91Ó°ÊÓ

In mathematics, parametric equations allow us to express coordinates like \( x \), \( y \), and \( z \) as continuous functions of an independent parameter, usually denoted as \( t \). This approach enables us to define curves in a more flexible and intuitive way rather than relying on a single equation. A parametric equation for a vector in space might look like \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \).This means that as \( t \) changes, the vector \( \mathbf{r}(t) \) traces out a path in space.

• For example, in our exercise: \( \mathbf{r}_1(t) = 2 e^{-t} \mathbf{i} + \cos t \mathbf{j} + \left(t^{2}+3\right) \mathbf{k} \) and \( \mathbf{r}_2(t) = (1-t) \mathbf{i} + t^{2} \mathbf{j} + \left(t^{3}+4\right) \mathbf{k} \).
• These equations give paths based on the value of \( t \).

In the exercise, the goal was to find a point \((2,1,3)\) where both curves\( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \) intersect, which involves solving for \( t \) in both sets of parametric equations.

A tangent line to a curve at a particular point is a straight line that touches the curve at that point without crossing it. It only "touches" the curve providing a linear approximation of the curve at that point. In vector calculus, we determine tangent lines using derivatives.When analyzing curves represented by parametric equations, we find the tangent vector by differentiating the parametric function with respect to \( t \).

• For instance, the derivative of \( \mathbf{r}_1(t) \) results in the tangent vector \( \mathbf{r}_1'(t) = -2 e^{-t} \mathbf{i} - \sin t \mathbf{j} + 2t \mathbf{k} \).
• Similarly, the derivative of \( \mathbf{r}_2(t) \) gives \( \mathbf{r}_2'(t) = - \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

At specific values of \( t \)—such as \( t=0 \) for \( \mathbf{r}_1(t) \) and \( t=-1 \) for \( \mathbf{r}_2(t) \)—these derivatives provide the direction vectors for the tangent lines at those intersection points. The tangent line's direction is crucial because it helps us find angles between curves, as seen in the next section.

The dot product is a mathematical operation that takes two vectors and returns a single scalar. It provides a means to relate the vectors' magnitudes with their directional relationship. The dot product is denoted as \( \mathbf{a} \cdot \mathbf{b} \), where \( \mathbf{a} \) and \( \mathbf{b} \) are vectors. This operation is particularly useful for finding the angle between two vectors.Here's the formula: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \] where

• \(|\mathbf{a}|\) and \(|\mathbf{b}|\) are magnitudes (lengths) of vectors \( \mathbf{a} \) and \( \mathbf{b} \),
• \( \theta \) is the angle between them.

In the exercise:

• Given vectors are \( \mathbf{a} = -2 \mathbf{i} \) and \( \mathbf{b} = - \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} \).
• The dot product \( \mathbf{a} \cdot \mathbf{b} \) gives 2.

Finding magnitudes, we get \( |\mathbf{a}| = 2 \) and \( |\mathbf{b}| = \sqrt{14} \). Thus, the acute angle between these vectors is approximated as \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{14}}\right) \), showcasing the use of dot product in identifying angles in vector calculus.