# Optimizing hiker's walk

1. Aug 27, 2015

### PAstudent

1. The problem statement, all variables and given/known data

A hiker starting at point P wants to get to a forest cabin, which is two miles from the road at point Q (See diagram). The hiker can walk at 6.00 mph on the road and 2.50 mph through the forest. Suppose the hiker walks a distance X down the road before cutting through the forest, straight to the cabin. Determine the value of X for which the trip take the least time. Consider the distances to be good to three significant figures.

2. Relevant equations
Trigonometry and Derivatives

3. The attempt at a solution
I attempted to draw the lines again and label points to simplify the triangle. However, I am getting stuck on how to optimize the walk.

Last edited by a moderator: Aug 27, 2015
2. Aug 27, 2015

### Staff: Mentor

You really need to make an effort when posting schoolwork questions here.

You are given the speed for both paths. Write an equation for the total time taken based on those speeds and different values of x. Please show your work.

3. Aug 27, 2015

### PAstudent

I'm sorry I am awful at physics and my 70 year old professor doesn't teach anything. Just assigns chapters to read and gives us the basic formulas. I hope you have a great day

4. Aug 27, 2015

### Staff: Mentor

I'm sorry you are giving up so easily. It just takes some practice to get comfortable with these types of problems.

In case you can give it one more try, use my advice from my previous post about writing the equation for the time taken based on 2 paths and 2 speeds. Then plug in some numbers to see what the total time looks like for a few sample choices of x. Like if x=0, then it's just the diagonal path length at the woods-walking speed. And if x = 3mi, then the total time is formed by walking 3mi at the path speed and 2mi at the woods speed. The quickest time is probably somewhere in the middle of those two choices...

5. Aug 27, 2015

### ZenchiT

It can be really difficult when you have a teacher who doesn't give much apart from chapters in a book. The effect teachers can have on someones passion for a subject is huge

6. Aug 27, 2015

### PAstudent

That is the issue. I just freeze up when trying to write equations. I just don't know what to do. I know you guys hate hearing that and see it as a lack of effort.

7. Aug 27, 2015

### haruspex

Add two more labels to the diagram. Label the point where the walker leaves the road A. The point on the road that is nearest to point Q label point R.
I want you to use symbols for the given values, not the numbers. Use PR instead of '3 miles' and QR instead of '2 miles'. Use u instead of 6 mph and v instead of 2.5mph. This is a useful habit to get into. You already have x for the (unknown) distance walked on the road.

Try to answer these questions in sequence:
In terms of these symbols, how long will the walker be on the road?
How far will the walker go through the forest? How long will that take?

8. Aug 27, 2015

### PAstudent

So the how long walk on the road= (PR-x)/u?

9. Aug 27, 2015

### billy_joule

time = distance/velocity

So
time = x/u

Last edited: Aug 27, 2015
10. Aug 27, 2015

### PAstudent

So to find how far walked through the forest would it be [sqrt(QR^2+AR^2)]/v which would find that straight diagonal line? Or would I have to use angles?

11. Aug 27, 2015

### billy_joule

Pythagoras works fine. No need for trig.

So now you have expressions for the time of each leg, the total time is just the sum of these.
Note that you'll need to express AR in terms of x.
You mention derivatives in the OP so I assume you know about minimisation problems?

12. Aug 27, 2015

### PAstudent

I understand derivatives, but I never really used them with minimization problems. How would you go about using a derivative to do that?

13. Aug 27, 2015

### haruspex

If you have an independent variable x and a function of it f(x) that you want to find either a local minimum or local maximum of, you look at the derivative f'(x) of f (df/dx). At a point that is locally a minimum, a small change in x either way will increase f, so f' will be negative on one side of x and positive on the other. Right at the minimum, f' will be zero.

14. Aug 27, 2015