I am asked to find the length of the parametric curve and given:
$x = 5t,\ y=4t^2,\ z = 3t^2, 0 < t < 2 $
Following the formula for length of a parametric curve I got :
$L = \int_0^2 \sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2 }$
With:
$f'(t) = 5$ $g'(t) = 8t$ $h'(t) = 6t$
placing this back in the equation gives
$L = \int_0^2 \sqrt{(5)^2 + (8t)^2 + (6t)^2 } = $ $\int_0^2 \sqrt{25 + 64t^2 + 36t^2 }$ $ = \int_0^2 \sqrt{25 + 100t^2 }$
I cant seem to figure out the integral at this point.
$ \int_0^2 \sqrt{25 + 100t^2 } dt$
= $ 5\int_0^2 \sqrt{4t^2 + 1 } dt$
Let $x = \frac{\tan(u)}{2} \rightarrow u= \tan^{-1}(2x) \rightarrow\frac{dx}{du}= \frac{sec^2(u)}{2}$
So, the integral is equal to $ \int_0^2 \frac{\sec^2(u)\sqrt{\tan^2(u) + 1}}{2} du$
Using the trignometric identity $\tan^2(u) + 1 = \sec^2(u)$, we simplify even further:
$ \frac{5}{2}\int_0^2 \sec^3(u) du $
Using the reduction formula, we have that this is equal to
= $ \frac{5}{2} \int_0^2 sec(u) du$ + $\frac{sec(u)tan(u)}{2}$
$= \frac{5}{2}(\frac{\ln(\tan(u)+\sec(u)}{2} + \frac{\sec(u)\tan(u)}{2} \left.\right\vert_{0}^{2}$)
Now we must plug $u$ back in:
= $\frac{5\ln(\sqrt{4x^2 + 1} + 2x)}{4} + \frac{5x\sqrt{4x^2 + 1}}{2} \left.\right\vert_{0}^{2}$
When we evaluate this expression from $0$ to $2$, we obtain the answer of $23.234$.
Note that a hyperbolic sin inverse substitution could have been used on $sec^3(x)$, however, I did not do this as it would have made typing the solution more tedious.
The hyperbolic form of the integral, which gives the same answer, is:
$\frac{5arcsinh(2x)}{4} + \frac{x\sqrt{100x^2+25}}{2}$