Unit tangent vector and curvature with arc length parameterization

[songoku]

Homework Statement

Please see below

Relevant Equations

ds/dt = |r'(t)|

T(t) = r'(t) / |r'(t)|

K = |dT/ds|

(a)
$$\frac{ds}{dt}=|r'(t)|$$
$$=\sqrt{(x(t))^2+(y(t))^2+(z(t))^2}$$
$$=\frac{2}{9}+\frac{7}{6}t^4$$

$$s=\int_0^t |r'(a)|da=\frac{2}{9}t+\frac{7}{30}t^5$$

Then I think I need to rearrange the equation so $t$ is the subject, but how?

Thanks

Edit: wait, I realize my mistake. Let me redo

 

[songoku]

I think I can do (a). I got $s=t^2$

For (b), I just want to ask about the correct approach. I think I need to use the chain rule to find r'(t) so $r'(t)=\frac{dr}{ds}\times \frac{ds}{dt}$

Am I correct? Thanks