First Graded Homework Assignment in Multivariate Calculus for Lydia Hyman

$3.$ Let's be circular. Throughout all these problems, simplify your calculations thoroughly, especially using the Pythagorean Idendity that, $AAinR,si{n}^2+co{s}^2=1.$

Let $p(t)=[cos(t),sin(t){]}^T$

$($a$)||p(t)||=$

$($b$)p^{\text{'}}(t)=$

$($c$)p^{\text{'}\text{'}}(t)=$

$($d$)||p^{\text{'}\text{'}}(t)||=$

$4.$ Let's go nowhere, fast. Using the same position function as in Problem $3,$ compute these values. Tell the correct genus$-$set of your outcome. One is done for you as an example. Notice that getting the genus right can be the most important part of your answer.

Genus$-$set

$($e$.$g$.)p(0)$

$=[1,0{]}^T$

$R^2$

$($a$)p(2)-p(0)=$

$($b$)p(0)-p^{\text{'}\text{'}}(0)=$

$($c$)p(t)-p^{\text{'}\text{'}}(t)=$

$($d$)\frac{d}{dt}||p^{\text{'}\text{'}}(t)||=$

$5.$ Oh this is good. If you understand rotary motion, then you can come up with these answers without much of any computation. You may, though, grind away to get them.

$($a$)p()-p(0)=$

$($b$){\displaystyle \underset{0}{\overset{}{}}}{p}^{\text{'}}(t)dt=$

$($c$)$ Note the bound of integration is changed from above.

${\displaystyle \underset{0}{\overset{2}{}}}||p^{\text{'}}(t)||dt$

$($d$)$ Your genus must be right to get any credit!

$||{\displaystyle \underset{0}{\overset{2}{}}}{p}^{\text{'}}(t)dt||=$