<p>for 5b this is what i did:
ln ly-1l = -1/x + k
e^(lnly-1l) = e^(-1/x) times e^(k)
let C = e^(k)
ly-1l = C(e^(-1/x))
then l0-1l=C(e^(-1/2)
so C = e^(1/2)
so y = e^(-1/x) times e^(1/2) + 1= e^(.5 - 1/x) +1</p>
<p>then for 5c: you'd get lim x-> inf. of the above
so e^(-1/infinity) equals 0 so the limit would be e^(.5 - 0) equal e^(.5)
and therefore the final answer is e^(.5) +1 or = sqrt(e) + 1</p>
<p>Since the initial condition is indeed (2, 0), then the absolute value sign cannot simply be removed, and therefore, when the absolute value sign is removed, the sign of the right-hand side of the equation will have to become negative.</p>
<p>lost a point there. they'll probably give the points for everything else though, like they did on the past exams. hope so! i wonder if they'll give credit for doing part c with the wrong f(x)</p>
<p>what did u guys get for numbers 1 and 2?</p>
<p>how do you do # 6</p>
<p>wendyling, I would imagine they would give the point for a part (c) limit, so long as the function you came up with in part (b) is exponential.</p>
<p>Sixthsense, I got the following:</p>
<h1>1</h1>
<p>Let f(x) = sin(pi<em>x) - (x^3 - 4x)
a) Area = integral(0,2) f(x) dx = 4
b) Area = integral(.539,1.675) (-2 - (x^3 - 4x)) dx
c) Volume = integral (0, 2) [f(x)]^2 dx = 9.978
d) Volume = integral (0, 2) [f(x)]</em>(x-3) dx = 8.369 or 8.370</p>
<h1>2</h1>
<p>a) L'(5.5) = [L(7) - L(4)]/(7-4) = 8 people/hour
b) (156 + 120)/2 + 2*(176 + 156)/2 + (176 + 126)/2 = 621 = integral (0, 4) L(t) dt, therefore 1/4 L(t) dt = 155.25
c) Three, one where L changes from increasing to decreasing on [1, 4], one where L changes from decreasing to increasing on [3, 7], and one where L changes from increasing to decreasing on [4, 8].
d) integral (0, 3) r(t) dt = 973</p>
<p>for #6 i got:
a) y-(2/e^2)=(-1/e^4)(x-e^2)
b) x=e, relative maximum, because f'x changes from positive to negative
c) x=e^(3/2)
d) got wrong lol dont want to say what i wrote</p>
<p>For number 2 this I what I think I got</p>
<p>a) 8 people/hr
b) 151.75
c) 3
d) 973</p>
<p>I agree with wendyling's answers for parts (a)-(c), and part (d) has been previously answered.</p>
<p>wow i didn't see average and forgot to divide by 4.. i feel like an idiot :(</p>
<p>wow for number two part b i assumed the intervals were all of one, so i did the b-a/2n instead of doing it the way u did MathProf. Do you think I will still receive any credit?</p>
<p>btw what did u guys get for number 4?</p>
<p>My guess is that if you did the average value, you'd probably get one, maybe two points. I suspect that there are three points on this question: one for doing the trapezoidal rule, one for the average value, and one for the solution. Assuming that point distribution, it's the first point that I'm not sure about whether you can earn it or not without the different subinterval widths (I would be inclined to say yes.)</p>
<p>For 2(d), was the answer really as simple as
</a>, or 973 tickets? That seemed way too easy for a part d question but I couldn't think of any more complicated way to do it.</p>
<p>Also, I wish we could embed images.</p>
<p>Pretty sure you would not get the points if you did the b-a/n method.</p>
<p>4a. t=3, position = -10
4b. 3 times
4c. Speed is decreasing. V<0, but increasing. speed = |V|, so speed decreases.
4d. acceleration is negative when velocity is decreasing, so 0<x<1 or 1.5, idr, and 4<x<6</p>
<p>ya thats all i wrote for 2d as well. i was surprised as well when i was doing it during the actual test.</p>
<p>seanbow, it seems that way to me.</p>
<p>Sixthsense, I have this for #4:</p>
<p>(a)
x(0) = -2
x(3) = -2 + integral (0,3) v(t) dt = -10
x(5) = -10 + integral (3,5) v(t) dt = -7
x(6) = -7 + integral (5,6) v(t) dt = -9
The particle is furthest left at time 3. Its position is x = -10.</p>
<p>(b) 3 times, by the Intermediate Value Theorem.</p>
<p>(c) v is negative and increasing. Since v is increasing, the acceleration is positive. Since v<0 and a>0, the speed is decreasing.</p>
<p>(d) Acceleration is negative when v is decreasing, which occurs on [0,1] and [4,6].</p>
<p>^ Yep my answers as well.</p>
<p>what was the integral for the question with the oil thing.. i think i put - 2000 instead of + 2000.. will i get any credit?</p>
