1. ## Parametric Curves

I'm currently doing Parametric Curves where we have to fine the length of the curve using L = the integral of sqrt((dx/dt)^2+(dy/dt)^2) dt, but I'm having a lot of trouble simplifying the equations to make it easier to take the integral of them. Can I get some help on these few question?

2. Originally Posted by iebi
I'm currently doing Parametric Curves where we have to fine the length of the curve using L = the integral of sqrt((dx/dt)^2+(dy/dt)^2) dt, but I'm having a lot of trouble simplifying the equations to make it easier to take the integral of them. Can I get some help on these few question?

$x=\frac{t}{4+t} \implies \frac{dx}{dt}=\frac{4+t-t}{(t+4)^2} = \frac{4}{(t+4)^2}$
$y=\ln(4+t) \implies \frac{dy}{dt}=\frac{1}{t+4}$
$0\leq t\leq 5$

$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2} = \sqrt{\frac{16}{(t+4)^4}+\frac{1}{(t+4)^2}}$ $= \sqrt{\frac{16}{(t+4)^4}+\frac{(t+4)^2}{(t+4)^4}} = \frac{\sqrt{(t+4)^2+16}}{(t+4)^4}$

$\int_0^5\frac{\sqrt{(t+4)^2+16}}{(t+4)^2}\,dt$

This is a really ugly integral. Let $u=t+4$ and $du=dt$

Now you have $\int_0^5\frac{\sqrt{u^2+16}}{u^2}\,du$

Let $u=4\tan(v)$ and $du = 4\sec^2(v)\,dv$

Now you have $\int_0^5\frac{\sqrt{16\tan^2(v)+16}}{16\tan^2(v)}\ cdot4\sec^2(v)\,dv = \int_0^5\frac{\sec^3(v)}{\tan^2(v)}\,dv$

I have to go. Can you finish from here? If not, I'm sure someone else will help you. Hint: Split $\sec^3(v)$ into $\sec(v)(1+\tan^2(v))$.

3. Thanks. I got to the part where I combine the equations, but didn't know what to do after that. I can finish that problem. Can anyone start me off with the other 3 problems?

4. Originally Posted by iebi
Thanks. I got to the part where I combine the equations, but didn't know what to do after that. I can finish that problem. Can anyone start me off with the other 3 problems?
8.) $\frac{dx}{dy} = \frac{y^4}{2}-\frac{1}{2y^4} = \frac{y^8-1}{2y^4}$

Square it: $\frac{(y^8-1)^2}{4y^8} = \frac{y^{16}-2y^8+1}{4y^8}$

Add 1: $\frac{y^{16}-2y^8+1}{4y^8}+1 = \frac{y^{16}-2y^8+1+4y^8}{4y^8} = \frac{y^{16}+2y^8+1}{4y^8} = \frac{(y^8+1)^2}{4y^8}$

Take the square root: $\sqrt{\frac{(y^8+1)^2}{4y^8}} = \frac{y^8+1}{2y^4}$

Integrate: $\int_1^4\frac{y^8+1}{2y^4}\,dy = \int_1^4\frac{y^4}{2}+\frac{1}{2y^4}\,dy = \left[\frac{y^5}{10}-\frac{1}{6y^3}\right]_1^4 = \boxed{\frac{65577}{640} \approx 102.464}$

5. Can anyone show me how to start the last two problems?