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To grasp the concept of a tangent vector, we must first compute the derivative of a vector function. Vector functions describe the position of a point along a curve in space. The derivative of a vector function provides a new vector that shows the rate and direction at which the point moves along the curve.

For a given vector function \( \mathbf{r}(t) = \langle e^t, e^{3t}, e^{5t} \rangle \), we compute the derivative by independently differentiating each component with respect to the parameter \( t \). This gives:

• For \( e^t \), the derivative is \( e^t \).
• For \( e^{3t} \), using the chain rule, the derivative is \( 3e^{3t} \).
• For \( e^{5t} \), again applying the chain rule, the derivative is \( 5e^{5t} \).

This leads us to the derivative of the vector function \( \frac{d \mathbf{r}}{dt} = \langle e^t, 3e^{3t}, 5e^{5t} \rangle \).

Understanding this derivative is crucial, as it forms the basis for identifying a tangent vector. This derivative represents the instantaneous motion along the curve.

When we talk about curve tangency, we're referring to the concept of a tangent line or vector that just "touches" a curve at a particular point. Unlike a secant line, which cuts through a curve, a tangent vector doesn't cross the curve but more so grazes it. This property gives the tangent vector its immediate direction at the specified point.

The process of finding a tangent vector involves two primary steps. Firstly, we locate the derivative of the vector function. This provides a vector that indicates the direction we would travel if we moved infinitesimally from our current point along the curve.

Secondly, we evaluate that derivative at a specific point—in this case, where the parameter \( t \) equals a given value. This specific step gives us the exact tangent vector at that particular instance along the curve. It succinctly encapsulates the direction and speed of movement along the curve at that point.

Evaluating a derivative at a specific point transforms a general rate of change into a specific tangent vector. This step is crucial for understanding precise behaviors of functions at exact locations.

For the vector function \( \mathbf{r}(t) = \langle e^t, e^{3t}, e^{5t} \rangle \), we first computed its derivative to be \( \frac{d \mathbf{r}}{dt} = \langle e^t, 3e^{3t}, 5e^{5t} \rangle \).

By substituting \( t = 0 \) into this expression, we evaluate each component at this specific point:

• \( e^0 = 1 \)
• \( 3e^{3 \cdot 0} = 3 \times 1 = 3 \)
• \( 5e^{5 \cdot 0} = 5 \times 1 = 5 \)

This results in the tangent vector \( \langle 1, 3, 5 \rangle \) at \( t = 0 \).

Thus, evaluating derivatives at a point solidifies the tangential direction at that precise instance, providing insights into both the magnitude and orientation of motion on the curve at that point.