Problem 53 - Calculus Volume 3

Question

Find a tangent vector at the indicated value of \(t\). \(\mathbf{r}(t)=3 e^{t} \mathbf{i}+2 e^{-3 t} \mathbf{j}+4 e^{2 t} \mathbf{k} ; \quad t=\ln (2)\)

Step-by-Step Solution

Answer

The tangent vector at \( t = \ln(2) \) is \( 6\mathbf{i} - \frac{3}{4}\mathbf{j} + 32\mathbf{k} \).

Step 1: Differentiate the vector function

To find the tangent vector, first compute the derivative of the vector function \( \mathbf{r}(t) \) with respect to \( t \). This gives us the tangent vector function \( \mathbf{r}'(t) \). Differentiate each component separately: - The \( \mathbf{i} \) component: \( \frac{d}{dt}(3e^t) = 3e^t \).- The \( \mathbf{j} \) component: \( \frac{d}{dt}(2e^{-3t}) = -6e^{-3t} \).- The \( \mathbf{k} \) component: \( \frac{d}{dt}(4e^{2t}) = 8e^{2t} \).Thus, \( \mathbf{r}'(t) = 3e^t \mathbf{i} - 6e^{-3t} \mathbf{j} + 8e^{2t} \mathbf{k} \).

Step 2: Evaluate the derivative at the indicated t-value

Evaluate the tangent vector function \( \mathbf{r}'(t) \) at \( t = \ln(2) \). Substitute \( t = \ln(2) \) into \( \mathbf{r}'(t) \).- The \( \mathbf{i} \) component becomes: \( 3e^{\ln(2)} = 3 \times 2 = 6 \).- The \( \mathbf{j} \) component becomes: \( -6e^{-3\ln(2)} = -6 \times 2^{-3} = -\frac{6}{8} = -\frac{3}{4} \).- The \( \mathbf{k} \) component becomes: \( 8e^{2\ln(2)} = 8 \times 2^2 = 8 \times 4 = 32 \).Thus, the tangent vector at \( t = \ln(2) \) is \( 6 \mathbf{i} - \frac{3}{4} \mathbf{j} + 32 \mathbf{k} \).

Key Concepts

Vector Differentiation

Vector differentiation is a useful technique in calculus, extending the concept of differentiating functions to vector-valued functions. Here, we deal with vectors that change with respect to a variable, often denoted as \( t \), which could signify time or another parameter.
To differentiate a vector function \( \mathbf{r}(t) \), simply differentiate each component of the vector independently:

• The \( \mathbf{i} \)-component is handled with regular function rules.
• The \( \mathbf{j} \)-component follows the same principle.
• The \( \mathbf{k} \)-component likewise.

Hence, when you compute the derivative, you're determining how each directional component changes. This derivative vector \( \mathbf{r}'(t) \) gives the rate of change at any point \( t \), also known as the tangent vector to the curve represented by \( \mathbf{r}(t) \). This concept becomes invaluable in physics where motion and trajectory are analyzed.

Evaluating Derivative at a Point

Once the vector derivative is determined, evaluating it at a specific point \( t \) gives specific insight into behavior at that moment. In terms of our exercise, we evaluated the derivative at \( t = \ln(2) \).
This means we substitute \( \ln(2) \) into \( \mathbf{r}'(t) \) and compute each component:

• Convert expressions like \( e^{\ln(2)} \) into a more manageable number like \( 2 \).
• Perform basic operations like multiplication or division where necessary.
• Combine these results to get the full vector at that point.

As a result, this pinpoints the direction and rate at which the curve passes through that point. Knowing this makes it easier to draw graphs and estimates of motion and further computations in physics and engineering.

Vector Calculus

Vector calculus is an essential aspect of mathematics focused on multi-dimensional spaces, extending traditional calculus to vector fields or vector-valued functions. It provides tools to study curves, surfaces, and scalar fields defined in space.
This type of calculus is not limited to lines but extends into higher dimensions through:

• Understanding path and line integrals, vital for computing the work done by a force field.
• Differential operations like the gradient, divergence, and curl which explain field change rates.
• Applying principles to fluid dynamics, electromagnetism, and computer graphics among others.

Analyzing vector functions with derivatives as in our example illustrates changes along a path defined in three-dimensional space. This lays groundwork for grasping complex dynamic systems and their behaviors.