Question

attempt 2: one or more answers is incorrect. you will be told if you get them all right. for the vector function r(t), compute the tangent vector function vector r(t) for the given value of t. a) let r(t) = (cos t, sin t). then r($\frac{\pi}{4}$) = (-$\frac{sqrt{2}}{2}$, $\frac{sqrt{2}}{2}$)? b) let r(t) = (t², t³). then r(2) = (4, 27)? c) let r(t) = e^t i + e^(-2t) j + t k. then r(-1) = e^(-1) i+ -2e² j+ 1 k?

Explanation:

Step1: Differentiate component - by - component for A

If $\mathbf{r}(t)=(\cos t,\sin t)$, then $\mathbf{r}'(t)=(-\sin t,\cos t)$. Substitute $t = \frac{\pi}{4}$, we have $\mathbf{r}'(\frac{\pi}{4})=(-\sin\frac{\pi}{4},\cos\frac{\pi}{4})=(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

Step2: Differentiate component - by - component for B

If $\mathbf{r}(t)=(t^{2},t^{3})$, then $\mathbf{r}'(t)=(2t,3t^{2})$. Substitute $t = 2$, we get $\mathbf{r}'(2)=(2\times2,3\times2^{2})=(4,12)$.

Step3: Differentiate component - by - component for C

If $\mathbf{r}(t)=e^{t}\mathbf{i}+e^{-2t}\mathbf{j}+t\mathbf{k}$, then $\mathbf{r}'(t)=e^{t}\mathbf{i}-2e^{-2t}\mathbf{j}+\mathbf{k}$. Substitute $t=-1$, we obtain $\mathbf{r}'(-1)=e^{-1}\mathbf{i}-2e^{2}\mathbf{j}+\mathbf{k}$.

Answer:

A. $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$
B. $(4,12)$
C. $e^{-1}\mathbf{i}-2e^{2}\mathbf{j}+\mathbf{k}$