1. (11 pts) As always, give answers in simplified exact form. (a) Consider the vector function $$ \mathrm{r}(t)=\left\langle 6+t, 2 an ^{-1}(t), 3 t+e^{t^{2}} ight angle $$ Find the tangential component of acceleration at $$ t=0 $$.

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1. (11 pts) As always, give answers in simplified exact form. (a) Consider the vector function $$ \mathrm{r}(t)=\left\langle 6+t, 2 	an ^{-1}(t), 3 t+e^{t^{2}}ightangle $$ Find the tangential component of acceleration at $$ t=0 $$.

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Given the vector function
$$
\mathbf{r}(t)=\langle 6+t,\; 2\tan^{-1}(t),\; 3t+e^{t^2} \rangle,
$$
we want to find the tangential component of the acceleration at $$ t=0 $$. The tangential component $$ a_T $$ is given by
$$
a_T = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{\lVert \mathbf{r}'(t) \rVert}.
$$

### Step 1. Compute $$\mathbf{r}'(t)$$
Differentiate each component of $$\mathbf{r}(t)$$:
- First component: $$\frac{d}{dt}(6+t)= 1$$.
- Second component: $$\frac{d}{dt}\left(2\tan^{-1}(t)\right)= \frac{2}{1+t^2}$$ (using that $$\frac{d}{dt}\tan^{-1}(t)= \frac{1}{1+t^2}$$).
- Third component: $$\frac{d}{dt}\left(3t+e^{t^2}\right)= 3 + \frac{d}{dt}\left(e^{t^2}\right)= 3 + 2te^{t^2}$$ (applying the chain rule where $$\frac{d}{dt}e^{t^2}= 2te^{t^2}$$).

Thus,
$$
\mathbf{r}'(t)=\left\langle 1,\; \frac{2}{1+t^2},\; 3+2te^{t^2}\right\rangle.
$$
At $$ t=0 $$,
$$
\mathbf{r}'(0)= \langle 1,\; \frac{2}{1+0^2},\; 3+2\cdot0e^{0^2} \rangle = \langle 1,\, 2,\, 3 \rangle.
$$

### Step 2. Compute $$\mathbf{r}''(t)$$
Differentiate each component of $$\mathbf{r}'(t)$$:
- For the first component: $$\frac{d}{dt}(1)= 0$$.
- For the second component:
  $$
  \frac{d}{dt}\left(\frac{2}{1+t^2}\right)= 2\frac{d}{dt}\left((1+t^2)^{-1}\right)
  = 2(-1)(1+t^2)^{-2}(2t)= -\frac{4t}{(1+t^2)^2}.
  $$
- For the third component, differentiate $$3+2te^{t^2}$$ using the product rule on $$2te^{t^2}$$:
  $$
  \frac{d}{dt}\left(2te^{t^2}\right)= 2e^{t^2}+2t\cdot(2te^{t^2})
  = 2e^{t^2}+4t^2e^{t^2}
  = 2e^{t^2}(1+2t^2).
  $$
  The derivative of the constant $$3$$ is $$0$$.

Thus,
$$
\mathbf{r}''(t)= \left\langle 0,\; -\frac{4t}{(1+t^2)^2},\; 2e^{t^2}(1+2t^2) \right\rangle.
$$
At $$ t=0 $$,
$$
\mathbf{r}''(0)= \left\langle 0,\; -\frac{4\cdot 0}{(1+0^2)^2},\; 2e^{0}(1+2\cdot0^2) \right\rangle
= \langle 0,\, 0,\, 2 \rangle.
$$

### Step 3. Compute the Dot Product $$\mathbf{r}'(0) \cdot \mathbf{r}''(0)$$
$$
\mathbf{r}'(0) \cdot \mathbf{r}''(0) 
= \langle 1,\, 2,\, 3 \rangle \cdot \langle 0,\, 0,\, 2 \rangle 
= 1\cdot 0 + 2\cdot 0 + 3\cdot 2 
= 6.
$$

### Step 4. Compute the Magnitude $$\lVert \mathbf{r}'(0) \rVert$$
$$
\lVert \mathbf{r}'(0) \rVert = \sqrt{1^2+2^2+3^2} = \sqrt{1+4+9} = \sqrt{14}.
$$

### Step 5. Compute the Tangential Component of Acceleration
$$
a_T = \frac{\mathbf{r}'(0) \cdot \mathbf{r}''(0)}{\lVert \mathbf{r}'(0) \rVert} = \frac{6}{\sqrt{14}}.
$$

Thus, the tangential component of acceleration at $$ t=0 $$ is
$$
\boxed{\frac{6}{\sqrt{14}}}.
$$
The tangential component of acceleration at $$ t=0 $$ is $$ \frac{6}{\sqrt{14}} $$.

1. (11 pts) As always, give answers in simplified exact form. (a) Consider the vector function \( \mathrm{r}(t)=\left\langle 6+t, 2 \tan ^{-1}(t), 3 t+e^{t^{2}}\right\rangle \) Find the tangential component of acceleration at \( t=0 \).

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