To find the tangent vector of the given vector function $\mathbf{r}(t) = 3e^t \mathbf{i} + 4e^{-3t} \mathbf{j} + 2e^{4t} \mathbf{k}$ at $t = \ln(2)$, we need to:

1. Differentiate $\mathbf{r}(t)$ with respect to $t$ to get the velocity (tangent) vector.
2. Evaluate the result at $t = \ln(2)$.

Step 1: Differentiate $\mathbf{r}(t)$

Given $\mathbf{r}(t) = 3e^t \mathbf{i} + 4e^{-3t} \mathbf{j} + 2e^{4t} \mathbf{k}$, we differentiate each component with respect to $t$.

$\frac{d}{dt} (3e^t) = 3e^t$ $\frac{d}{dt} (4e^{-3t}) = 4(-3)e^{-3t} = -12e^{-3t}$ $\frac{d}{dt} (2e^{4t}) = 2(4)e^{4t} = 8e^{4t}$

Thus, the derivative (tangent vector) is:

$\mathbf{r}'(t) = 3e^t \mathbf{i} - 12e^{-3t} \mathbf{j} + 8e^{4t} \mathbf{k}$

Step 2: Evaluate at $t = \ln(2)$

Now substitute $t = \ln(2)$ into $\mathbf{r}'(t)$.

1. $e^{\ln(2)} = 2$
2. $e^{-3\ln(2)} = 2^{-3} = \frac{1}{8}$
3. $e^{4\ln(2)} = 2^4 = 16$

Substitute these into the components of $\mathbf{r}'(t)$:

$\mathbf{r}'(\ln(2)) = 3(2) \mathbf{i} - 12\left(\frac{1}{8}\right) \mathbf{j} + 8(16) \mathbf{k}$

Simplifying:

$\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - \frac{12}{8} \mathbf{j} + 128 \mathbf{k}$

$\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - 1.5 \mathbf{j} + 128 \mathbf{k}$

Final Answer:

The tangent vector at $t = \ln(2)$ is:

$\mathbf{r}'(\ln(2)) = 6 \mathbf{i} - 1.5 \mathbf{j} + 128 \mathbf{k}$

Would you like more details or have any questions?

Related Questions:

1. How do you find the unit tangent vector of a vector function?
2. What is the significance of the tangent vector in 3D space?
3. How do you compute the magnitude of a tangent vector?
4. How does the exponential function behave under differentiation?
5. What are the differences between velocity and acceleration vectors?

Tip:

When differentiating vector functions, apply the derivative to each component separately, as if they were scalar functions.