<p>Can someone please do these problems?
I really want to know the solutions.
<a href="http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_ab.pdf%5B/url%5D">http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_ab.pdf</a></p>
<p>Problem #1</p>
<p>A)G’(5)=-24.586 tons per hour^2. The amount of unprocessed gravel arriving per hour is decreasing at this rate.</p>
<p>B)Integration G(t) from t=0 to t=8 is equal to 825.551 tons</p>
<p>C)Amount of Gravel=G(5)-100=-1.8529. Thus the amount of unprocessed gravel is decreasing.</p>
<p>D)Critical points of G(t)-100 are at t=4.923, where t is a relative maximum. 500+Integration of G(t)-100 from t=0 to t=4.923 gets you the total amount of unprocessed gravel of 635.376 tons.</p>
<p>Problem #2</p>
<p>A) |2|=v(t). Solving for t=3.128 and t=3.473</p>
<p>B) s(t)=10+Integration of v(t) from 0 to t. When t=5, s(t) is equal to -9.207.</p>
<p>C) 0=v(t). Solving for t=0.536 and t=3.318. Point where velocity either goes from positive to negative or negative to positive.</p>
<p>D)Increasing because a(4)<0 and v(4)<0. </p>
<p>Problem #3</p>
<p>A) [C(4)-C(3)]/(4-3)=1.6 ounces per minute</p>
<p>B) Yes. According to the mean value theorem: Because the average rate of change from the interval (2,4) is equal to 2, there must be some time t in the interval (2,4) in which the instantaneous rate of change,C’(t), is also equal to 2.</p>
<p>C) (1/6)<em>2</em>(5.3+11.2+13.8)=10.1 ounces of coffee has dripped into the large cup since t=0.</p>
<p>D) B’(5)=6.4e^(-0.4*5t)=6.4/(e^2) ounces</p>
<p>Problem #4</p>
<p>A) x=6</p>
<p>B) 4+Integration of f’(x) from x=8 to 0 is equal to -12.</p>
<p>C) (0,1)U(3,4); f’(x)>0 signifies the graph is increasing, and slope of the line tangent to f’(x)<0 signifies the graph is concave down. These intervals satisfy both of these conditions.</p>
<p>D) g’(3)=3<em>f’(3)</em>f(3)^2=3<em>4</em>(-5/2)^2=75</p>
<p>Problem #5</p>
<p>A) Integration of g(x)-f(x) from x=0 to 2=16/PI-4/3</p>
<p>B) PI*Integration of (4-f(x))^2-(4-g(x))^2 from x=0 to 2</p>
<p>C) Integration of [g(x)-f(x)]^2 from x=0 to 2</p>
<p>Problem #6</p>
<p>a)f’(x)=e^0<em>(3</em>1^2-6*1)=-3. f(x)=-3(x-1); f(1.2)=0±3(0.2)=-.06</p>
<p>b)dy/(e^y)=(3x^2-6x)dx
(e^-y) dy=(3x^2-6x)dx
-e^(-y)=x^3-3x^2+c
-e^(0)=(1)^3-3(1)^2+c
c=1
e^(-y)=-x^3+3x^2-1
-y=ln(-x^3+3x^2-1)
y=-ln(-x^3+3x^2-1)</p>
<p>Thanks alot!</p>
<p>Wait for your solution for 3c isn’t the 10.1 ounces the average amount of coffee entering the cup per second due to the 1/6 multiplied to the integral. So 10.1 is only 1/6 of the total coffee in the cup but the integral shows is that coffee was going into the cup 10.1 ounces/second</p>
<p>Do you concur?</p>
<p>Yea you’re right.</p>
<p>why isnt #5C’s function cubed? it asked for volume correct?</p>