{"id":307,"date":"2014-08-07T15:55:48","date_gmt":"2014-08-07T19:55:48","guid":{"rendered":"http:\/\/advancedmathyoungstudents.com\/blog\/?p=307"},"modified":"2014-08-07T20:24:47","modified_gmt":"2014-08-08T00:24:47","slug":"making-the-best-of-things","status":"publish","type":"post","link":"https:\/\/advancedmathyoungstudents.com\/blog\/?p=307","title":{"rendered":"Making the Best of Things"},"content":{"rendered":"

A Very Useful Thing\u00a0About Slope Formulas<\/strong><\/p>\n

As you have worked through the previous posts and exercises, I hope you have noticed something special about places on a graph where the derivative is zero.\u00a0 These places have horizontal tangent lines.\u00a0 And in their own neighborhoods, they are often a local maximum or local minimum.\u00a0 This piece of information is very helpful when you are trying to solve an \u201coptimization\u201d problem.<\/p>\n

For example\u2026<\/p>\n

Suppose you are firing a projectile on a horizontal surface at a fixed launch speed, v.\u00a0 And let\u2019s also suppose that air friction is negligible.\u00a0 What angle will maximize the range (horizontal distance) of the projectile?<\/i><\/strong><\/p>\n

You may remember the answer from your first year of physics.\u00a0 It is intuitive and there was a way to get the answer without calculus.\u00a0 But we can still use this example to illustrate the calculus method.<\/p>\n

\"post20pic11\"<\/a><\/p>\n

(Graphic from http:\/\/outreach.phas.ubc.ca\/phys420\/p420_00\/darren\/web\/range\/range.html<\/a>\u00a0where you can also see a review of the projectile physics)<\/p>\n

STEP 1: Express the quantity to be optimized as a function of a single variable.<\/strong><\/p>\n

In this case, a review of that website or last year\u2019s notes will remind you that the distance varies with the launch angle by the rule:<\/p>\n

\"post20pic1\"<\/a><\/p>\n

In brief, we resolved the launch velocity into components, used the vertical component to determine time in air and then multiplied that time by the horizontal component of the velocity.\u00a0 But now that we have this expression, for our current purposes it would be helpful to make use of the double angle identity for the sine function.\u00a0 We can write:<\/p>\n

\"post20pic2\"<\/a><\/p>\n

Don\u2019t worry if you would not have remembered how to do all that without notes.\u00a0 The point is to\u00a0learn what we can do with this expression now that we have it.<\/p>\n

 <\/p>\n

STEP 2: Find the derivative<\/strong><\/p>\n

This is one we know how to do: \u00a0we know the derivative of sine functions and we know the special case of the chain rule.<\/p>\n

\"post20pic3\"<\/a><\/p>\n

STEP 3: What is the domain of interest to us?<\/strong><\/p>\n

When a math book asks for the domain of a function, they usually mean the largest, most inclusive domain possible.\u00a0 For a sine function, that would be all real numbers.\u00a0 But that is not what we are after here.\u00a0 We need to think about the portion of the domain that is relevant.\u00a0 To fire this cannonball as far as we can, clearly we will choose an angle between 0 and \u03c0\/2 radians.<\/p>\n

STEP 4: Where in that domain is the derivative equal to zero?<\/strong><\/p>\n

In this case\u2026<\/p>\n

\"post20pic4\"<\/a><\/p>\n

(OK, call it 45 degrees if you prefer.)<\/p>\n

\u00a0STEP 5: (Your math teacher may not\u00a0like this)\u00a0Convince ourselves that we are done.<\/strong><\/p>\n

In math class, once you have found where the derivative is zero (or undefined), you are not nearly done.\u00a0 You have to investigate each of these \u201ccritical points\u201d to see what kind of extreme it is and to justify your conclusion with such things as sign charts and 2nd<\/sup> derivative tests.\u00a0 But in physics class, you are going to get away a little easier because of the following claim:<\/p>\n

If a \u201csmooth\u201d function has only one critical point on some interval and the value of the function is higher at that\u00a0critical point\u00a0than at either of the endpoints, then that value is the \u201cabsolute maximum\u201d value.<\/i><\/p>\n

So in the example at hand, we know the cannonball\u2019s range is zero when the angle is zero, and its zero when the angle is 90 degrees (as the ball goes straight up and down).\u00a0 But it is NOT zero when the launch angle is 45 degrees.\u00a0 But the tangent line is horizontal at 45 degrees.\u00a0 So that\u2019s the \u201cbest\u201d angle.<\/p>\n

You still have to learn to do things the math-class way.\u00a0 There are lots of functions with multiple critical points.\u00a0 And there are functions that have horizontal tangent lines at places that turn out not to maxima or minima.\u00a0 So you do have to know how to investigate further.<\/p>\n

