<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d14085554\x26blogName\x3dAP+Calculus+AB\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://apcalc.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://apcalc.blogspot.com/\x26vt\x3d-7891817501038621673', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Monday, December 19, 2005

Yule Time

Need not to worry, I got this day covered. Prince told me last Sunday that Im the scribe. Ok? good! To start off today's class we had a group work about optimization. It is about Topher and Sully. The question goes like this:

Topher and Sully decided to go cut down a Christmas tree together, so they ventured out into the snowy woods. After wandering for close to three hours, the two found the perfect tree but also decided that they were lost. Luckily Sully found a map nailed to a tree which showed that they were four miles due north of a point on a perfectly straight road. Six miles east of that point was the park that the two called home. If Topher and Sully can walk and drag the tree at the rate of two miles per hour through the snow, and three miles per hour on the road, what is the minimum amount of time that they would need to get home?


T(x) = ((x^2+16)^1/2 ) / 2 +(6-x)/3
T(x) = 1/2 (x^2 +16)^1/2 +2 - x/3
T ' (x) = (1/2)(1/2)(x^2+16)^-1/2 (2x) - 1/3
T ' (x) = (3x - 2(x^2 +16)^1/2)/ (6(x^2+16)^1/2

After finding the derivative we find where there are zeros or where the derivative is undefined. The denominator will never equal zero so we just focus on the numerator.

3x - 2(x^2+16)^1/2 = 0
3x= 2(x^2+16)^1/2
(3x)^2= (2(x^2+16)^1/2)^2
9x^2 = 4(x^2 +16)
4x^2 + 64 -9x^2 = 0
(5^.5x-8)(5^.5x+8) = 0

therefore x = 8 / (5^1/2)
then we do the 1st derivative test
- +
T'<---------------8 / (5^1/2)------------->

by the first derivative test 8 / (5^1/2) is a min

then we plug in the root to the parent function to get 3.894
therefore they need three hours to minimize the time they take to walk

After the group work Mr. K gave us derivative functions which we need to find its parent function.


x^3.................x^4/4 + C
-cos(x)............-sinx + C
sinx..................-cosx + C
sec^2x............. tanx + C
1/x....................lnx + C
e^x....................e^x + C
2^x...................2^x/ln2 + C
x^1/2................2/3x^3/2 + C

Well thats the class...

This December,
That love weighs more than gold

the next scribe is steve

Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch


Post a Comment

Links to this post:

Create a Link

<< Home