# Find the arc length for the given interval (parametric curve)

1. Aug 9, 2009

### wh00ps

Find the arc length.

x = sqrt(t)
y = 6t - 2

Interval from 0 to 5 inclusive.

Whenever I do this, I get a long answer with big numbers in the numerator all divided by 48. Can someone walk me through the steps? THanks.

2. Aug 9, 2009

### mathman

It would be helpful if you could show what you did to get the "long answer ". In any case the derivative of arc length with respect to t (I assume the interval from 0 to 5 is a t interval) is infinite at t=0, so it is not surprising you get a big number.

3. Aug 10, 2009

### boris1907

Given parametric equation implies that $$y=6x^2 -2$$ , then you should use arc length formula: $$L=\int\limits_a^b {\sqrt {1 + [f'(x)]^2 } dx}$$ where $$f(x)=y=6x^2-2$$, a=0, b=$$\sqrt5$$ , f'(x)=12x
$$L=\int\limits_0^{\sqrt 5 } {\sqrt {1 + 144x^2 } dx}$$ ,after $$x = \frac{1}{{12}}\tan \theta$$ substution
$$\frac{1}{{12}}\int {\sec ^3 \theta d\theta } = \frac{1}{{24}}(\ln \left| {\tan \theta + \sec \theta } \right| + \tan \theta \sec \theta ) \]$$by using integration by parts
$$$\left. {\frac{1}{{24}}(\ln \left| {12x + \sqrt {1 + 144x^2 } } \right| + 12x\sqrt {1 + 144x^2 } )} \right|_0^{\sqrt 5 }$$$
$$$L = \frac{1}{{24}}(\ln \left| {12\sqrt 5 + \sqrt {721} } \right| + 12\sqrt 5 \sqrt {721} )$$$
approximately L=30.19 units