
Re: An Algebra 2 Test
Posted:
Sep 30, 2012 11:14 PM


By the way, if you missed the step as to how I arrived at the minimum, it was simply looking at the speed on the beach as being a vector (Vb) and thus the rate of closing on the ball is Vb * Sin(DBC) where DBC is the angle formed by DBC. We are at the minimum when this rate is equal to the swimming speed (Vw) because running any further would close the distance to the ball slower than swimming directly towards the ball. Thus...
Vb * Sin(DBC) = Vw Sin(DBC) = Vw/Vb DBC = ArcSin(Vw/Vb)
Knowing DBC we can then find DC...
DC = BC * Tan(DBC)
And putting it all together...
DC = BC * Tan(ArcSin(Vw/Vb))
Bob Hansen
On Sep 30, 2012, at 2:07 PM, Robert Hansen <bob@rsccore.com> wrote:
> Here is a problem I found on Dy/Dan's thread, believe it or not. > > http://blog.mrmeyer.com/?p=15160 > > Also, here is another teacher's take on this problem title "Do Dogs Know Calculus?" > > http://www.maa.org/features/elvisdog.pdf > > And here is my analysis (which actually turns out to be simple enough for a dog)... > > After reading the article ?Do Dogs Know Calculus?? I tried looking at this problem again from the dog?s point of view. So your master has thrown a ball down the beach and into the water and you want to fetch it as fast as you can. To put this another way, you want to close the distance between you and the ball as quickly as possible. Since the ball is down the beach (and out in the water), at the beginning, running on land closes this distance faster. At some point though, as the angle between you, the beach and the ball widens, running along the beach does not close this distance as fast as swimming directly to the ball. Thus, the dog is simply making a judgement call and changing course when swimming directly to the ball closes this distance faster than running at an angle to it (along the beach). People do this as well when navigating crowds. Rather than plotting a direct path through the crowd we move along side it until a direct path is the faster course. > > In any event, the distance DC is simply BC * Tan(ArcSin(Vw/Vb)) where Vw is the speed in water (swimming) and Vb is the speed on the beach (running). In this problem, DC = 5 * Tan(ArcSin(1/4)) = 1.29m. > > Note: It doesn?t matter how far down the beach the dog starts, assuming of course that it is farther than 1.29m. The point at which it makes more sense to start swimming towards the ball depends only on how far out the ball is and the speed in water versus on land. In fact, if the dog is anywhere within that 1.29m mark, then jumping straight into the water will be the quickest course of action, and probably the one the dog would choose. > > Bob Hansen

