3. the derivative of $mathbf{r}(t)=langle\tan(t) - 2,-5tsin(t),\frac{1}{2 - t}\rangle$ is $mathbf{r}(t)=langl…

3. the derivative of $mathbf{r}(t)=langle\tan(t) - 2,-5tsin(t),\frac{1}{2 - t}\rangle$ is $mathbf{r}(t)=langlesec^{2}(t),-5sin(t)-5tcos(t),-\frac{1}{2(2 - t)^{\frac{3}{2}}}\rangle$.

3. the derivative of $mathbf{r}(t)=langle\tan(t) - 2,-5tsin(t),\frac{1}{2 - t}\rangle$ is $mathbf{r}(t)=langlesec^{2}(t),-5sin(t)-5tcos(t),-\frac{1}{2(2 - t)^{\frac{3}{2}}}\rangle$.

Answer

Explanation:

Step1: Differentiate first - component

The derivative of $\tan(t)-2$ with respect to $t$ is $\sec^{2}(t)$ since the derivative of $\tan(t)$ is $\sec^{2}(t)$ and the derivative of a constant ($- 2$) is $0$.

Step2: Differentiate second - component

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-5t$ and $v = \sin(t)$. $u^\prime=-5$ and $v^\prime=\cos(t)$. So $(-5t\sin(t))^\prime=-5\sin(t)-5t\cos(t)$.

Step3: Differentiate third - component

Rewrite $\frac{1}{2 - t}=(2 - t)^{-1}$. Using the chain - rule, if $y = u^{-1}$ and $u = 2 - t$, then $\frac{dy}{du}=-u^{-2}$ and $\frac{du}{dt}=-1$. So $\frac{d}{dt}(2 - t)^{-1}=-(2 - t)^{-2}\times(-1)=\frac{1}{(2 - t)^{2}}$. The given answer for the third - component in the problem is incorrect.

Answer:

$\mathbf{r}^\prime(t)=\left\langle\sec^{2}(t),-5\sin(t)-5t\cos(t),\frac{1}{(2 - t)^{2}}\right\rangle$