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A vector equation represents a line, plane, or space using vectors, which allow us to describe geometry in terms of direction and magnitude. In this context, the vector equation of a tangent line gives a mathematical description of the line that just "touches" a curve at a certain point.

This line can be found using a simple formula:

• \(\mathbf{R}(t) = \mathbf{R}_0 + t \mathbf{D}\)

Here,

• \(\mathbf{R}_0\) is the position vector of a known point on the line.
• \(\mathbf{D}\) is the direction vector of the line.

For example, in the solution given, after calculating the appropriate components, you can express a tangent line with the help of our vector equation. By plugging in the point where the tangent occurs (known as \(\mathbf{R}_0 \) - the origin point), and combining it with the direction vector identified from the derivative, we can encapsulate the line behavior using this equation.

• The resulting equation resembles the function of the original curve you are dealing with, emphasizing the direction and magnitude.

A tangent line is a straight line that contacts a curve at a single point and does not cross the curve at that point. It offers an "instantaneous" view of the curve's behavior.

Mathematically, a tangent line is essential for many computations because:

• It provides an approximation that allows for easier handling of curves.
• It represents the best linear approximation of the curve at that point.

In our exercise, we find the tangent line at the point \(P_{0}(0,1,0)\). By evaluating the derivative of the original curve function at \(t=0\), we get the direction of the tangent line. We then use the vector equation to clearly define this line.

With the tangent line formulated as \(\mathbf{R}(t) = t \mathbf{i} + \mathbf{j} + t \mathbf{k}\), you can then predict the behavior of other points along this linear path that closely represents the curve at the point of tangency.

In calculus, the derivative measures how a function changes as its input changes. It's the slope of the function at any point and an essential tool in understanding the curve's behavior at that point.

To find a tangent line, you must first determine the derivative of the curve function. The derivative tells us the direction in which the curve is heading as well as how steeply it is climbing or descending. Let's break down the computation step by step:

• Differentiate each component of the vector function \(\mathbf{r}(t)=\sin t \mathbf{i}+\cosh t \mathbf{j}+\left(\tan ^{-1} t\right) \mathbf{k}\).
• Use known rules of differentiation for trigonometric, hyperbolic, and inverse trigonometric functions.

1. For \(\sin t\), the derivative is \(\cos t\); 2. For \(\cosh t\), the derivative is \(\sinh t\); 3. For \(\tan^{-1} t\), the derivative is \(\frac{1}{1+t^2}\).
Combining these, the vector derivative comes out to be \(\mathbf{r}'(t) = \cos t \mathbf{i} + \sinh t \mathbf{j} + \frac{1}{1+t^2} \mathbf{k}\). Evaluating this at \(t=0\) provides us the specific direction vector for the tangent line, guiding us in writing the final vector equation.