Question

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{u}(t) \cdot \mathbf{v}(t)$$

Short answer

Answer: The derivative is \(-2t\).

Step-by-step solution

Compute the dot product of the two vectors

To find \(\mathbf{u}(t) \cdot \mathbf{v}(t)\), we need to multiply the corresponding components of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) and then sum the products. Since \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle\) and \(\mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\), the dot product can be calculated as follows: $$\mathbf{u}(t) \cdot \mathbf{v}(t) = 1 \cdot t^2 + t \cdot (-2t) + t^2 \cdot 1$$

Simplify the dot product

Now, we need to simplify the expression obtained in Step 1, which is: $$\mathbf{u}(t) \cdot \mathbf{v}(t) = 1(t^2) - 2(t^2) + 1(t^2)$$ $$\mathbf{u}(t) \cdot \mathbf{v}(t) = -t^2.$$

Calculate the derivative

Next, we will compute the derivative with respect to \(t\) of the simplified expression for the dot product: $$\frac{d}{dt}(\mathbf{u}(t) \cdot \mathbf{v}(t)) = \frac{d}{dt}(-t^2)$$ $$\frac{d}{dt}(\mathbf{u}(t) \cdot \mathbf{v}(t)) = -2t.$$ That is, the derivative of the dot product of the given vector functions is \(-2t\).

Key concepts

Dot Product

The dot product is a fundamental operation in vector calculus, essential for understanding relationships between vectors. It is also called the scalar product because the result is always a scalar. To compute the dot product of two vectors, we multiply their corresponding components and then add the results.

In our exercise, we have two vector functions: \( \mathbf{u}(t)=\langle 1, t, t^2 \rangle \) and \( \mathbf{v}(t)=\langle t^2, -2t, 1 \rangle \). To find the dot product \( \mathbf{u}(t) \cdot \mathbf{v}(t) \), we perform the following operations:

• Multiply the first components: \( 1 \cdot t^2 = t^2 \)
• Multiply the second components: \( t \cdot (-2t) = -2t^2 \)
• Multiply the third components: \( t^2 \cdot 1 = t^2 \)

Now, we sum these individual products:
\[ t^2 - 2t^2 + t^2 = -t^2 \]
Thus, the dot product \( \mathbf{u}(t) \cdot \mathbf{v}(t) \) simplifies to \( -t^2 \). Easy, right? Understanding dot products is vital because they help analyze the angle between vectors and perform projections.

Derivative Calculation

Calculating derivatives is a crucial part of calculus, allowing us to understand the rate of change of functions. When dealing with vector functions, it's common to calculate the derivative of a scalar function derived from them, like the dot product we just discussed.

For our specific case, we have the simplified expression from the dot product \( -t^2 \). To find the derivative with respect to \( t \), we apply the basic rules of differentiation:

• The power rule: If \( f(t) = t^n \), then \( \frac{d}{dt}[t^n] = nt^{n-1} \)

Applying this to our expression, \( \frac{d}{dt}(-t^2) \):
According to the power rule, the derivative of \( t^2 \) is \( 2t \). Multiplying by the coefficient \(-1\) in front of \( t^2 \) gives us:
\[ \frac{d}{dt}(-t^2) = -2t \]
Therefore, the derivative of the dot product \( \mathbf{u}(t) \cdot \mathbf{v}(t) \) is \( -2t \). Derivatives like this help in understanding how the interaction between two vector functions changes over time.

Vector Functions

Vector functions are functions whose outputs are vectors. These functions play an essential role in physics and engineering, as they can describe quantities that have both magnitude and direction, such as velocity and force.

For example, in this exercise, we have two vector functions: \( \mathbf{u}(t)=\langle 1, t, t^2 \rangle \) and \( \mathbf{v}(t)=\langle t^2, -2t, 1 \rangle \). Each vector function gives a vector based on the input \( t \).

• \( \mathbf{u}(t) \) represents a point in 3D space at any time \( t \).
• \( \mathbf{v}(t) \) also represents another point in 3D space, but its components change differently with \( t \).

Operations on vector functions, such as dot products and derivatives, further our understanding by providing insights about their relationships and how they evolve with time. By understanding vector functions, we can model and analyze complex systems efficiently.