<h1>4 a) i got g(-3)=9pi/4 - 6 , and g’(-3)=2</h1>
<pre><code> b) i got absolute max at x=3/2
c) i got point of inflection at x=0
d) i didnt do
</code></pre>
<p>Can someone show work for 1.b. i can’t get the 1.9494 everybodies talking about…
for 1.c you are right, i did that on the test just forgot. I just did integral from 0 to 5.195 then (-)integral from 5.195 to 6. same thing 12.573.
@BadGirlsClub im not sure if you should integrate the function - 1 or 1 - function. Anybody comment on this?</p>
<p>oh yes my bad. i think i did put minus on the test.
(hopefully)</p>
<p>you do x(0) + integral from o to 6 of the absolute value of v(t)
which should give you 2 + 12.573
which equals 14.573</p>
<p>i think its 1- the function since 1 is above the function</p>
<p>@mattgtodude
I believe 1.d is 1-function</p>
<p>For 1.b you integrate velocityfrom 0 to 6 and divide by six</p>
<ol>
<li> .
a. g(-3) = -6 + integral from -3 to 0 of f(x)
g(-3) = -6 + ¼ * pi * (3^2) = 1.0685
g’(x) = 2 + f(x)
g’(-3) = 2 + f(-3) = 2
b. 2 + f(x) = 0
F(x) = -2 at x = 3/2
c. g’’(x) = f’(x) = 0 at nowhere. Nowhere is there a zero slope.
d. F(3) – F(-4) / 3- (-4) = -2/7
MVT does not have to apply because of the discontinuities in the slope</li>
</ol>
<p>On 1.b. everybody else is correct, I used average rate of change instead of average value. shoudl be 1.949. My Bad.
Please post anything you got different to my answers</p>
<p>1a is increasing because v < 0 and a < 0</p>
<p>kwokings is right because it asks for speed and not velocity.</p>
<p>does anyone know what raw score you need to get a 4 and 5?</p>
<p>but if the velocity is negative wouldnt it be decreasing?</p>
<p>nope. for speed, it is only increasing if v AND a are BOTH less than zero or more than zero</p>
<p>Basically the velocity is decreasing but it is going more in the negative direction. It is still going faster because the aceleration is pushing it faster into the negative direction.</p>
<p>ugh -_- i never knew that</p>
<p>will 68+ be a 5?</p>
<p>is this true for the 2011 ap test?
5 (68-108)
4 (52- 67)
3 (39-51)
2 (27-38)
1 (0-26)</p>
<ol>
<li> .
a. F(0) + F(0)(.25-0)
1400 + (1/25)(1400-300) * (.25) = 1411
b. I did this wrong I think. I said second derivative = (1/25) so it is concave up everywhere. Because it is concave up, derivative is increasing, thus the guess is lower than the actual, because the derivative at t = .25 is higher than if you are guessing at t = 0.
c. Solve for W and dW on one side and t on the other
dW / (W-300) = (1/25)dt
integrate
ln(W-300) = (1/25)(t) + C
e both sides
W-300 = e^(1/25t + C)
Bring down C, add 300 to both sides
W = Ce^(1/25t) + 300
Plug 0 in for t, W = 1400, so C = 1100
W = 1100e^(1.25t) + 300</li>
</ol>
<p>for the multiple choice question about related rates, what did you get?
it was the question about a cylindrical pool filling with water at a rate of 2 m/min, the radius was 3, the height was 7 (but this part was extra unnecessary info), and they wanted you to find the rate at which the height was increasing at h=2.</p>
<p>i got 0.071</p>
<p>Got the same badgirlsclub</p>