Official AP Calculus Thread for 2009

<p>Does anyone know how to do this problem?</p>

<p>“A particle is moving along the x-axis so that at any time t>0, its velocity is given by v(t)=4-6t^2. If the particle is at position x=7 at time t=1, what is the position of the particle at time t=2?”</p>

<p>^ integrate v(t) and plug in x=7 when t=1 to solve for C…then solve for t=2</p>

<p>Our teacher gave us a BC Calculus Mock AP Exam as our final with the whole 44 MC’s and 6 FRQ’s. I got a 96/108 (High 5 lol) :smiley: so I was really happy, especially since the average and median scores were around a 45ish. Hopefully I’ll get the same result tomorrow.</p>

<p>good luck everyone! i’m taking AB tomorrow…and i’m going to sleep now :)</p>

<p>Thanks for the help.</p>

<p>is there temperature on past FRQs??</p>

<p>There was a FRQ concerning changes of temperature, but I haven’t seen one about Newton’s Law of Cooling or anything.</p>

<p>There have been some where it gives you a table of t values and their corresponding H(t) (temperature) outputs, then they asked you approximate the derivate at a certain time using just the secant between to points near that time. Then they asked as question on the rate at which the function of heat was increasing such as whether the function was concave up or down based on the progression on the slopes as you increased the time. That’s the only temperature one that I remember. Though I’m sure there could be one on Newton’s law of cooling and heating (Differential Equations!!)</p>

<p>Ugh, I sooo regret signing up for this test.</p>

<p>Can anyone explain speed to me?
I don’t understand how to make conclusions about speed based on velocity graphs.</p>

<p>speed = |velocity|</p>

<p>For the function f, f’(x) = 2x + 1 and f(1) = 4. What is the approximation for f(1.2) found by using the line tangent to the graph of f at x = 1?</p>

<p>Seems like a really simple question, but I can’t for the life of me figure it out.</p>

<p>(Answer is 4.6)</p>

<p>find the derivative at 1</p>

<p>f’(1) = 2(1) + 1 = 3</p>

<p>since f(1) = 4…tangent line is y - 4 = 3 (x-1)
plug in 1.2 for x and you have your answer</p>

<p>i really need to go to sleep…lol</p>

<p>^
omg, so that’s what you’re suppose to do. I think I’m brain dead. Thank you!</p>

<p>well you can find the slope of the tangent to the curve at x=1 so f’(1) = 2(1) + 1 = 3. The line tangent to the curve at x = 1 is y = 3(x-1) + f(1) so y(tangent) = 3x + 1. Then f(1.2) can be approximated using this tangent line so y(approximation at 1.2) = 3(1.2) +1 which is 3.6 + 1 = 4.6</p>

<p>

wow, why has no one told me this before? thanks so much</p>

<p>^Have you ever bothered taking a physics class? lol =)</p>

<p>for BC calc when it asks you for the speed based on the vectors, the equation is sqrt ( xsquared + ysquared) - sorry for crappy notation :stuck_out_tongue:
that came up quite a bit in our FR practice so it might be a handy one to know…</p>

<p>Okay, i need help you guysss.</p>

<p>Taking the AB Calculus test tomorrow.
I know how to do all the math, but…</p>

<p>Sometimes I don’t know what their asking!
or the multiple choice will trick me!</p>

<p>:[ </p>

<p>Are there any set tips for FRQs?</p>

<p>I need to get a 4 or 5… im afraid ill get a 3…</p>

<p>

haha no, physics next year. perhaps that explains it.</p>