Question

I need help with this... im lost and i really would appreciate any help.
the below spiral curves are generated by the parametric equations.

x1=tcos(t), y1=tsin(t), pi = 13 years ago

Answer 1

Well, the arc length equation is
\[\begin{align}
\int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt
\end{align}\]
So you need to evaluate that for the two curves.
For part (b), note that spiral curves are generally very easy to express in polar form, so you'll need to convert them and then subtract their integrals as you would to find the area between any two curves.

Answer 2

formula for arc length is following
\[\Large L=\int\limits_{a}^{b} \sqrt{|r'(t)|}\]
where
r=x i+y j
so
\[\Large r(t)=tcos(t)i+tsin(t)j\]
take derivative
\[\Large r'(t)=(cos(t)-tsin(t)) i+(sin(t)+tcos(t)) j \]
now take magnitude
\[\Large |r'(t)|=\sqrt{((cos(t)-tsin(t))^2 +(sin(t)+tcos(t))^2}\]
simplify this the use the first formula with limitas a=-pi ,b=pi

Answer 3

i dont understand what the i and j are in ur solution sami-21 and thats still a little confusing. are you saying x1 and y1 are in the same to form r(t)?. and i dont understand why the use of magnitude. Many thanks for the help though

Answer 4

i and j are the x and y components in the vector notation.
Magnitude is used in formula . yes i am saying x1 and y1 are in the same to form r(t).

Answer 5

thank you very much :) im glad you know your multivariate calculus XD. u just saved me