Calculate the length of parameterized curve (Part 2)

Question

This question is related to question i asked before. link. This time with slightly different parametric curve and slightly different problem.

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Problem

Calculate the length of parameterized curve which is defined as:

$$ r(t) = (\frac{t^3}{\sqrt 3},t^2) $$

Attempt to solve

We can express our parameterized curve in form of vector:

$$ r(t)=\begin{bmatrix} \frac{t^3}{\sqrt{3} \\t^2} \end{bmatrix} $$

where $r(t) \in R^2$

Tangent vector or velocity (as suggested in post earlier) vector can be defined as:

$$ r'(t)= \begin{bmatrix} \frac{1}{\sqrt{3}}\frac{d}{dt}(t^3) \\ \frac{d}{dt} (t^2) \end{bmatrix} $$

$$ r'(t)=\begin{bmatrix} \sqrt{3}t^2 \\ 2t \end{bmatrix} $$

And velocity vector length:

$$ ||r'(t)||=\sqrt{(\sqrt{3}t^2)^2+(2t)t^2} $$ $$ ||r'(t)||=\sqrt{3t^4+4t^2} $$

Now by integrating velocity vector length from 0 to 5 we can get the length for the curve.

$$ L=\int_{1}^{5}\sqrt{3t^4+4t^2} dt $$ $$ L=\frac{79\sqrt{79}}{9}-\frac{7\sqrt{7}}{9} $$ $$ L\approx 75.960789 $$

To this point this solution is almost identical to one in the other post. So i thought why you couldn't just try to calculate the length of the curve vector. This is without integrating the velocity vector length in the first place. We can try:

$$ ||r(t)||=\sqrt{(\frac{t^3}{\sqrt{3}})^2+(t^2)^2} $$ $$ L=||r(5)||-||r(1)||$$ $$ L=\sqrt{(\frac{5^3}{\sqrt{3}})^2+(5^2)^2}-\sqrt{(\frac{1^3}{\sqrt{3}})^2+(1^2)^2}$$ $$ L\approx 75.221561 $$

Now this is fairly close to the length we got by integrating the velocity vector length. Now the problem is i don't understand why these two results are different. What is difference between these two ?

$$\text{difference }= 1-\frac{75.221561}{75.960789}\approx 0.97\% $$

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I made some plots from this parametric curve using matlab (in hopes for better understanding). I added the plot and the source code used for plotting.

Matlab code for plot

  t= linspace(1,5);
  A= t.^3/sqrt(3);
  B= t.^2;
  C= sqrt(3)*t.^2;
  D= 2*t;
  E= sqrt(3*t.^4+4*t.^2);
  F= sqrt((t.^3/sqrt(3)).^2+t.^2);
  fun= @(t) sqrt(3*t.^4+4*t.^2);
  G= integral(fun,1,5);

  length= sqrt((5.^3/sqrt(3)).^2+5.^2)-sqrt((1.^3/sqrt(3)).^2+1.^2);

  plot(A,B,C,D,'--',t,E,t,F)
  xlim([1 5])
  title("curve $r(t)=(\frac{t^3}{\sqrt{3}},t^2)$","Interpreter",'latex')
  legend("Curve","Curve Tangent","Tangent vector length","Curve vector 
  length","location","northwest")
  dim = [.15 .40 .3 .3];
  dim2 = [.15 .20 .3 .3];
  str = {sprintf('Curve length with integration ≈  %f ', G),sprintf('Curve 
  length with vector length ≈ %f', length)};
  annotation('textbox',dim,'String',str,'FitBoxToText','on');
  saveas(gcf,'Plot.png')

Answer 1

Because in general $$ \|r'(t)\|\ne\|r(t)\|'. $$ $\|r(5)\|-\|r(1)\|$ is the length of the vector from the initial to the end point of the curve. If you try the second method with a closed curve, you will get that its length is $0$, which is clearly wrong.