<p>@asc3nd:</p>
<p>Is this the correct way of writing your problem? <a href="http://i30.tinypic.com/13z0gi0.jpg%5B/url%5D">http://i30.tinypic.com/13z0gi0.jpg</a></p>
<p>EDIT: If so, the answer is 20. See Solutions</a> to the Limit Definition of a Definite Integral for an idea as to how you can solve it. :)</p>
<p>I broke my link in the above post by putting it in parentheses.</p>
<p>Let's try this:</p>
<p>hey guys. i was wondering is Newton's method on the calculus ab test because i saw a newtwon's method question on the 1993 test</p>
<p>i doubt it.</p>
<p>Wow.. </p>
<p>the 1998 test was a friggin breeze compared to the 3 barron's tests i took over the weekend.</p>
<p>Barron's tests are at least twice as difficult as the real thing IMO. :D</p>
<p>Two more days..... Good luck guys =]</p>
<p>I have a question about a FRQ. I got parts a, c, and d. The general question is</p>
<p>The velocity of a particle moving along the x axis is given by v(t) = [e^(t-1)]/(2t^2+1) - t^2 for t on [0,12]. At t = 2, x(t) = 3.</p>
<p>Part b asks: What is the position of the particle when it is farthest to the left. </p>
<p>Obviously it's farthest to the left when it's traveled the longest distance at a negative velocity. Graph v(t) (this is a calculator allowed problem) and you find that for [0,12], v(t) has zeros at .62something and 11.something (too lazy to look it up exactly lol). It's below the x axis from .62 to 11, so greatest distance from the left is at t = 11.whatever. With the equation
New position = initial position + distance traveled, I get</p>
<p>x(t) = x(0) + fnint(v(t),t,0,11....)
All I need is x(0). But I don't have it. I figure you would use the initial condition (2,3) and integrate v(t) to find x(t), then plug in 0, but v(t) is not integrable (at least by AB standards). Some help would be appreciated. If you need to check, x(0) = 2 (yay for back of book).</p>
<p>anyone else nervous for the test besides me? can someone list the most important equations/formulas we need to know for the test? thanks</p>
<p>oh and by the way...is it to our advantage to skip problems we don't know, or should we just do everything?</p>
<p>For ^,
for the new general equation for x(t), I used X(t)=x(2)+fnint(v(t),t,2,t). That way, when t=2, x(2)=3. When you plug 11~ in, you get -330~, which would be the answer I suppose. The only thing that doesn't work is when you plug 0 in for t. You get 5.</p>
<p>@Ohsnap422:</p>
<p>Your calculus book should have a page somewhere (usually front/back cover) with most of the formulas you need (differentiation/integration tables in addition to some trig stuff in mine). Aside from those, I'd suggest skimming through your book and looking at MVT, FTC1/2, and all of that fun volume of revolution jazz. I'm probably forgetting a bunch, but that's the gist of it.. I think? :)</p>
<p>Alright. But my book, the Pearson Prentice Hall edition, doesn't have a front/back cover with formulas...lol. So basically MVT and FTC? Any others? =)</p>
<p>I found what looks to be a pretty good study guide for AB; I suggest looking it over - <a href="http://www.eecs.berkeley.edu/%7Ecelaine/apcalc/apcalc.pdf%5B/url%5D">http://www.eecs.berkeley.edu/~celaine/apcalc/apcalc.pdf</a></p>
<p>Lozz,</p>
<p>I get about 336.717. But the answer is supposed to be like -336.69 units to the left of the starting point. Your solution seems OK except for the sign, but I also want to understand the book's solution. They do (exactly)</p>
<p>x(t) = x(0) + fnint(v(t),t,0,11.448)
~ 2 -335.690 = -333.69</p>
<p>The particle is 336.69 units to the left of the starting point.</p>
<p>I still don't get where they get x(0) to be 2 and how they get 336.69 from -333.69.</p>
<p>thanks!! those topics help a lot!</p>
<p>On this website: <a href="http://www.eecs.berkeley.edu/%7Ecelaine/apcalc/apcalc.pdf%5B/url%5D">http://www.eecs.berkeley.edu/~celaine/apcalc/apcalc.pdf</a>
If u scroll all the way to the bottom where it says Calculator programs, one of them is a trapezoid formula. However, when I put it in it doesnt work (error). Is there a specific button for Y naught (initial Y).</p>
<p>If f'(x) = sin ((pi* e^x)/2) and f(0) = 1, then f(2) =
how would you solve this??</p>
<p>also guys what does it mean when it asks when is the particle closest to the left...what would you have to do</p>
<p>I think you have to find the zero's of the V(t) equation, then solve for the min.</p>
<p>how do you know when to apply the FTC or the MVT?</p>
<p>people help me with this: If f'(x) = sin ((pi* e^x)/2) and f(0) = 1, then f(2) =
how would you solve this??</p>
