<p>the "acceleration" of a particle moving along a straight line is given by
a = 10(e^(2t))</p>
<p>during the time that the velocity increases from 5 to 15 how far does the particle travel?</p>
<p>so, v = 5(e^(2t))</p>
<p>when v =5 , t=0
when v=15, t= ln(sq root of 3)</p>
<p>did i solve this problem right so far and what do i do next?</p>
<p>youre right so far but for v=15, t=(ln 3)/2, not what u stated. now, u find the integral of the velocity equation, which will give u the displacement equation x=2.5(e^2t). now just plug in the final time t=(ln 3)/2 to the equation minus the initial time t=0 like this:
xf-xi=2.5(e^(2((ln 3)/2))) - 2.5(e^(2X0) which equals:
3-2.5=.5
i hope this is right.</p>
<p>your right about my mistake but i think i found another mistake in your calculation</p>
<p>xf-xi=2.5(e^(2((ln 3)/2))) - 2.5(e^(2X0) which equals:</p>
<p>2.5(3) - 2.5 = 5</p>
<p>please let me know if this is correct</p>
<p>yeah i just totally forgot that 2.5 at the front for some reason. i guess we caught eachother's mistakes! the answer u have is right.</p>
<p>On the apcentral website, you are asked to find the area enclosed by the leminscate or r^2 = 18cos(2theta), I found the points -pi/4 and pi/4 as the endpoints. I used the symmetry 2 * from 0 to pi/4. (points of integration) [to integrate]
2*int(pi/4,0).5[18cos(2theta)] =18int(pi/4 , 0) [cos2theta] = 18 sin[2theta]/2 = 18 sin(pi/2) / 2 = 9.
The answer is 18, but I get 9. Hope you can read that, I diddnt know what to use for symbols.</p>