Question

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}.\) 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

Answer: No, there are no points on the curve at which \(\mathbf{r}(t)\) and \(\mathbf{ristonelfile}'(t)\) are orthogonal for \(t>0\).

Step-by-step solution

Derivative of \(\mathbf{r}(t)\)

To find the derivative of the curve, we will take the derivative of each component with respect to \(t\). So we have: $$ \mathbf{r}'(t) = \left\langle\frac{d}{dt}(\sqrt{t}), \frac{d}{dt}(1), \frac{d}{dt}(t)\right\rangle = \left\langle\frac{1}{2\sqrt{t}}, 0, 1\right\rangle. $$

Dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}'(t)\)

Compute the dot product: $$ \mathbf{r}(t) \cdot \mathbf{r}'(t) = \langle\sqrt{t}, 1, t\rangle \cdot \left\langle\frac{1}{2\sqrt{t}}, 0, 1\right\rangle = (\sqrt{t})(\frac{1}{2\sqrt{t}}) + (1)(0) + (t)(1). $$ Simplifying the dot product expression, we have: $$ \mathbf{r}(t) \cdot \mathbf{r}'(t) = \frac{1}{2} + t. $$

Solve for \(t\)

Equating the dot product to zero, we have: $$ \frac{1}{2} + t = 0. $$ Solving for \(t\), we get: $$ t = -\frac{1}{2}. $$ However, since it is given that \(t > 0\), there are no valid solutions for \(t\) in this problem.

Find the corresponding points

Since there are no valid solutions for \(t\), we cannot find any points on the curve \(\mathbf{r}(t)\) at which \(\mathbf{r}\) and \(\mathbf{r}'\) are orthogonal.

Key concepts

Derivative of Vector Functions

When we talk about the derivative of vector functions, we are essentially looking at how a vector-valued function changes over time or another variable. In the case of vector functions, the derivative is computed component-wise, meaning each part of the vector is differentiated independently with respect to the variable.

For a vector function like \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle\), we take the derivative of each component, which gives us \(\mathbf{r}'(t)\). In this example, the component \(\sqrt{t}\) becomes \(\frac{1}{2\sqrt{t}}\), the constant '1' has a derivative of zero, and for \(t\), it's simply '1'. The resulting derivative vector is \(\mathbf{r}'(t) = \left\langle\frac{1}{2\sqrt{t}}, 0, 1\right\rangle\).

This step is crucial because the derivative tells us the velocity of the object at any point in time if we were referring to a motion along the curve. The derivative can also help us understand the rate of change in various applications like physics, economics, and other fields.

Dot Product

Moving on to another key concept, the dot product comes into play when we want to determine the angle between two vectors. Particularly, if two vectors are orthogonal, their dot product is zero. The dot product is a scalar value found by multiplying corresponding components of two vectors and then summing up those products.

In this exercise, we calculate \(\mathbf{r}(t) \cdot \mathbf{r}'(t)\) using the components from each vector. We have \(\sqrt{t}\) times \(\frac{1}{2\sqrt{t}}\) for the first components, the middle terms cancel out since one component is zero, and for the last components, we have \(t\) times '1' which is just \(t\). Simplifying yields \(\frac{1}{2} + t\).

For vectors to be orthogonal at any point on the curve, this dot product must equal zero. However, after setting this up in Step 3, we find that there are no positive values of \(t\) that satisfy this condition. This outcome indicates that \(\mathbf{r}(t)\) and \(\mathbf{r}'(t)\) cannot be orthogonal for any \(t>0\), as specified in the exercise parameters.

Parametric Equations

Lastly, let's delve into parametric equations, which are a powerful way to describe a curve in space by defining its components as separate equations that depend on a common parameter, typically \(t\), the independent variable. Parametric equations break free from the constraints of the typical \(y\) versus \(x\) format and allow us to model more complex curves and motions.

For the curve \(\mathbf{r}(t)=\langle\sqrt{t}, 1, t\rangle\), each value of \(t\) corresponds to a point in three-dimensional space, with \(\sqrt{t}\) as the x-coordinate, '1' as the y-coordinate, and \(t\) as the z-coordinate. The use of parametric equations enables us to trace the path of an object or plot the graph of a function that isn't explicitly solvable for \(y\) in terms of \(x\).

In this scenario, we use parametric equations to analyze characteristics of the curve like direction, curvature, and, as seen with the dot product, orthogonality to its derivative vector. However, as previously mentioned, the nonexistence of a positive \(t\) value satisfying the orthogonality condition means we won't find any specific points that meet this criteria on the given curve.