<p><a href=“http://www.viewdocsonline.com/document/zhpjqx”>http://www.viewdocsonline.com/document/zhpjqx</a></p>
<p>Fixed mistakes. I still think the answer to 4(d) is 160 though. What is the strategy other than sqrt(vA^2 + vB^2)?</p>
<p>SuperN0va, you must use distances not velocities. d^2 = A^2 + B^2</p>
<p>Differentiating and cancelling the twos we get:</p>
<p>dd’ = AA’ + BB’</p>
<p>Where a prime = velocity and nonprime = distance.</p>
<p>That is how you solve for d’, you cannot use Phythagorean theorem with velocities. Although you get a very close answer, it is not entirely correct.</p>
<p>For a nice explanation w/ example:</p>
<p><a href=“Related Rates: Two Cars at a Crossroads - dummies”>dummies - Learning Made Easy;
<p>Ok…so this is how my Calc teacher had us do rate problems… and this is how I solved the train one without dealing with square roots and stuff.</p>
<p>So three sides of the triangle are 300, 400, and 500 (hypotenuse). The rate of change for the 300m side is 100m/min, the rate of change for the 400m side is 125m/min (plugged t=2 in velocity equation) and the unknown rate of change is for the 500m hypotenuse.</p>
<p>Pythagorean theorem: a^2 + b^2 = c^2
Derive: 2a<em>a’ + 2b</em>b’ = 2c *c1
Plug in: 2(300) * 100 + 2(400) * 125 = 2(500) * c’
60,000 + 100,000 = 1000c’
c’ = 160</p>
<p>So I’m super mad at myself because I made an arithmetic error in the plugging in step b/c I was rushing and got a final answer of 156…but hopefully I won’t be deducted many points for that!</p>
<p>Hmm…So we get 500d’ = 300(100) + 400(125) and d’ = 160?</p>
<p>Yeah, you’re right. That’s it.</p>
<p>megans113, that is the proper way of doing intersection questions. That’s how every textbook I’ve read explains it as well.</p>
<p>If you isolate d and not d^2, you must use the chain rule which is a pain. And megans113, you’ll only loose 1pt for an incorrect answer. If your work is correct, you’ll get most points.</p>
<p>I fixed the pdf. Is the work right now? <a href=“http://www.viewdocsonline.com/document/cq889f”>http://www.viewdocsonline.com/document/cq889f</a></p>
<p>Perfect.</p>
<p>Good. So are there any other answers we dispute? </p>
<p>@kingofxbox99
On 6©,
I got up to this party…
ln(3-y) = ln(2) - sinx</p>
<p>But I solved the rest a bit different,
e^(ln(3-y))=e^(ln(2))-e^(sinx)
3-y=2-e^(sinx)
-y=-e^(sinx)-1
Y=e^(sinx) +1</p>
<p>Is this the same?</p>
<p>If you plug in x = 0 for your answer, you get e^0 + 1 = 2, which does not equal the initial point, so no it’s incorrect. :(</p>
<p>You’ll probably get the majority of the points since you did the majority of the work correct.</p>
<p>SuperN0va, from all the AB questions I had they look good! 
btw, I sent you a PM. :)</p>
<p>Yeah… I’ll probably lose just 1 point for not getting that final answer… No big deal</p>
<p>When are scoring guidelines released? </p>
<p>July, approximately same time as when scores are released.</p>
<p>the links say the file has been deleted for AB answers. can someone upload them again please? </p>
<p>I never posted a link for AB answers – unless you are talking about a different one. </p>
<p>I apologize for my ignorance, (and fantastic job on the answers by the way SuperN0va), but on question #3, part b, I believe the question asked for open intervals, and also, the second derivative is not defined at that point, so I don’t think -5 should be included. Sorry if I am wrong, and please let me know if I am!</p>
<p>Also, do you think I will lose points on question #5 if for part C, I was a doofus, and first of all, didn’t think to solve the bottom length for sqrt(5) and instead included 2 integrals (one with y=-2x) - in addition, I didn’t actually take the derivative, I just included d/dx(xe^x^2).</p>
<p>You will not lose points for not solving for sqrt(5). You probably will lose a point for not taking the derivative of
xex^2. I wouldn’t worry – you will get some credit for knowing HOW to do it, even if you didn’t actually do it right.</p>