<p>88.4 degrees is about 1.543 in rad. so that’s kinda close.</p>
<p>heres my contribution since ive nothing better to do right now…someone verify…</p>
<h1>1)</h1>
<p>a) a = (.2 - .3)/(8 - 7) = -1/10 miles/min^2</p>
<p>b) The total distance Caren traveled from t=0 to t=12: 1.8 miles</p>
<p>c) t=2, since v(2) = 0 and changing from + to -.</p>
<p>d) Larry: fnInt(p/15sin(pt/12),t,0,12) = 1.6 miles
Caren : 1.4 miles -> closer to school</p>
<p>wouldnt the acceleration be v(t)/t which would be .3/7?</p>
<p>iambored: No, you are looking for the instantaneous acceleration, not average over the interval.</p>
<p>no im saying .3 over 7 is the instantaneous… because her velocity was .3 and the time was 7?</p>
<p>It’s delta V / delta t. Delta t is 1 because that is where they are looking for.</p>
<p>EDIT: delta V in this case is -0.1. You are only looking at the interval between [7,8].</p>
<p>alright got that wronggg</p>
<p>Yep, [f(8) - f(7)] / [8 - 7].</p>
<p>Basic slope problem, lol.
And make sure you guys had your units right for full credit. It was MILES PER MINUTE SQUARED.</p>
<p>Here’s my answers:
1.
a) -.1 m/s^2
b) 1.35
c) t=2
d) Caren
2.
a) 980
b) t=1.363
c)
d)21.052
3.
a)
b)
c)
d)
4.
a) 4.25
b)
c)
5.
a) -3
b) 8
c)
d)
6.
a)
b)
c)
d) one point</p>
<p>This is what you all did for number 3d right:</p>
<p>tan (theta) = -dy/dx and then go on from there b/c dy/dx = dy/dt / dx/dt</p>
<p>join the calc bc chat…[AIM</a> Chat](<a href=“http://chat-beta.aim.com/chats/special-interests/ccapcalcbc-]AIM”>http://chat-beta.aim.com/chats/special-interests/ccapcalcbc-)</p>
<p>You got 1 POI? I got none. After finding the taylor expansion for just e^(x-1)^2, you get 1+(x-1)^2+(x-1)^4/2! + (x-1)^6/3!+…+(x-1)^(2n-2)/(n-1)!+… from n=1 to infinity. Then the first term, 1, was canceled out by the -1 in f(x)=(e^(x-1)^2 - 1)/(x-1)^2 so you have (x-1)^2 +(x-1)^4/2! + (x-1)^6/3! + (x-1)^8/4! on top. Then you can factor out and (x-1)^2 in the numerator to cancel out the (x-1)^2 factor in the denominator so f(x) = 1+(x-1)^2/2! + (x-1)^4/3! + (x-1)^6/4!+…(x-1)^(2n-2)/(n!) from n = 1 to infinity. Then when you take the second derivative, you eventually get f’'(x)=1+12(x-1)^2/3! + 30(x-1)^4/3! + 56(x-1)^6/4! +… and since (x-1) is always raised to an even integer, that value will always be positive and also, since the first term of the series is 1, the second derivative will always be equal to or greater than 1 so there are no POI’s since no value of x can ever get a value for the second derivative of 0.</p>
<p>I’m pretty sure I got everything right (straight 9s, wooo! provided i showed the setups well enough)</p>
<p>1a) -.1 miles/min^2
1b) 1.8 miles, total distance travelled on time interval in miles
1c) t=2, velocity changes sign
1d) Caren 1.4 miles < Larry 1.6 miles</p>
<p>2a) 980 people
2b) t=1.363, second derivative test
2c) 387.5 hours (first fundamental theorem)
2d) .776 hours</p>
<p>3a) dy/dt = 0 @ t=.3673, so 12.061m
3b) A = 1.936 seconds
3c) s = 12.946m
3d) arctan(dy/dt/dx/dt) = 1.519 radians</p>
<p>4a) 4.25
4b) 2+4(x+1)-6(x+1)^2
4c) y = -4/e^(1/3) * e^(-x^3/3) + 6</p>
<p>5a) -3
5b) 8
5c) 18
5d) y+2 = 3(x-5) tangent line; y-3 = 5/3(x-8) secant line, just plugin and state since the graph is always concave down, tangent line over approximates, secant line underapproximates</p>
<p>6a) 1 + (x-2)^2 + (x-1)^4 / 2 + (x-1)^6 / 3! + … + (x-1)^(2n) / n!
6b) 1 + (x-1)^2 / 2 + (x-1)^4 / 3! + (x-1)^6 / 4! + … + (x-1)^(2n) / (n+1)!
6c) -inf < x < inf
6d) no points of inflection since original function is an always positive even function, so second derivative is an always positive even function, indicating concavity never changes. if you show the second derivative, you’ll see that it’s always >=1. </p>
<p>That’s all.</p>
<p>yes, laplast i agree with you on like everything!</p>
<p>i forgot what i put for 3c, but i was doing it now and im getting 15.394</p>
<p>dx/dt = .8 and dy/dt = 3.6 - 9.8t right?</p>
<p>You need to do arclength for 3c. integral from 0 to A of sqrt((dx/dt)^2 + (dy/dt)^2)dt</p>
<p>laplast: for 2c…</p>
<p>isn’t it the second fundamental theorem? the first one just states that the integral of a function is an antiderivative.</p>
<p>the second one states that integral of f’(x) from a to b = f(b) - f(a)</p>
<p>i used the second fundamental theorem and ended up with a different answer:
w’(t) = (2-t) R(t)
w(2) - w(1) = integral of w’(t)dt from 1 to 2 = 291.25</p>
<p>how did you arrive at your answer??</p>
<p>whoops calculator error. disregard my last post!!!</p>
<p>no one’s in the aim chat…
is it dead?</p>
<p>@plumsnow: Second Fundamental Theorem states that the derivative of an integral from a constant to u of f(x)dx is just f(u)*u’. First Fundamental theorem states that the integral from a to b of f(x)dx is F(b) - F(a) where F(x) is the antiderivative of f(x).</p>
<p>Laplast, for 2d how did you get .776 for the min/person? don’t you do 387.5/980? which equls .39</p>