Scroll down a bit to see the problem.
Simple derivative.
You have a formula that shows the speed for a combination of water and land. Take the first derivative, set it to zero, solve for time, and then compare that to the answers you already calculated for “all water” and “straight across.”
Easy peasy. The minimum point on a curve is the point where the slope equals zero.
I wonder what grade level the kids taking this are? If they know a bit of Calc one, it’s very easy. If not, not.
Article says 17 and 18 year olds.
And…
Tears? Really?
OK, so let’s say I don’t know derivatives. It’s still easy, just a bit more time consuming.
Start with the endpoints. Then calculate the time in the middle (crocodile swims halfway between the giraffe and the straight-across point.) If it’s higher than either endpoint, you’re done: th elower endpoint is the optimal time. If it’s lower, then pick the middle of the remaining segments. Eventually you find the lowest time. It’s only a second-degree equation so there is only one absolute minimum. Slightly more work, but still easy peasy and all done with algebra.
Trial and error can solve a lot of problems, but it’s pretty tedious. I don’t think they’d put an optimization problem in a test for people who hadn’t studied the best way to solve those problems.
Agreed that it’s tedious. But with a calculator, it’s not all that time consuming, and you’re right that these kids probably have studied at least derivatives of quadratics.
This reminds me of Feynman’s observation about knowledge being fragile. He relates the story of being in class with the then-classic plastic protractor with the French curve on top, and one of the guys says, “I wonder why they put this curve on this thing?”
Feynman immediately responds, “Didn’t you know? It’s a special curve: no matter how you turn it, the slope at the lowest point of the curve is horizontal.”
They all start turning the curve different ways and putting their pencils tangent to the lowest point, and are amazed that it’s true.
As Feynman observes, they did this even though they had all learned that the first derivative to any curve is the slope at that point, and where the slope is zero the curve is at its lowest. And a line with a slope of zero is horizontal.
I figured someone would be upset about the crocodile going after the zebra.
It’s an INCREDIBLY poorly phrased problem, although I don’t think any of the math is particularly hard (assuming the students have had calculus).
Assuming I’m understanding the question properly, and the fact that I’m not quite confident I am is a pretty damning indictment, it’s basically saying:
(1) Here’s a formula that tells you how long it will take the crocodile to reach the zebra if he first swims across, landing X feet upstream from where he currently is, then walks on land the rest of the way to the zebra
(2) (this is a(i) ) Using that formula, how long will it take the crocodile to swim directly to where the zebra is standing? (ie, if X = 20)
(3) (this is a(ii) ) Using that formula, how long will it take the croc to swim straight across the river, emerging precisely opposite where he started, then walk to the zebra? (ie, if X = 0)
(4) (this is b ) If the croc wishes to minimize the total time it takes him to reach the zebra, there’s a particular point which is p feet upstream from where he currently is that he should swim to. Calculate p. (ie, find X such that formula(x) is minimized.)
But it’s phrased kind of circularly, where we mention p, as if it were going to be an input to further steps in the problem. Then it isn’t. Then the third step (I believe) wants us to calculate p, but doesn’t actually say so.
Right, but that doesn’t tell us whether or not the 17 and 18 year olds would have been expected to have studied Calculus.
As noted, it’s a pretty standard type of Calculus problem: I bet you could find similar exercises in pretty much any Calc textbook.
[QUOTE=Bricker]
Start with the endpoints. Then calculate the time in the middle (crocodile swims halfway between the giraffe and the straight-across point.) If it’s higher than either endpoint, you’re done: th elower endpoint is the optimal time. If it’s lower, then pick the middle of the remaining segments. Eventually you find the lowest time. It’s only a second-degree equation so there is only one absolute minimum. Slightly more work, but still easy peasy and all done with algebra.
[/QUOTE]
This method isn’t guaranteed to give an exact answer, but will come arbitrarily close given enough iterations. (In this case, the answer turns out to be a whole number, but if you tweak the parameters a bit it could easily be irrational.)
