A circular helix in $xyz$-space has the following parametric equations, where $\theta \in \mathbb{R}$.
$x(\theta)= 5\cos(\theta)$
$y(\theta)=5\sin(\theta)$
$z(\theta)=\theta$
Let $L(\theta)$ be the arc length of the helix from the point $P(\theta)=(x(\theta), y(\theta), z(\theta))$ to the point (5, 0, 0), and let $D(\theta)$ be the distance between $P(\theta)$ and the origin. If $L(\theta_0)=26$ then $D(\theta_0)=$
I understand that the arc length of a parametric curve is the integral of the square root of the squared derivatives of each parametric equations.
So here is my attempt:
$L(\theta_0)=\int_0^{\theta_0} \sqrt{(-5\sin(\theta)^2+(5\cos(\theta))^2+1))} d\theta=26$
$=\sqrt{26}*\theta_0=26$
$\rightarrow \theta_0=\frac{26}{\sqrt{26}}$
Applying the distance formula we know
$\sqrt{(5\cos(\theta_0))^2+(5\sin(\theta_0))^2+(\theta_0)^2)}$
plugging everything in gives me
$\sqrt{\frac{701}{26}}$
Which is none of the options provided.
The derivatives are: $$x'(\theta)=(5\cos(\theta))'=-5\cdot\sin(\theta)$$ $$y'(\theta)=(5\sin(\theta))'=5\cdot\cos(\theta)$$ $$z'(\theta)=(\theta)'=1$$ Notice also that the point $(5, 0, 0)$ is achieved by $\theta=0$, hence the integral for $L(\theta_0)$ is:
$$L(\theta_0)= \int_{0}^{\theta_0}\sqrt{(-5\cdot\sin(\theta))^{2}+(5\cdot\cos(\theta)^{2}+(1)^{2}}d\theta= \int_{0}^{\theta_0}\sqrt{25(\cos^{2}(\theta)+\sin^{2}(\theta))+1}d\theta$$ $$=\int_{0}^{\theta_0}\sqrt{26}d\theta=\sqrt{26}\theta_0$$ By the given, this implies: $$\sqrt{26}\theta_0=26 \implies \theta_0=\frac{26}{\sqrt{26}} \implies \theta_0=\sqrt{26}$$
Now we are left with finding the distance of $(x(\theta_0), y(\theta_0), z(\theta_0))$ from the origin. By the distance formula:
$$D(\theta_0)=\sqrt{(5\cos(\theta_0))^{2}+(5^{2}\sin(\theta_0))^{2}+\theta_0^{2}}=\sqrt{25+\theta_0^{2}}=\sqrt{25+26}=\fbox{$\sqrt{51}$}$$
Your solution seems to be correct except for the final plugging-in part, in which I don't know what went wrong.