Otdfbt

Find the length of the following parametric curve.

$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$

I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.

My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$

$$frac23cdot 17^3/2+4-frac23$$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$

Which gives us:

$$int_0^2 24sqrtt^6+t^10dt$$

Which, when integrated, gives us: $$68sqrt17-4$$

I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.

Line 5 is correct.

Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.

You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$

Alternatively:
$$begincasesx=5+6t^4\ y=5+4t^6endcases Rightarrow begincasest^2=left(fracx-56right)^1/2\ t^2=left(fracy-54right)^1/3endcases Rightarrow y=4left(fracx-56right)^3/2+5\
0le tle 2 Rightarrow 5le xle 101$$
Hence:
$$S=int_a^b sqrt1+y'(x) dx= int_5^101 sqrt1+left(fracx-56right) dx=\
=int_5^101 sqrtfracx+16 dx=4cdot fracx+16cdot sqrtfracx+16bigg_5^101=\
=68sqrt17-4.$$