Answer to Question #100153 in Calculus for Joseph Yep

Question

1. Let L be the length of the curve y= x^5 + 8/ 16x^2 from x=2 to x=3, an integral expression for L is ________.

2. Let L be the length of the curve x= y^4/8 + 1/ 4y^2 from y= 1 to y=4, an integral expression for L is______.

Answer 1

Length of the curve is $\int\limits_{t_1}^{t_2}\sqrt{(x'(t))^2+(y'(t))^2}dt$ , where $(x(t),y(t)), t\in[t_1,t_2]$ is parametrization of the curve.

1)We have $x(t)=t$ and $y(t)=\frac{t^5+8}{16t^2}$ , $t\in[2,3]$

So $x'(t)=1$ and $y'(t)=\frac{5t^4\cdot 16t^2-(t^5+8)\cdot 32t}{256t^4}=\frac{3t^5-16}{16t^3}$

$L=\int\limits_2^3\sqrt{1+\bigl(\frac{3t^5-16}{16t^3}\bigr)^2}dt$

2)We have $x(t)=\frac{t^4}{8}+\frac{1}{4t^2}$ and $y(t)=t$ , $t\in[1,4]$

$x'(t)=\frac{t^3}{2}-\frac{1}{2t^3}=\frac{t^6-1}{2t^3}$, $y'(t)=1$

$L=\int\limits_1^4\sqrt{1+\bigl(\frac{t^6-1}{2t^3}\bigr)^2}dt=\int\limits_1^4\frac{t^6+1}{2t^3}dt$

Answer: 1)$\int\limits_2^3\sqrt{1+\bigl(\frac{3t^5-16}{16t^3}\bigr)^2}dt$ , 2)$\int\limits_1^4\frac{t^6+1}{2t^3}dt$