Find the limit.
-$$r ( t ) = \sin t j - \cos t k ; 0 \leq t \leq \frac { \pi } { 2 }$$
A) Figure 7
B) Figure 4
C) Figure 1
D) Figure 2

Name: Find the limit. -\[r ( t ) = \sin t j - \cos t k ; 0 \leq t \leq \frac { \pi } { 2 }\] A) Figure 7 B) Figure 4 C) Figure 1 D) Figure 2
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The vector function $ r(t) = \sin t \, \mathbf{j} - \cos t \, \mathbf{k} $ traces a quarter-circle in the $ jk $-plane from $ (0, -1) $ to $ (1, 0) $ as $ t $ goes from 0 to $\frac{\pi}{2}$. This corresponds to Figure 2.

Find the Limit r(t)=sin⁡tj−cos⁡tk;0≤t≤π2r ( t ) = \sin t j - \cos t k ; 0 \leq t \leq \frac { \pi } { 2 }r(t)=sintj−costk;0≤t≤2π​

Find the Limit $r ( t ) = \sin t j - \cos t k ; 0 \leq t \leq \frac { \pi } { 2 }$