<p>Ehh I can do the first one at least. </p>
<p>1.
a) W’(12) = (W’(15) - W’(9)) / (15 - 9)
= (67.9 - 61.8) / 6
= 1.017 degrees Fahrenheit per minute</p>
<p>b) W(20) - W(0)
=71 - 55
=16 degrees Fahrenheit
The temperature of the water in the tub raises 16 degrees Fahrenheit from t=0 to t=20 minutes</p>
<p>c) (1/20)[(55<em>4) + (57.1</em>5) + (61.8 * 6) + (67.9*5)] degrees Fahrenheit
=60.79 degrees Fahrenheit</p>
<p>This approximation is an underestimate because W(t) is concave up. (This was my explanation, but I’m not sure whether or not it’ll suffice.)</p>
<p>d) 55+ ∫W’(t)dt on the interval [0,25]
=74.0773 degrees Fahrenheit at t=25 minutes</p>
<p>Your part D is incorrect because it is set up incorrectly. I don’t have a calculator, so I can’t tell you the right answer, but W’(t) is only applicable for time 20<t<25.</p>
<p>The correct setup is W(20) + integral of W’(t) from [20,25] = W(25)</p>
<p>For part c, your answer explanation is incorrect as well. W(t) is increasing, but not necessarily concave up. Thus, the left reimann sum is an underestimate. I almost missed that!</p>
<p>I’ll do second one then.
2.
a) intersection point is (3.69344, 1.30656)
R= ∫ lnx dx from x=1 to 3.69344 + ∫ 5- x dx from x=3.69344 to 5.</p>
<p>b)Volume with square slices have the form V= ∫ y^2 dx. Substitute y with given equations.
V= ∫ (ln x)^2 dx [from x=1 to 3.69344] + ∫ (5- x)^2 dx [from x=3.69344 to 5].</p>
<p>c) Turn the graph so you are slicing dy instead of dx. Make sure y is used instead of x.
∫ 5 -y -e^y dy [from y=0 to k] = ∫ 5 -y -e^y dy from [y=k to 1.30656]</p>
<p>Ohhhh that makes sense. I knew I did something wrong. :</p>
<p>I got the third.</p>
<p>a) g(2)= integral from 1 to 2 = -1/4
g(-2)= integral from 1 to -2 = -((-pi/2)+(1.5)) = (-3+pi)/2</p>
<p>b) g’(x)=f(x)
g’(-3)= 2
g’’(x)=f’(x)
g’’(-3)=1</p>
<p>c) At x=-1 and 1
x=-1 is a maximum because g’(x) changes from positive to negative.
x=1 is neither because g’(x) does not change sign.</p>
<p>d) At x=-2,0,1
At these points g’(x) changes from increasing to decreasing, decreasing to increasing and increasing to decreasing respectively.</p>
<p>You could specify that it was generally concave up [0,20], which could probably suffice as an answer. I sketched a small graph to support it as well, but not sure if they will accept it.</p>
<p>But it isn’t entirely concave up, you only need to say that W(t) is increasing to get the explanation point.</p>
<p>4.
a) f’(x) = (-x)/(sqrt(25-x^2))</p>
<p>b) y-4=(3/4)(x+3)</p>
<p>c) Yes, g is continuous at x = -3 because the limit as x goes to - 3 is 4 on both sides and the limit as x goes to -3 is equal to g(-3) </p>
<p>d) ∫x(25-x^2)^(1/2)dx [0,5]</p>
<p>u= 25-x^2
du = -2xdx</p>
<p>-(1/2)∫u^(1/2)du [0,5]</p>
<p>-(1/2)[(2/3)u^(3/2) [0,5]</p>
<p>-(1/2) [(2/3)(25-x^2)^(3/2) [0,5]</p>
<p>I rewrote this to read -(1/2)times [(2/3) times the square root of (25-x^2)^3] </p>
<p>When you plug in 5 for x it’s merely 0, and when you plug in 0 for x you get 125/3</p>
<p>-(1/2)[0 - (250/3)] = 125/3</p>
<p>Oh yes, you’re right. I’ll edit that right away. I did this right on the exam, it’s just difficult for me to do it on the computer I guess. Haha</p>
<p>I’ll do five, you can do six.</p>
<p>5a) 40 grams. Plug in for B=40 and 70. dB/dt at 40 =12. dB/dt at 70 =6. Thus, the bird is gaining weight faster at 40 grams. </p>
<p>b) Second derivative = (-1/5)<em>(dB/dt) = (-1/25)</em>(100-B). For any value B between 0<B<100, B’’(t) is negative. Thus, the original graph is always concave down and cannot switch from concave up to concave down as shown in the incorrect drawing. </p>
<p>c)Kind of a pain!</p>
<p>Put B’s on one side and t’s on the other. Integrate both sides. You will then get:
-ln(abs(100-B))=(t/5)+C
Put negative on (t/5)+C side
e both sides
100-B = e^(-t/5 +C).
Bring constant down
100-B = Ce^(-t/5)
Solve for C
100-20= C(1)
C=80
Solve for B
100-B= 80e^(-t/5)
B=100-80e^(-t/5)</p>
<p>You can justify this answer because the original graph will always be concave down. Since e^x is increasing concave up, -e^-x will be increasing concave down, which correlates to our justification in part a and part b.</p>
<p>im pretty sure i got a 5 on the CAlc AB it was easy as hell</p>
<p>6.
a) v(t) = 0 when t=3 and t=9
The particle’s moving to the left on the interval [3,9] </p>
<p>b) ∫|v(t)|dt [0,6]</p>
<p>c)v’(t) = a(t)
a(t) = -(pi/6)sin((pi/6)t)
a(4) < 0
v(4) < 0
The speed of the particle is increasing at t=4 because both v(t) and a(t) are < 0 at this time. </p>
<p>d) -2 + ∫v(t)dt [0,4] will give you the position I think.</p>
<p>did you guys get the related rates problem on the multiple choice with the six foot man walking with a shadow</p>
<p>Yeah it was 2.667</p>
<p>Answer to question #5 is on the previous page!</p>
<p>Ahahaha I just realized I made the stupidest mistake on question 5 for the second derivative. I put -(1/10) instead of -(1/25). I guess it’s cause I saw two 5s and my brain just said multiply by 2. Ah well, I did the separable equation part right at least, and that’s probably where most of the points for that question are.</p>
<p>how did you guys do part c of number 2 on the free response because it was one of the few parts that gave me trouble</p>
<p>That sucks!
Usually 5/9 of the points on the differential equations questions are on part c, so you’ll be fine!</p>
<p>Explanation for #2 is on the previous page.</p>
<p>@heyouthere</p>
<p>c) Turn the graph so you are slicing dy instead of dx. Make sure y is used instead of x.
∫ 5 -y -e^y dy [from y=0 to k] = ∫ 5 -y -e^y dy from [y=k to 1.30656]</p>
<p>are you ****ing serious 5/9 damn
but at least i pretty confident about my multiple choice
i expect at least a 35/45</p>