This is also the kind of thing that could be done with a graphing calculator by someone proficient in its use.
I took calculus in high school and we could solve this with derivatives and algebra. The hard part of the problem is the crappy diagram and the fact that you have to deduce the geometry of the problem from the results.
Tx = 5 * (36+x^2)^(1/2) + 4 * (20-x)
So time is in seconds (s) and speed is in meters/second (m/s) well now we need to know some very basic physics to know that time it takes to get from a to b is distance/time. Most calculus students probably know that but it’s not a certainty.
Look at Tx and you can figure out the aspects of geometry not shown on the diagram. The width of the river for example and the diagonal that the crocodile swims is figured from the 5 * (36+x^2)^(1/2) term. And the horizontal land distance time is from the 4 * (20-x) term.
But first lets figure out these units real quick. (36+x^2)^(1/2) and (20-x) each are meters. 5 and 4 have the units s/m. This might trip up high school calculus students who had no physics and aren’t used to working with units.
So to answer
a. i. No travel on land? The crocodile is swimming a distance of ((6m)^2+(20m)^2)^0.5 with a time per meter of 5s/m. So 104s.
a. ii. Minimum time in water? That is straight across the 6m wide river. So 6m5s/m+20m4s/m = 110s.
Now the trickier part. Does it require calculus? I’m not 100% sure it does. But I think solving it with calculus would be the easiest method for someone who has had high school calculus. And that is taking the first derivative of the time equation and setting that equal to 0 and solving for x. With a calculator or with proper study like for an AP test the calculus portion is straightforward. It’s been a few years so I’ll use electronic help.
In which case x = 8m. I wouldn’t have had an issue with the calculus or algebra back in the day. Nor the physics. It would have taken me a few minutes to puzzle out the geometry from the term because typically the questions we had to work with in our high school physics and calculus were a bit more straightforward in terms of presenting a clear diagram.
T(8m) would be 98s.
Bricker said it was a giraffe.
It does appear that such questions can be solved without calculus, however I’m not sure about the algebra involved. Any of you folks use the method described here?http://totalgadha.com/mod/forum/discuss.php?d=4040
Why the fuss about calculus? Calculus is on the syllabus for Scottish Highers. Certainly it was on the syllabus for my A levels (taken at 18) and AO levels (16).
Previous discussoin of this problem A few questions about the crocodile math problem that made the Scottish kids cry - In My Humble Opinion - Straight Dope Message Board
It was on the syllabus as well for my 2º de BUP, 16. We did limits and derivatives in 2º, integrals in 3º.
I’d have to actually think about it, since the braincells where I stored derivatives haven’t been used in over a decade, but as a math book I taught from said “this book doesn’t have problems, it has exercises. Something is a problem if you don’t know how to solve it; since this book begins by teaching how to solve the questions given, those questions are exercises”.
I admit, I can’t get past the wording, and the diagram is the opposite of helpful. How does the crocodile not travel on land when it’s clearly on sodding land and so’s the zebra?
Oh, and crocodiles don’t stalk zebras. Run this crap past the biology department first please, crocodiles are sit and wait predators, there’s no way in hell a crocodile is running down a zebra on land unless the zebra is dead.
Yes, it looks like an easy exercise in calculus. The answer to the last part is x = 8, incidentally. The question is simple if you have had done calculus, nearly impossible if you haven’t. A good illustration of the power of calculus.
I was reflecting on this problem just the other day, and I agree that it’s a very simple problem using the first derivative of the given equation (and ignoring the terrible wording and everything that’s biologically unrealistic about the situation). However, the fragility of the knowledge (as discussed above, which I think is true) might be due to the fact that you have to take what might be an unfamiliar derivative, dT / dx, where T is the time taken and x is the independent variable. I wonder if the students were just so used to derivatives of the form dx/dt that they tried to recast the equation for T in terms of x, and then got into a mess.
Of course, all that proves is that they haven’t really learnt calculus, but rather how to apply a set of equations to solving a problem without thinking about what the problem is really asking for.