# Thread: Arc Length and parametric equations

1. ## Arc Length and parametric equations

I'm having a few problems with these questions... any help is appreciated! Thanks!

find $\displaystyle d^2y \over{dx^2}$ of $\displaystyle x = \sqrt{t}$; $\displaystyle y = 2t + 4$ at $\displaystyle t = 1$

Find the arc length of the curve $\displaystyle x = e^t cos(t)$, $\displaystyle y = e^tsin(t)$, $\displaystyle 0 <= t <= {\pi\over{2}}$

2. Originally Posted by akwingzero
I'm having a few problems with these questions... any help is appreciated! Thanks!

find $\displaystyle d^2y \over{dx^2}$ of $\displaystyle x = \sqrt{t}$; $\displaystyle y = 2t + 4$ at $\displaystyle t = 1$
[snip]
The easy road: Eliminate the parameter to get the Cartesian equation:

$\displaystyle x = \sqrt{t} \Rightarrow x^2 = t$ .... (1)

$\displaystyle y = 2t + 4$ .... (2)

Substitute (1) into (2): $\displaystyle y = 2x^2 + 4$

Getting the double derivative should be simple. For the record, you'd then note that $\displaystyle t = 1 \Rightarrow x = 1$ and do the substitution. But there's no need to substitute anything on this occassion.

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$\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = (2) (2 \sqrt{t}) = 4 \sqrt{t}$ since $\displaystyle \frac{dx}{dt} = \frac{1}{2 \sqrt{t}}$.

Therefore:

$\displaystyle \frac{d^2 y}{dx^2} = \frac{d}{dx} \left[ \frac{dy}{dx} \right] = \frac{d}{dx} [4 \sqrt{t} ] = \frac{d}{dt}[4 \sqrt{t}] \times \frac{dt}{dx} = \left( \frac{2}{\sqrt{t}} \right) \, \left( 2 \sqrt{t}\right) = 4$.

Originally Posted by akwingzero
[snip]
Find the arc length of the curve $\displaystyle x = e^t cos(t)$, $\displaystyle y = e^tsin(t)$, $\displaystyle 0 <= t <= {\pi\over{2}}$
I assume you're familiar with the arc length formula and I assume you can successfully differentiate both x and y with respect to t. That doesn't leave much to have trouble with.

You might note that $\displaystyle \left(e^t \cos t - e^t \sin t \right)^2 + \left(e^t \sin t + e^t \cos t \right)^2 = e^{2t} \left( [\cos t - \sin t ]^2 + [\sin t + \cos t]^2 \right) = 2 e^{2t}$

after expanding and simplifying. Therefore the integrand in the arc length formula boils down to $\displaystyle \sqrt{2} e^t$. Now it should be blue sky all the way.

3. Originally Posted by akwingzero
I'm having a few problems with these questions... any help is appreciated! Thanks!

find $\displaystyle d^2y \over{dx^2}$ of $\displaystyle x = \sqrt{t}$; $\displaystyle y = 2t + 4$ at $\displaystyle t = 1$

Find the arc length of the curve $\displaystyle x = e^t cos(t)$, $\displaystyle y = e^tsin(t)$, $\displaystyle 0 <= t <= {\pi\over{2}}$
For the second one, I would just note that this is the paramerization of the curve

$\displaystyle r=e^{\theta}$

Then we can uset that arc length in polar coordinates is

$\displaystyle \int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^ 2}$

So we would have

$\displaystyle \int_0^{\frac{\pi}{2}}\sqrt{(e^{\theta})^2+(e^{\th eta})^2}d\theta=\int_0^{\frac{\pi}{2}}\sqrt{2}e^{\ theta}d\theta$

Same as Mr. F, just a little less messy.