But the problems you will see in AP Physics don\u2019t<\/span> turn out that way.\u00a0 Most often, there will be only one critical point and it will be the one we were looking for.\u00a0 Here is another example I have always liked.<\/p>\n

\"post20pic12\"<\/a><\/p>\n

Suppose you have a large cylindrical tank full of water, open at the top.\u00a0 You are going to punch a hole in the tank somewhere along its side and you want the resulting jet of water to land as far from the tank as possible.\u00a0 Where should you make the hole?\u00a0 And where will that initial jet of water land?<\/i><\/strong><\/p>\n

Let\u2019s say you make the hole y units from the ground and the initial jet lands x units from the tank.<\/p>\n

STEP 1: Determine x as a function of y<\/strong><\/p>\n

First we need the \u201cspeed of efflux\u201d which you can think of as the horizontal launch velocity of the water coming out of the hole.\u00a0 There is a neat result from fluid dynamics that shows that the water leaves the tank at the same speed as if it had free-fallen from the top of the tank.\u00a0 It’s called Torricielli’s Law.<\/p>\n

\"post20pic5\"<\/a><\/p>\n

You can see more about this calculation here:<\/p>\n

http:\/\/en.wikipedia.org\/wiki\/Torricelli’s_law<\/a>\u00a0 (where I also got the graphic above)<\/p>\n

Now, we just find the time in air and multiply by that launch speed.<\/p>\n

For time in air, we can use the fact that<\/p>\n

\"post20pic6\"<\/a><\/p>\n

and solve for time to get:<\/p>\n

\"post20pic7\"<\/a><\/p>\n

So for the distance as a function of the height of the hole we get:<\/p>\n

\"post20pic8\"<\/a><\/p>\n

 <\/p>\n

STEP 2: Find the derivative<\/strong><\/p>\n

Well, it seems we need the more advanced version of the chain rule. But there is a trick to make things easier.\u00a0 The distance x will be at its maximum when the term inside the square root is at its maximum.\u00a0 So let\u2019s ignore the square root and just maximize the innards.<\/p>\n

\"post20pic9\"<\/a><\/p>\n

STEP 3: Domain of interest?\u00a0<\/strong> We can\u2019t make a hole below the ground or above the height of the cylinder so we only have to consider y values between 0 and h.<\/p>\n

STEP 4: Set the derivative equal to zero<\/strong><\/p>\n

\"post20pic10\"<\/a><\/p>\n

We need a hole in the middle!<\/p>\n

\u00a0STEP 5:\u00a0 Convince ourselves that we are done<\/strong><\/p>\n

There was only one critical point.\u00a0 And the endpoints both lead to a distance of zero.\u00a0 So the hole at y=h\/2 is the one that gives us the farthest landing point.<\/p>\n

And what is that landing point? Substitute y=h\/2 into our original function, which was:<\/p>\n

\"post20pic8\"<\/a><\/p>\n

I will leave the substitution and clean-up to you.\u00a0 I will say that it cleans up quite nicely and has a memorable result.\u00a0 The first of my students to post the correct\u00a0result in a comment wins some silly thing or other.<\/p>\n

Quick Recap and Closing Remarks<\/strong><\/p>\n

In AP Physics, when you want to find the maximum or minimum possible value of some quantity, most of the time it will just be a matter of taking a derivative, then setting it equal to zero and then waving your hands around to convince us that you are done.<\/p>\n

The hardest part of the process is not going to be the calculus.\u00a0 It’s going to be setting up the variables and finding the expression for the quantity you hope to optimize.\u00a0 That’s why you had all those homework problems back in pre-calc that said “Express this thing in terms of that thing”.<\/p>\n

I guess I should mention that if you are having trouble either finding the derivative or solving after you set it equal to zero, you may find your TI82 to be of some use.\u00a0 But honestly, sometimes it feels like the calculator only gives intelligible answers to people who already know what the solution looks like.<\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

A Very Useful Thing\u00a0About Slope Formulas As you have worked through the previous posts and exercises, I hope you have noticed something special about places on a graph where the derivative is zero.\u00a0 These places have horizontal tangent lines.\u00a0 And … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false},"categories":[1],"tags":[],"jetpack_featured_media_url":"","jetpack_shortlink":"https:\/\/wp.me\/p4uvY7-4X","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/307"}],"collection":[{"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=307"}],"version-history":[{"count":9,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/307\/revisions"}],"predecessor-version":[{"id":328,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/307\/revisions\/328"}],"wp:attachment":[{"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=307"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=307"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/advancedmathyoungstudents.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=307"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}