Problem 212 - Calculus Volume 3

Question

Find the derivatives of \(\mathbf{u}(t), \quad \mathbf{u}^{\prime}(t), \quad \mathbf{u}^{\prime}(t) \times \mathbf{u}(t),\) \(\mathbf{u}(t) \times \mathbf{u}^{\prime}(t),\) and \(\mathbf{u}(t) \cdot \mathbf{u}^{\prime}(t) .\) Find the unit tangent vector. \(\mathbf{u}(t)=\left\langle e^{t}, e^{-t}\right\rangle\)

Step-by-Step Solution

Answer

The derivatives are \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \), all cross products are zero, \( \mathbf{u}(t) \cdot \mathbf{u}'(t) = e^{2t} - e^{-2t} \). The unit tangent vector is \( \mathbf{T}(t) = \left\langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \right\rangle \).

Step 1: Find Derivative of \( \mathbf{u}(t) \)

We begin by differentiating each component of \( \mathbf{u}(t) = \langle e^t, e^{-t} \rangle \). The derivative of \( e^t \) with respect to \( t \) is \( e^t \), and the derivative of \( e^{-t} \) is \( -e^{-t} \). Therefore, \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \).

Step 2: Compute \( \mathbf{u}'(t) \times \mathbf{u}(t) \)

In 2D, the cross product of vectors \( \langle a_1, a_2 \rangle \) and \( \langle b_1, b_2 \rangle \) is zero as they are confined to a plane. So, \( \mathbf{u}'(t) \times \mathbf{u}(t) = 0 \).

Step 3: Compute \( \mathbf{u}(t) \times \mathbf{u}'(t) \)

As in the previous step, in 2D, the cross product for vectors is zero, thus \( \mathbf{u}(t) \times \mathbf{u}'(t) = 0 \).

Step 4: Compute \( \mathbf{u}(t) \cdot \mathbf{u}'(t) \)

The dot product is computed as \( e^t \cdot e^t + e^{-t} \cdot (-e^{-t}) = (e^{2t}) + (-e^{-2t}) = e^{2t} - e^{-2t} \).

Step 5: Find the Magnitude of \( \mathbf{u}'(t) \)

The magnitude is calculated as \( \| \mathbf{u}'(t) \| = \sqrt{(e^t)^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}} \).

Step 6: Determine the Unit Tangent Vector

The unit tangent vector is given by \( \mathbf{T}(t) = \frac{\mathbf{u}'(t)}{\| \mathbf{u}'(t) \|} = \left\langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \right\rangle \).

Key Concepts

derivative of vector functions

In vector calculus, derivatives of vector functions play a crucial role in understanding how vector quantities change. When you have a vector function like \( \mathbf{u}(t) = \langle e^t, e^{-t} \rangle \), you find its derivative by differentiating each of its components individually.
This step is akin to finding the derivative of a function and measures the rate of change of each component.
For \( e^t \), the derivative is straightforward: it remains \( e^t \). For \( e^{-t} \), applying the chain rule results in \( -e^{-t} \). Therefore, the derivative of \( \mathbf{u}(t) \) with respect to \( t \) is \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \).
Understanding derivatives of vector functions is essential as it provides information about the velocity and acceleration of moving objects.

unit tangent vector

The unit tangent vector is derived from a vector function's derivative and it provides the direction of the function at a particular point.
To find it, you take the derivative of the vector function — the result being \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \) in our case — and you normalize it.
This is achieved by dividing the derivative by its magnitude.
The magnitude, calculated as \( \sqrt{e^{2t} + e^{-2t}} \), gives you the length of the derivative vector. The unit tangent vector, representing direction only, is then \( \mathbf{T}(t) = \frac{\mathbf{u}'(t)}{\|\mathbf{u}'(t)\|} = \left\langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \right\rangle \).
This is a pivotal concept because it describes how paths change direction without considering how the speed changes.

cross product of vectors

When dealing with vectors in 3D space, the cross product is a method to find a vector perpendicular to two input vectors. However, in 2D space, like our example, the result is different.
The cross product of any two vectors confined to a plane will be zero. In our case, both \( \mathbf{u}'(t) \times \mathbf{u}(t) \) and \( \mathbf{u}(t) \times \mathbf{u}'(t) \) are zero, since the vectors operate within the same plane.
In 3D, a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) crossed with \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) yields \( \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \), which does not apply here because our vectors lack the third component.

dot product of vectors

The dot product is a scalar resulting from multiplying corresponding components of two vectors and summing them.
This operation helps determine the angle between vectors if they are non-zero and whether the vectors are orthogonal.
For our specific vectors \( \mathbf{u}(t) = \langle e^t, e^{-t} \rangle \) and \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \), the dot product is \( e^t \cdot e^t + e^{-t} \cdot (-e^{-t}) = e^{2t} - e^{-2t} \).
This result provides a measure of how parallel the vectors are at any given time \( t \), showing the degree of alignment or opposition in their respective directions.