<p>It isn’t. It’s from 1 to infinity. But you take it from 0 to infinity so it can be considered a geometric series and you can use the a/1-r formula to find convergence. </p>
<p>Here’s another question: </p>
<p>The third degree Taylor polynomial about x=0 of ln(1-x) is ___ ?
A) -x - x^2/2 - x^3/3
B) -x + x^2/2 - x^3/3</p>
<p>A. f’(0) is -1. The coefficient of the x^2 term is -1, so it must be A.</p>
<p>^
But isn’t it f’‘(0)?
f(0) = 0
f’(0) = -1
f’‘(0) = 1
f’‘’(0) = -2</p>
<p>The second derivative evaluated at 0 equals 1.</p>
<p>So it would be 0 -x + x^2/2 - x^3/3 right?</p>
<p>Yes, I think it’s B.</p>
<p>Also, hotinpursuit, what was the answer to your question at the bottom of page 4 here:</p>
<p><a href=“http://talk.collegeconfidential.com/ap-tests-preparation/871862-2010-ap-physics-c-study-thread-4.html[/url]”>http://talk.collegeconfidential.com/ap-tests-preparation/871862-2010-ap-physics-c-study-thread-4.html</a></p>
<p>^
Okay that answer was 7.93 m/s</p>
<p>Also, you guys agree that the answer is B? Because I am getting that, but the CB website’s sample question has an answer of A.
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
go to page 32, answer is on page 38</p>
<p>anyone care to explain why?
I have a bad feeling I’m doing something really dumb though.</p>
<p>
I agree.
f’(x)=-(1-x)^-1
f"(x)=-(1-x)^-2</p>
<p>Don’t forget chain rule.</p>
<p>Oh wow, I tried to do it in my head but made a stupid mistake.</p>
<p>f(x) = ln (1-x), so f(0) = 0
f’(x) = -1/(1-x), so f’(0) = -1
f’‘(x) = -1/(1-x)^2, so f’'(0) = -1</p>
<p>When you take the derivitave, you multiply by negative one using the power rule, but then you multiplly by negative one again because of the chain rule.</p>
<p>f’(x)=-(1-x)^(-1)
f’(0)=-1</p>
<p>f’‘(x)=-(1-x)^(-2) [Power rule cancels out the negatives, but chain rule makes it negative again]
f’'(0)=-1</p>
<p>Both negatives, so it’s A.</p>
<p>wowwwwww
I cant take derivatives or use the chain rule
wooooooowwwww</p>
<p>my bad guys</p>
<p>Hey, I have a question about speed</p>
<p>How does speed relate in terms of velocity and acceleration?</p>
<p>I know that speed is the magnitude of velocity, but what does that mean? When do you know if speed is positive even though velocity is negative?</p>
<p>My Barron’s AP Calc review book said that “if a and v are both positive or both negative, then the speed of P is increasing or that P is accelerating; if a and v have opposite signs, then the speed of P is decreasing or P is deceleration.”</p>
<p>That didn’t make much sense when I read it. Would be so kind to explain that to me?</p>
<p>Thanks in advance.</p>
<p>Speed is always positive. Explaining the Barron’s quote - just think about it.</p>
<p>If something is moving in the positive direction AND accelerating in the positive direction, it’s going to speed up, right (think about a car with your foot on the gas)? However, say it’s moving in the positive direction but it begins to have a negative acceleration (using the previous example, think that you just hit the brakes). Speed will decrease.</p>
<p>If something is moving to the left, and you accelerate it to the left, it’s going to start moving even faster. If it’s already moving to the left and you accelerate it to the right, it’s going to begin to slow down until velocity equals zero.</p>
<p>Hopefully you get my explanation. If not, just remember that speed is increasing when a and v have the same sign, and speed is decreasing when they have opposite signs. The AP test will probably ask something like “When is speed increasing?”</p>
<p>Tip: When integrating v(t) to find distance, take the absolute value of the function because distance is the total distance traveled which is always increasing.</p>
<p>My question is does the same thing apply for velocity/speed?
So is this true?
When integrating a(t) to find speed, take the absolute value of the function because speed is always positive.</p>
<p>yeah ChemE14, that makes perfect sense.</p>
<p>thanks a lot.</p>
<p>Well, speed is just the absolute value of velocity, so just find velocity like you normally would. Say you get v = -5.0 m/s, then speed = 5.0 m/s. Of course, that’s just for one dimension. If your velocity is a vector (2-dimensions), then</p>
<p>v(t) = ( dx/dt , dy/dt )
speed = ||v(t)|| = sqrt ( (dx/dt)^2 + (dy/dt)^2 )</p>
<p>Is displacement always positive or can it be negative? I understand that speed is the absolute value of velocity, but can someone explain the relationship with displacement (like what is the absolute value or whatever)?</p>
<p>Displacement is just the definite integral, so it can most definitely be negative.</p>
<p>Distance, however, involves absolute values.</p>
<p>Does the BC Calc test ever have questions relating integrals and water pressure? I never quite got the hand of those problems.</p>
<p>Can someone clarify cross sectional volume for me real quick? Is is just the integral of the base*height (where base = region bound by given functions and height is in terms of x or base)?</p>
<p>No. Cross sectional volume is the integral of the cross sectional area.</p>
<p>So. umm. I’m a lazy bum and don’t feel like memorizing all the series tests, especially since I need to cram for other exams.</p>
<p>I’m self-studying this exam (kinda), and I just took a practice test and got all the questions right except for the series ones (which came out to be 49/55). </p>
<p>Do you guys think I can still get a 5 with rudimentary knowledge of series? I obviously konw basic stuff like p-series, harmonic series, etc but not much more.</p>
<p>How many MCQs/FRQs are generally on series?</p>