their corresponding components. This operation yields a scalar rather than a vector.

To apply this to our curve example, compute the dot product of the vector \(\mathbf{r}(t)\) and its derivative \(\mathbf{r}'(t)\):

\[\mathbf{r}(t) \cdot \mathbf{r}'(t) = (\sqrt{t})(\frac{1}{2\sqrt{t}}) + (1)(0) + (t)(1) = \frac{1}{2} + t.\] This reveals how close the two vectors are to being perpendicular (orthogonal). If this product equals zero, the vectors form a right angle. However, in this specific case, setting the result to zero \(\frac{1}{2} + t = 0\) and solving gives \(t = -\frac{1}{2}\), outside the valid range for \(t\).

Parametric Equations

Parametric equations describe geometry, like curves, by expressing the coordinates of the points as functions of a separate parameter \(t\). Unlike regular equations, parametric forms allow for more flexibility and greater control over the shape and orientation of the curve.

The given curve \(\mathbf{r}(t) = \langle \sqrt{t}, 1, t \rangle\) precisely defines the path in three-dimensional space, where each component depends directly on 't'.

• The first component, \(\sqrt{t}\), varies with \(t\), showing the horizontal stretching as the parameter changes.
• The second component is constant, contributing to the alignment of the curve along the vertical axis.
• The last component, \(t\), ensures the curve spans linearly with 't' along the third dimension.

Parametric equations like this allow us to map complex shapes that might be hard to represent with a single equation, offering a versatile approach to exploring and understanding curves in a multidimensional space.