Finding the unit tangent vector+length

[mindy88]

Sry guys, I have so many questions. I have a midterm next week

The question is
1. r(t)=ti+(2/3)t^(3/2)k 0<t<8 (this is less than equal to or greater or equal to)
a. find the curve's until tangent vector
b. find the length of the indicated portion of the curve

This is how i did it
a. r'(t)=i+t^1/2k (i took the derivative)
lvl= square root of 1+t^2
T=v/(lvl)
which is 1/square root 1+t^2 i + square root t/square root 1+t^2k

b. i took the integral from o to 8 for square root of 1+t^2
this is where i get stuck, i'm not sure how to integrate that



i'm sorry if this is hard to read, i don't have any programs which can do math symbols.


thanks for the help

 

[soroban]

Hello, Mindy!

A small but critical error . . .

1. \(\displaystyle \;{\bf r}(t)\:=\:t{\bf i}\,+\,\frac{2}{3}t^{\frac{3}{2}}{\bf k}\;\;\; 0\,\leq\,t\,\leq\,8\)

a) Find the curve's unit tangent vector
b) Find the length of the indicated portion of the curve


This is how i did it

\(\displaystyle {\bf r}'(t)\:=\:{\bf i}\,+\,t^{\frac{1}{2}}{\bf k}\;\;\) (i took the derivative) . . . . Yes!

\(\displaystyle |{\bf v}|\:=\:\sqrt{1\,+\,t^2}\;\;\) . . . no

\(\displaystyle \text{We have: }\;{\bf v}(t) \:=\:1\cdot{\bf i} \,+ \,t^{\frac{1}{2}}\cdot{\bf k}\)

\(\displaystyle \text{Then: }\:|{\bf v}| \:=\:\sqrt{1^2\,+\,\left(t^{\frac{1}{2}}\right)^2} \:=\:\sqrt{1\,+\,t}\)

 

[mindy88]

oh i see, i squared it wrong. thanks