Find the derivative. $$ \vec{r}(t)=4 \sqrt{t} \vec{\imath}+t^{2} \sqrt{t} \vec{\jmath}+\ln \left(t^{2}\right) \vec{k} $$ $$ \vec{r}^{\prime}(t)=\frac{[?]}{\sqrt{t}} \vec{\imath}+\frac{[]}{[]} t^{\frac{3}{2}} \vec{\jmath}+\frac{[]}{[]} \vec{k} $$

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UpStudy AI · Accepted Answer

To find the derivative of the vector function
$$
\vec{r}(t) = 4 \sqrt{t} \, \vec{\imath} + t^{2} \sqrt{t} \, \vec{\jmath} + \ln \left(t^{2}\right) \, \vec{k},
$$
we differentiate each component with respect to $$ t $$.

### 1. $$ \vec{\imath} $$ Component:
$$
4 \sqrt{t} = 4 t^{1/2}
$$
$$
\frac{d}{dt} \left(4 t^{1/2}\right) = 4 \cdot \frac{1}{2} t^{-1/2} = \frac{2}{\sqrt{t}}
$$

### 2. $$ \vec{\jmath} $$ Component:
$$
t^{2} \sqrt{t} = t^{2} \cdot t^{1/2} = t^{5/2}
$$
$$
\frac{d}{dt} \left(t^{5/2}\right) = \frac{5}{2} t^{3/2}
$$

### 3. $$ \vec{k} $$ Component:
$$
\ln(t^{2}) = 2 \ln t
$$
$$
\frac{d}{dt} \left(2 \ln t\right) = 2 \cdot \frac{1}{t} = \frac{2}{t}
$$

### Putting It All Together:
$$
\vec{r}\,'(t) = \frac{2}{\sqrt{t}} \, \vec{\imath} + \frac{5}{2} t^{3/2} \, \vec{\jmath} + \frac{2}{t} \, \vec{k}
$$

**Final Answer:**
$$
\vec{r}^{\prime}(t) = \frac{2}{\sqrt{t}}\,\vec{\imath} + \frac{5}{2}\, t^{3/2}\,\vec{\jmath} + \frac{2}{t}\,\vec{k}
$$
$$ \vec{r}^{\prime}(t) = \frac{2}{\sqrt{t}} \vec{\imath} + \frac{5}{2} t^{3/2} \vec{\jmath} + \frac{2}{t} \vec{k} $$

Find the derivative.

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