Calculate the the length of the curve $\gamma(t)=(\frac{t^{2}}{4},\frac{t^{3}}{3},\frac{t^{4}}{4})$

Question

I just discovered that I have a Calculus 3 test on Thursday and my teacher posted a list of exercises for the test. As I'm a little desperate I thought about asking the question here and start studying in the process. The question is the following:

Find the length:

$\int^{}_{\gamma} dt|\dot{\gamma}(t)|$

where $\gamma(t)=(\frac{t^{2}}{4},\frac{t^{3}}{3},\frac{t^{4}}{4})$

The biggest problem is that right now I'm not really familiar with the notation used or even the theory involved in the solution of the question. Any help is appreciated during the moment of desperation that I find myself in right now

Answer 1

From Paul's Online Notes and other sources:

From Calculus II, $||\gamma'(t)||$ represents the arc length of the curve. By converting $\gamma(t)$ into the parametric form $x = f(t), y = g(t), z = h(t)$, we can extend this concept into three dimensions with the formula:

$$||\gamma'(t)|| = \int \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}$$

The notation $\int_{\gamma}$ means to find the arc length function from $0$ to $t$, so in your case, you have to find:

$$||\gamma'(t)|| = \int_0^t \sqrt{\left[\frac{t^2}{4}'\right]^2 + \left[\frac{t^3}{3}'\right]^2 + \left[\frac{t^4}{4}'\right]^2}$$

Answer 2

Being sure that you did follow what Taby Mak answered, after computing $x'$, $y'$ and $z'$, you should arrive to $$||\gamma'(t)|| = \int_a^b \sqrt{t^6+t^4+\frac{t^2}{4}}\,dt$$ and factoring, you should notice that $$t^6+t^4+\frac{t^2}{4}=t^2 \left(t^2+\frac{1}{2}\right)^2$$ which then makes the problem vey simple.