Very difficult calculus related rates question I was looking at a math problem from a few years ago that I could not solve. I was wondering if anyone knows where to even begin. I have the answer along with the question, however, I do not know how to arrive at this answer -
Here is the question:

The answer is:

 A: To simplify  this problem, we can change the perspective by noting that climbing a mountain with decreasing velocity is equivalent to climb with constant velocity a mountain that grows larger as we rise up. In particular, based on the data of the problem, we can see our progressively enlarging mountain as a cylinder: in fact, since at any height  $z $ the corresponding radius of the cone is  $r_0\,(h-z)/z\,\, \,$, its circumference is $2\,\pi\,  r_0(h-z)/z \,\,\,$ and the velocity is $ v_0\,(h-z)/z\,\, \,$, the time needed by a climber to cover the circumference is $2\,\pi \, r_0/v_0\,\,\,$, i.e. is independent by the height $z $. In other words, we   can simplify this problem by imagining a man climbing a cilindrical mountain having radius $r_0$ with constant velocity  $v_0$. The problem therefore reduces to that of calculating the position of our original target point on such a cylinder. 
To do this, we can note that if we call $L$ the slant of the initial cone and $x$ the distance from its top at a given instant of our ascent, the instantaneous velocity is $v_0 \cdot \frac{x}{L} \,\,$. So, at any instant of our ascent, to cover an infinitesimal distance $dx$ we need a time equal to $\frac{L}{v_0\,x} dx \,$. Integrating in the range between $L$ and $L/2$ (i.e., from the beginning of the ascent to the height corresponding to our target point) and multiplying to $v_0$ to get the distance covered on the cylinder, this leads to a distance of 
$$\displaystyle \int_{L/2}^L \frac{L}{x} dx=L \left[\log L - \log (L/2) \right]=L \log 2$$  
Therefore, to reach the original target point climbing on the cylinder, unwrapping the lateral surface of the cylinder on to a plane to see it as a rectangle, we have to cover a distance equal to the hypothenuse of a right triangle whose legs are $L \log 2$ and $ \pi \, r_0$. This directly yields a distance of 
$$\sqrt {(L \cdot \log 2)^2 + (\pi r_0)^2}$$
which divided by the velocity $v_0$ (remind that climbing on our cylinder we have assumed the velocity to be constant) gives a time $T$ equal to
$$T= \sqrt {\frac {(L \cdot \log 2)^2}{v_0^2} + \frac{(\pi r_0)^2}{v_0^2}}$$
Because the slant $L$ of the initial cone is equal to $\sqrt{h^2 + r_0^2} \,\,\,$, we obtain
$$T=\sqrt {\frac {(h^2 + r_0^2) \cdot \log^2 2}{v_0^2} + \frac{(\pi r_0)^2}{v_0^2}}$$
and substituting $h=4 \,\,$, $r_0=100\,\,$, and $v_0=2\,\,$, we obtain for $T $ the value
$$\sqrt {\frac {(4^2 + 100^2) \cdot \log^2 2}{2^2} + \frac{(100 \pi)^2}{2^2}} \\ = \sqrt {2504 \cdot \log^2 2 + 2500 \pi^2}  $$
A: I got a different answer, with terms that look remarkably similar. I didn't use the trick of transforming the problem into climbing a growing mountain with constant velocity. I started by noticing that the time to walk from one side the mountain to the opposite point at any height is constant - namely,  $\frac{πr_0}{v_0}$, or, in this case, $50π$. 
Walking up and toward the point on the opposite side halfway up has horizontal and vertical components. At any point on our way up we cover a horizontal distance around the mountain $dx$ and distance up the slope of $ds$ in an interval $dt$. 
The time to cover each $dx$ up the mountain is the same since the time to go around at any height is the same. So, the time to complete the horizontal movement, as long as we don't head back down on our way to the other side, is $50\pi$. The sum of the $ds's$ for all the points on our walk - again, assuming we keep heading up and towards our goal, is just the distance up the side if we were to head straight up - half the length of mountain side, or $\frac{1}{2}*\sqrt{10000+16}$. Let's call that $S$. 
The speed up $S$ varies by height, so $\frac{ds}{dt}=(8-2h)/4=(4-h)/2$. We'd really like to get to $\frac{dt}{dh}$ so we can integrate from $h=0$ to $h=2$ to get the time to get that far up the slope. The vertical component of $\frac{ds}{dt}$ is proportional to the ratio of height to length of the slope, so $\frac{dh}{dt}=(4/S)*ds/dt=(4/S)*(4-h)/2$. 
Then $\frac{dt}{dh}=(S/4)*(2/(4-h)=(S/2)/(4-h)$. 
Integrating, we get $t=-(S/2)*\ln(4-h)$ from $h=0$ to $h=2$, or $t=-(S/2)(ln2-ln4)=-(S/2)(\ln2- 2\ln2)=(S/2)\ln2$. 
So total time is time to cover horizontal distance plus time to cover vertical distance $= 50π+(S/2)\ln2$. But $S/2=(\sqrt{10016})/2=\sqrt{10016/4}=\sqrt{2504}$. 
So my answer is $\boxed{50π+\ln2*\sqrt{2504}}.$
You'll note if you square both the terms, and take the square root of the sum, you get the book answer. 
