deliredejoker7m

2021-11-28

Find the second derivative of the vector-valued function
$\to r\left(t\right)=\left(5t+7\mathrm{sin}\left(t\right)\right)i+\left(7t+2\mathrm{cos}\left(t\right)\right)j$

Befoodly

Step 1
Given that:
$r\left(t\right)=\left(5t+7\mathrm{sin}t\right)i+\left(7t+2\mathrm{cos}t\right)j$
So, 1st derivative of r:
$\frac{dr}{dt}=i\frac{d}{dt}\left(5t+7×\mathrm{sin}t\right)+j×\frac{d}{dt}\left(7t+2\mathrm{cos}t\right)$
$=i\left(5\frac{d}{dt}\left(t\right)+7\frac{d}{dt}\left(\mathrm{sin}t\right)\right)+j\left(7\frac{d}{dt}\left(t\right)+2\frac{d}{dt}\mathrm{cos}t\right)$
$=i\left(5+7\mathrm{cos}t\right)+j\left(7-2\mathrm{sin}t\right)$
So, $\frac{dr}{dt}=\left(5+7\mathrm{cos}t\right)i+\left(7-2\mathrm{sin}t\right)j$
2nd derivative of r:
$\frac{{d}^{2}r}{{dt}^{2}}=\frac{d}{dt}\left[\left(5+7\mathrm{cos}t\right)i+\left(7-2\mathrm{sin}t\right)j\right]$
$=\left(5\frac{d}{dt}\left(1\right)+7\frac{d}{dt}\mathrm{cos}t\right)i+\left(7\frac{d}{dt}\left(1\right)-2\frac{d}{dt}\mathrm{sin}t\right)j$
$=\left(0-7\mathrm{sin}t\right)i+\left(0-2\mathrm{cos}t\right)j$
$-7\mathrm{sin}t×i-2\mathrm{cos}t×j$
So, $\frac{{d}^{2}r}{{dt}^{2}}=-7\mathrm{sin}t×i-2\mathrm{cos}t×j$

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