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