AP Calc AB help!!

Test Preparation AP Test Preparation
1

<p>Can someone help me with these:</p>

<p>-The average value of cos x on teh interval [-3,5] is....
They said (sin 3) + (sin 5) all over 8...wouldn't it be (sin 5) - (sin -3) all<br>
over 8?</p>

<p>-lim as x approaches 1 of (x/ln x) is...
They said nonexistant, but isn't it 1 b/c of hopital's rule...speaking of which, can you ALWAYS use l'hopital's rule when there's a fraction or can you only do it when the lim isn't a hole (0/0)</p>

<p>-at which point on the graph of y=.5x^2 is the tangent line parallel to the<br>
line 2x-4y=3?
they said (1/2, 1/8)</p>

<p>Also...can someone please explain cross-sections? :) lol</p>

2

<p>for the first question, no, it would not. in fact, doing the exact same problem, i got that answer and my teacher had to explain. i'll find the paper to help you in a minute. it's in the massive pile on my desk.</p>

<p>l'hopital's rule is not on the ab exam, but i don't understand how to answer that problem...</p>

<p>also did this problem... in fact, did it a couple hours ago.
solve 2x-4y=3 for y. (you get y=.5x+.75) now that you have that, you know the slope has to be .5 for the equation y=.5x^2. so you differentiate and get y'=x. Plug .5 into x to get that y'=.5. Now plug .5 into the original equation for y (y=.5x^2). That should give you 1/8.</p>

<p>Sorry, can't help with cross sections.</p>

3

<p>I'm pretty sure -(sin -3) = sin(3)</p>

<p>ln 1 = 0</p>

<p>dy/dx = x</p>

<p>y = 1/2x - 3/4</p>

<p>slope of equation is 1/2 make the derivative 1/2 then plug 1/2 into
y=.5x^2. (1/2 , 1/8)</p>

4

<ol>
<li><p>they're the same thing</p></li>
<li><p>no, if you plugin 1 into (x/lnx) you get 1/0, which is nonexistant. and also you can't use lhopital's rule for that...you can only use that rule when its 0/0 or inf/inf.</p></li>
<li><p>i solved for y in the equation 2x-4y=3 and i found the slope to be .5.
and since you want to find the point on the graph of .5x^2 that has the same slope, you solve for thats deriv and just get y=x. so x=.5 because you can set the two derivatives equal to each other. then you just go back to equation y=.5x^2 and plugin .5 for x, and get 1/8. so (1/2, 1/8)</p></li>
<li><p>cross sections are usually vertical cut pieces of a volume. so if you had like y=ln(x), and they said they had cross sections of squares, and they wanted you to find the volume from 1 to 2, you would take the integral from 1 to 2 of the cross section's area (which is ln(x)^2) and that would give u the volume of the whole thing. remember, the derivative = surface area, volume = antiderivative</p></li>
</ol>

5

<p>punch number 1 into a calculator same answer either way.</p>

6

<p>didn't realize that said sin(5)-sin(-3), thought that said sin(3).</p>

<p>and is l'hopitals rule really that simple?</p>

7

<p>That first answer should be (sin 5) - (sin 3) all over 8, which is the same as (sin 5) - (sin -3) all over 8.</p>

<p>And you're right again--the answer to the second one should be 1, not nonexistent. You can use that rule when the lim goes to 0/0 (which it did when you plugged 1 in). </p>

<p>For the third one:
2x-4y=3, when rearranged, becomes y = 0.5x-0.75, the slope of which is 0.5 (y = mx + b, where m is the slope).</p>

<p>Since parallel lines means equal slopes, you just have to plug 0.5 into the original y=0.5x^2 to get 1/8, so I actually ended up with (1/2, 1/8).</p>

<p>Cross sections aren't too bad.
If squares: (Integral of region)^2
If rectangles: Integral of (length * width), where for example length is distance between curves and width might be given as 5 times the length, in which case you do Integral of (distance between curves)*5(distance between curves)
Those are really the main ones you have to worry about.</p>

<p>Also, I know I sometimes have trouble with knowing when to include pi in the area/volume problems, so here's a quick rundown:
Areas: Never use pi
Volumes: Only use pi when solids are revolved about axes</p>

<p>I hope this helps! Good luck! Your questions are helping me study too, so thanks! :)</p>

8

<p>yes, lhopital's rule is really that simple</p>

9

<p>a) its sin5 + sin3, (sin 5) - (sin -3) is equal to sin5 + sin3 since sin is an odd function.</p>

<p>b) it is nonexistent. the equation doesn't approach 0/0, when you plug one into the numerator you get 1, which isn't 0</p>

<p>and thanks for the cross section info.</p>

10

<p>

When you plug the value (1 in this case) into the overall function and it comes out to 0/0, you have to either factor more or use l'Hopital's rule. Once you do, you get 1/(1/x), which just comes out to x, which is 1.</p>

11

<p>the overall function is X / ln X</p>

<p>if you plug 1 into that, you get 1 / 0</p>

<p>that equals infinity</p>

<p>which means the limit is non-existant b/c it's not just a hole, it's a whole asymptote. </p>

<p>the answer IS nonexistent.</p>

12

<p>When you plug 1 into X/ln X , that is 1/0 , you can only use L'Hospital's rule when top and bottom are both 0 or infinity</p>

13

<p>That's not factoring... that's taking a derivative, which is not asked for in this question. Limits and derivatives are not the same thing. There is a limit definition of a derivative, but that question does not fall into that point.</p>

<p>Looking at the graph of the equation, I can tell you the limit is not 1. I plugged the equation into the calculator, and found the value at 1.003, which is close to one, but not 1. I got 251.50167. When I plug in .996, I get -248.50167. There is no limit as x-> 1.</p>

14

<p>OH crud! Sorry guys, in my haste to answer the question I didn't realize that the answer doesn't initially come out to be 0/0, it comes out to be 1/0 (in which case you can't use l'Hopital's rule) which doesn't exist. You're right.</p>

<p>Sorry for the confusion everybody...I think I need some sleep before my brain explodes XD</p>