Calculate length of a curve in 3d

Ive ploted a rising spiral with the following code.

clear all

syms t real

ezplot3((sqrt(9)).*cos(9./2.*t), (sqrt(16)).*sin(9./2.*t),t./pi.*(9./2) ,'animate')

But i have a hard time trying to calculate the length of the curve up to z = 9. I know that i 2d i use the formula int(sqrt(1+(dx/dy)^2)) But how do i do this in 3d? What code do i use?

[Ema]

According to http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx i could use the formula int(sqrt x'^2+y'^2+z'^2, t , 0 , 2*pi) In other words , integral of the addition for the derivatives of x,y,z raised with the power of two.

But i get the following answer : "int((324*cos((9*t)/2)^2 + (729*sin((9*t)/2)^2)/4 + 81/(4*pi^2))^(1/2), t, 0, 2*pi)" that doesnt make sense. Shouldnt the t be calculated from 0 to 2pi?

Answer 1

[Torsten]

http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx

Best wishes

Torsten.

[Ema]

Ive tried it but dont get it solved. According to your site i could use the formula int(sqrt x'^2+y'^2+z'^2, t , 0 , 2*pi) In other words , integral of the addition for the derivatives of x,y,z raised with the power of two.

But i get the following answer : "int((324*cos((9*t)/2)^2 + (729*sin((9*t)/2)^2)/4 + 81/(4*pi^2))^(1/2), t, 0, 2*pi)" that doesnt make sense. Shouldnt the t be calculated from 0 to 2pi?

[Torsten]

f=@(t)sqrt((sqrt(9)*9/2*sin(9/2*t)).^2+(sqrt(16)*9/2*cos(9/2*t)).^2+(1/pi*9/2)^2);
value=integral(f,0,2*pi);

Best wishes

Torsten.