Question

Find the arc length of the curve x= y^4/8+1/〖4y〗^2 from y=1 and y=4

Answer 1

We need the curve $x=\frac{y^4}8+\frac1{16y^2}$ from

$y=1$ to $y=4$. So let's show it on the picture.

To find the length of this curve we should use the formula $L=\int_a^b{\sqrt{1+(x')^2}}dy,$

where

$x'=\frac{y^3}2-\frac1{8y^3}=\frac{4y^6-1}{8y^3};$

$(x')^2+1=\frac{16y^{12}-8y^6+64y^6+1}{64y^6}=\\ {}=\frac{16y^{12}+56y^6+1}{64y^6};$

$L=\int_1^4\sqrt{\frac{16y^{12}+56y^6+1}{64y^6}}dy;$

There is no suitable analytical solution of this integral, but we can calculate it approximately. For this we use the trapezium method in Matlab. The following lines provide the code of calculations and the output:

Y=1:0.01:4;

X=sqrt((16*Y.^12+56*Y.^6+1)./(64*Y.^6));

I=trapz(Y,X);

disp(I);

Then, the result is given by $L\approx32.2323.$