What is the shortest time to reach Q on a circular lake?

[vasdueva]

Homework Statement

A man with a boat is located at point P on the shore of a circular lake of radius 5 miles. He wants to reach the point Q on the shore diametrically opposed to P as quickly as possible. He plans to paddle his boat at an angle t(0<t<pi/2)<or equal to** to PQ to some point R on the shore, then walk along the shore to his Q. If he can paddle 3.4 miles per hour and walk at 3.8 miles per hour, what is the shortest possible time it will take him to reachQ?

Homework Equations

The Attempt at a Solution

I've got one part of the problem, and I understand the concept of optimization, but how would i find the length of the paddle distance?

((t*10pi)/(2pi)) this is what i came up with to find the length of his walking distance. then divide that by 3.8 and divide whatever his paddle distance is by 3.4 to have total time. If I have the formula, I can easily differentiate it, so no need to do any of that.

 

[vasdueva]

I figure if i draw a line perpendicular to the center point I can find part of the length of the distance, but how would i go from there to the whole length

 

[HallsofIvy]

A geometry theorem: An angle of measure [/itex]\theta[/itex] with vertex on a circle of radius r cuts of arc with angular measure \theta/2 and so length \frac{\theta r}{2}. In this case, the lake has radius 5 mi., so the distance he must walk around the lake is \frac{5\theta}{2} mi
The straight line distance he must paddle is a little harder. The arc from P to the point where he lands has angular measure \pi- \theta/2 and that is the angle the two radii from those points measure at the center. Use the cosine law to determine the length of the third side of a triangle with two sides of length 5 and angle between them \pi- \theta/2.

 

[vasdueva]

thank you, when I get home I'll see if the online webwork takes the my answer.

 

[vasdueva]

gimme a sec let me look how to type in the itex thing, it doesn't seem to like my answer

 

[vasdueva]

\frac{\frac{5\theta}{2}}{3.8}