Problem 213 - 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. \(\mathbf{u}(t)=\left\langle t^{2}, 2 t+6,4 t^{5}-12\right\rangle\)

Step-by-Step Solution

Answer

Differentiate each component of \( \mathbf{u}(t) \) to find \( \mathbf{u}'(t) \), then solve for the cross product and dot product; the unit tangent is \( \mathbf{T}(t) = \frac{\mathbf{u}'(t)}{\|\mathbf{u}'(t)\|} \).

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

To find the derivative of \( \mathbf{u}(t) = \langle t^2, 2t + 6, 4t^5 - 12 \rangle \), differentiate each component with respect to \( t \). \[ \mathbf{u}'(t) = \left\langle \frac{d}{dt}(t^2), \frac{d}{dt}(2t + 6), \frac{d}{dt}(4t^5 - 12) \right\rangle = \langle 2t, 2, 20t^4 \rangle \]

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

The cross product \( \mathbf{u}'(t) \times \mathbf{u}(t) \) is computed as follows: \[ \mathbf{u}'(t) \times \mathbf{u}(t) = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2t & 2 & 20t^4 \ t^2 & 2t+6 & 4t^5-12 \end{array} \right| \] Calculate the determinant of the matrix to get the cross product vector.

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

By properties of the cross product, \( \mathbf{u}(t) \times \mathbf{u}'(t) = - (\mathbf{u}'(t) \times \mathbf{u}(t)) \). Therefore, this is just the negative of the vector obtained from Step 2.

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

For the dot product, multiply corresponding components and add them up: \[ \mathbf{u}(t) \cdot \mathbf{u}'(t) = t^2 \cdot 2t + (2t + 6) \cdot 2 + (4t^5 - 12) \cdot 20t^4 \] Simplify the expression to get the dot product scalar.

Step 5: Determine the Unit Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector \( \mathbf{T}(t) \) is given by the formula:\[ \mathbf{T}(t) = \frac{\mathbf{u}'(t)}{\|\mathbf{u}'(t)\|} \]Calculate \( \|\mathbf{u}'(t)\| \) as the magnitude of the vector \( \mathbf{u}'(t) = \langle 2t, 2, 20t^4 \rangle \) using the formula \( \sqrt{(2t)^2 + 2^2 + (20t^4)^2} \), then divide each component of \( \mathbf{u}'(t) \) by this magnitude.

Key Concepts

Derivative of Vector Functions

When working with vector functions, taking the derivative means differentiating each component individually. Let's take an example of the vector function \( \mathbf{u}(t) = \langle t^2, 2t + 6, 4t^5 - 12 \rangle \). The derivative, \( \mathbf{u}'(t) \), is found by calculating the derivative of each component function with respect to \( t \):

• The derivative of \( t^2 \) is \( 2t \)
• The derivative of \( 2t + 6 \) is \( 2 \)
• The derivative of \( 4t^5 - 12 \) is \( 20t^4 \)

Thus, the derivative is \( \mathbf{u}'(t) = \langle 2t, 2, 20t^4 \rangle \). This means every time \( t \) changes, the rate of change of the entire vector \( \mathbf{u}(t) \) can be captured by \( \mathbf{u}'(t) \), representing how swiftly and in which directions the vector is changing.
This concept applies powerfully in physics for describing motion along a path.

Cross Product

The cross product of two vectors in three-dimensional space results in another vector that is perpendicular to the plane containing the original vectors. Consider the vectors \( \mathbf{u}'(t) = \langle 2t, 2, 20t^4 \rangle \) and \( \mathbf{u}(t) = \langle t^2, 2t+6, 4t^5-12 \rangle \). To find \( \mathbf{u}'(t) \times \mathbf{u}(t) \), we use the determinant of a matrix formed by the unit vectors and the components of these vectors:
\[\mathbf{u}'(t) \times \mathbf{u}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2t & 2 & 20t^4 \ t^2 & 2t+6 & 4t^5-12 \end{vmatrix} \]This determinant computes to a vector perpendicular to both \( \mathbf{u}'(t) \) and \( \mathbf{u}(t) \). It's important to remember that the cross product is not commutative; meaning, \( \mathbf{u}(t) \times \mathbf{u}'(t) = - (\mathbf{u}'(t) \times \mathbf{u}(t)) \).
The cross product is widely used in physics and engineering to find torque and rotational axes.

Dot Product

The dot product of two vectors results in a scalar value and is often used to measure how 'aligned' two vectors are. It is calculated by multiplying corresponding components and summing those products. For the vectors \( \mathbf{u}(t) = \langle t^2, 2t + 6, 4t^5 - 12 \rangle \) and \( \mathbf{u}'(t) = \langle 2t, 2, 20t^4 \rangle \), the dot product \( \mathbf{u}(t) \cdot \mathbf{u}'(t) \) becomes:

• \( t^2 \cdot 2t = 2t^3 \)
• \( (2t + 6) \cdot 2 = 4t + 12 \)
• \( (4t^5 - 12) \cdot 20t^4 = 80t^9 - 240t^4 \)

Adding them gives a combined scalar expression.
The result tells us about the magnitude of projection of one vector onto another and is pivotal in determining angles between vectors using the cosine of the angle.

Unit Tangent Vector

The unit tangent vector, \( \mathbf{T}(t) \), represents the direction of a curve at a particular point and has a length of one. It's derived from the derivative vector \( \mathbf{u}'(t) \) of the original vector function \( \mathbf{u}(t) \). The formula to find it is:
\[ \mathbf{T}(t) = \frac{\mathbf{u}'(t)}{\|\mathbf{u}'(t)\|} \]This involves calculating the magnitude (or norm) of the derivative vector:
\[ \|\mathbf{u}'(t)\| = \sqrt{(2t)^2 + 2^2 + (20t^4)^2} \]Once you have the magnitude, divide each component of \( \mathbf{u}'(t) \) by this number to get \( \mathbf{T}(t) \).
The unit tangent vector is vital in calculus and physics as it provides insights into the direction and path of moving objects, ensuring that at any time \( t \), we know precisely the direction of motion.