Minimum problem involving rates and basic trig

In summary, the man at point A on the shore of a circular lake of radius 1mi wants to reach the opposite point C as soon as possible by walking 6mi/h and rowing his boat 3mi/h. To find the angle theta to the diameter AC at which he should row, he uses the equation T = AB/3 + 2theta/6 and calculates AB using the law of cosines. After some confusion, he realizes that AB = 2cos(theta) and sets T' equal to zero to find theta = pi/6, which is a maximum point for the shortest total time.
  • #1
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Homework Statement



A man at point A on the shore of a circular lake of radius 1mi wants to reach the opposite point C as soon as possible. He can walk 6mi/h and row his boat 3mi/h. At what angle theta to the diameter AC should he row?

The Attempt at a Solution



My diagram looks like this: I first drew the diameter of the circle AC with A being on the left. Next, I called the point to which the man should row B and I drew a line from A to B (B being somewhere between theta = 0 and theta = pi/2). Finally, I drew another line from O (the center of the lake) to B. Therefore, angle CAB = theta and angle COB = 2theta. Moreover, AC=2, so OB=1, AO=1, and OC=1.

My initial equation for the shortest total time that it will take the man to reach point C is:

T = AB/3 + 2theta/6

Now, I know that I need to calculate the derivative of T and set it equal to zero to minimize T. However, I'm not sure how to calculate AB.

By the law of cosines:

(AB)2=AO2 + OB2 -2ab(cosAOB), which equals:

(AB)2=12 + 12 - 2cos(AB) or 2-2cos(AOB)

This is where I get confused. I set cos(AOB) equal to cos(2theta - 2pi), but that isn't working out for me. The correct answer has AB = 2cos(theta). I've been working at this for a while and I don't understand how to get AB = 2cos(theta). I would greatly appreciate your help.
 
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  • #2
I figured this out, if anyone is interested. I set AOB equal to cos(pi-2theta) instead of cos(theta - 2pi).

(AB)2 = 2 - 2cos(pi - 2theta)
(AB)2=2-2(cos(pi)cos(2theta) + sin(pi)(2sin(theta)cos(theta))
Because cos(pi)=-1 and sin(pi)=0, we have
(AB)2 = 2-2(-cos(2theta))
(AB)2=2(1+cos(2theta))
(AB)2=2(2cos2(theta)
(AB)2=4cos2(theta)
AB=sqrt[4cos2(theta)]
AB=2cos(theta)

Therefore, the shortest total time is

T=(2cos(theta))/3 + (2theta)/6

T'=[-2sin(theta) +1]/3
Setting T' equal to zero yields sin(theta)=1/2

Therefore, theta equals pi/6.

However, T''=[-2cos(theta)]/3, so the critical point sin(theta)=1/2 is a maximum. Is this right? Shouldn't it be a minimum?
 
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FAQ: Minimum problem involving rates and basic trig

1. What is a minimum problem involving rates?

A minimum problem involving rates is a type of mathematical problem that involves finding the minimum value of a variable while considering the rate of change of another variable. These types of problems are often used in real-world scenarios to optimize a certain outcome.

2. How are rates and basic trigonometry related in minimum problems?

In minimum problems involving rates, trigonometry is often used to represent the relationship between variables. This can be done through creating a right triangle and using trigonometric functions such as sine, cosine, and tangent to represent the rates of change of the variables involved.

3. Can you give an example of a minimum problem involving rates and basic trigonometry?

Sure, an example could be a farmer trying to determine the optimal dimensions of a rectangular pen for their chickens. The farmer knows that the pen must have a perimeter of 100 feet and wants to minimize the amount of fencing used. This problem involves rates (perimeter) and basic trigonometry (using the Pythagorean theorem to represent the sides of the pen).

4. What are some strategies for solving minimum problems involving rates and basic trigonometry?

One strategy is to draw a diagram and label the variables involved. From there, you can use trigonometric functions and equations to represent the rates of change and set up an equation to find the minimum value. Another strategy is to use calculus techniques such as taking derivatives to find the minimum value.

5. How can minimum problems involving rates and basic trigonometry be applied in real life?

Minimum problems involving rates and basic trigonometry can be applied in various real-life scenarios such as optimizing production costs in manufacturing, maximizing profits in business, and determining the most efficient route for transportation. These types of problems help in making informed decisions and finding the best possible outcome.

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