A snail, who can travel at 3 km per hour (running), is 10 km from the nearest point P on a straight highway ... (Quetion continues)
A snail, who can travel at 3 km per hour (running), is 10 km from the nearest point P on a straight highway. She wishes to travel to a point Q, 50 km along the highway from P, and she can arrange to have a turtle pick her up anywhere along the highway. The turtle travels at 15 km per hour. What point on the highway should the snail head for so as to arrive at Q in the shortest possible time?
To solve this problem, we need a few parameters:
`d_(s)` = distance traveled by snail on its own
`d_(t) =` distance traveled by snail on turtle
`theta` = angle that snail travels toward road with respect to the line going from the start point to P
`t` = total time spent travelling by snail
It is pretty clear that the total time will be found by combining the distance travelled by the snail on its own divided by its own speed with the distance travelled by the snail with the turtle divided by the speed of the turtle:
`t = d_(s)/3 + d_(t)/15`
We can get the distance the turtle travels based on the distance between P and Q (50 km), the distance traveled by the snail, and the angle at which it approaches the highway through trigonometry. We just need to recognize a right triangle can be made by the three points P, the snail's start point, and the point at which the turtle and snail meet. This gives `d_(s)` as the hypotenuse and `theta` as the angle of approach:
`d_(t) = 50 - d_(s)sintheta`
We can express `d_(s)` now, also as a function of the angle `theta` and the distance between the snail's start point and P (10 km) in the following way:
`d_(s) = 10/costheta`
This also gives us a relation for `d_(t)` that is purely a function of theta! Keep in mind, sin(theta)/cos(theta) = tan(theta):
`d_(t) = 50 - 10tantheta`
We can now substitute `d_(s)` and `d_(t)` with functions of `theta`, the angle of approach to get a final equation for the total travel time of the snail:
`t = 1/3*10/costheta + 1/15*(50 - 10tantheta)`
`t = 10/3 sectheta + 10/3 - 2/3tantheta`
Now, to optimize, we take the derivative and set it to zero (you can find tables of derivatives for trig functions at the link below):
`dt/(d theta) = 10/3secthetatantheta -2/3sec^2theta`
`0 = 10/3secthetatantheta - 2/3sec^2theta`
To simplify, we'll multiply both sides by `cos^2theta` (recall, `sectheta = 1/costheta`)
`0 = 10/3sintheta - 2/3`
Now, we add `2/3` to both sides and divide by `10/3`:
`1/5 = sintheta`
Now, we take the arcsin of both sides. Keep in mind, theta must be between 0 and pi/2 in order for it to fit within the parameters we specified at the begining of the problem (Theta must be acute, or our assumption of a right triangle doesn't hold!):
`theta = 0.201`
We have our angle at which the snail must get to the road to minimize travel time!
Recall, now, that the distance the turtle must travel is given by the following (when we incorporate the formula we found for `d_(s)`):
`d_(t) = 50-10tantheta`
Now, if you examine this relation, you can see quickly that `10tantheta` will be the distance from P that the turtle will pick up the snail. Therefore, to find their meeting point, we just solve for `10tantheta`!
Here we go:
`10tantheta = 10tan(.201) = 2.04`
Therefore, the snail must aim to meet up with the turtle 2.04 km from point P.
Hope that helps!