Find the derivative. $$ \vec{r}(t)=\cos ^{3}(t) \vec{\imath}+\sin ^{3}(t) \vec{\jmath}+\vec{k} $$ $$ \vec{r}^{\prime}(t)=\left(-3[?]^{2}(t)[](t)\right) \vec{\imath}+\left(3[]^{2}(t)[](t)\right) \vec{\jmath} $$

Question

UpStudy AI · Accepted Answer

To find the derivative of the vector function $$ \vec{r}(t) $$, we'll differentiate each component with respect to $$ t $$.

Given:
$$
\vec{r}(t) = \cos^{3}(t) \, \vec{\imath} + \sin^{3}(t) \, \vec{\jmath} + \vec{k}
$$

**Step 1: Differentiate Each Component**

1. **Differentiating $$ \cos^{3}(t) $$ with respect to $$ t $$:**
   $$
   \frac{d}{dt}[\cos^{3}(t)] = 3\cos^{2}(t) \cdot (-\sin(t)) = -3\cos^{2}(t)\sin(t)
   $$

2. **Differentiating $$ \sin^{3}(t) $$ with respect to $$ t $$:**
   $$
   \frac{d}{dt}[\sin^{3}(t)] = 3\sin^{2}(t) \cdot \cos(t) = 3\sin^{2}(t)\cos(t)
   $$

3. **Differentiating $$ \vec{k} $$ with respect to $$ t $$:**
   $$
   \frac{d}{dt}[\vec{k}] = 0
   $$

**Step 2: Assemble the Derivative Vector**

Putting it all together:
$$
\vec{r}^{\prime}(t) = \left(-3\cos^{2}(t)\sin(t)\right) \vec{\imath} + \left(3\sin^{2}(t)\cos(t)\right) \vec{\jmath} + 0 \cdot \vec{k}
$$

Since the $$ \vec{k} $$ component is zero, it can be omitted:
$$
\vec{r}^{\prime}(t) = -3\cos^{2}(t)\sin(t) \, \vec{\imath} + 3\sin^{2}(t)\cos(t) \, \vec{\jmath}
$$

**Final Answer:**
$$
\boxed{\vec{r}^{\prime}(t) = -3\cos^{2}t\,\sin t\,\vec{\imath} + 3\sin^{2}t\,\cos t\,\vec{\jmath}}
$$
$$ \vec{r}^{\prime}(t) = -3\cos^{2}(t)\sin(t) \, \vec{\imath} + 3\sin^{2}(t)\cos(t) \, \vec{\jmath} $$

Find the derivative.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions