The image contains three vector calculus problems asking for the tangent velocity vector $\mathbf{r}'(t)$ for specific position vectors $\mathbf{r}(t)$ and given values of $t$. Here's a breakdown of each part:

Problem A

Given: $\mathbf{r}(t) = (\cos 2t, \sin 2t)$ Find: $\mathbf{r}'\left(\frac{\pi}{4}\right)$

Problem B

Given: $\mathbf{r}(t) = (t^2, t^3)$ Find: $\mathbf{r}'(3)$

Problem C

Given: $\mathbf{r}(t) = e^{2t} \hat{i} + e^{-3t} \hat{j} + t\hat{k}$ Find: $\mathbf{r}'(-5)$

Solving each part:

A) Differentiating $\mathbf{r}(t) = (\cos 2t, \sin 2t)$:

1. Derivative of $\cos 2t$ with respect to $t$ is $-2 \sin 2t$.
2. Derivative of $\sin 2t$ with respect to $t$ is $2 \cos 2t$.

Thus: $\mathbf{r}'(t) = (-2 \sin 2t, 2 \cos 2t)$ At $t = \frac{\pi}{4}$, $\mathbf{r}'\left(\frac{\pi}{4}\right) = (-2 \sin \frac{\pi}{2}, 2 \cos \frac{\pi}{2}) = (-2, 0)$.

B) Differentiating $\mathbf{r}(t) = (t^2, t^3)$:

1. Derivative of $t^2$ with respect to $t$ is $2t$.
2. Derivative of $t^3$ with respect to $t$ is $3t^2$.

Thus: $\mathbf{r}'(t) = (2t, 3t^2)$ At $t = 3$, $\mathbf{r}'(3) = (2 \cdot 3, 3 \cdot 9) = (6, 27)$.

C) Differentiating $\mathbf{r}(t) = e^{2t} \hat{i} + e^{-3t} \hat{j} + t \hat{k}$:

1. Derivative of $e^{2t}$ is $2e^{2t}$.
2. Derivative of $e^{-3t}$ is $-3e^{-3t}$.
3. Derivative of $t$ is $1$.

Thus: $\mathbf{r}'(t) = 2e^{2t} \hat{i} - 3e^{-3t} \hat{j} + \hat{k}$ At $t = -5$, $\mathbf{r}'(-5) = 2e^{-10} \hat{i} - 3e^{15} \hat{j} + \hat{k}$

Final Answers:

• A) $\mathbf{r}'\left(\frac{\pi}{4}\right) = (-2, 0)$
• B) $\mathbf{r}'(3) = (6, 27)$
• C) $\mathbf{r}'(-5) = 2e^{-10} \hat{i} - 3e^{15} \hat{j} + \hat{k}$

Would you like more details on any of these solutions?

Here are 5 questions related to these concepts:

1. How do you interpret the physical meaning of a tangent velocity vector?
2. What is the general formula for differentiating trigonometric functions in vector calculus?
3. How does the velocity vector change when the position vector includes exponential functions?
4. What is the significance of using parametric equations in finding velocity vectors?
5. Can you find the acceleration vector by differentiating the velocity vector?

Tip: Always differentiate each component of a vector function independently to find the velocity or acceleration.