notes / updates
May 2, 2025 | PAST FRQ Questions
AP PAST FRQ Questions
April 30, 2025 | Key to remaining FRQs - Packet "WUXI"
P5.
a) \( dy/dx = \frac{-2x-y}{x+6y} \)
b) \( y = 6x + 1 \)
c) \( y"(0)= -11/36 \)
d) The slope is zero at the given point, so the tangent line is horizontal and has equation: \( y = -2a \)
P6.
Set up the ratio of similar triangles to find \( r = h/5 \).
a) \( V = \frac{\pi}{75} h^3 \)
b) \( dV/dt = 8\pi \) cubic feet per second.
c) \(dA/dt = \frac{8\pi}{5} \) square feet per second. (Use: \( dr/dt = (1/5) dh/dt \text{ and } r=2 \text{ when } h=10 \).)
April 29, 2025 | Answer key to MC Packet "SEOCHO"
The answers to problems 1 through 10 were posted on the screen. Below is the rest of the answer key.
11 C 12 D 13 A 14 E 15 D 16 A 17 B 18 B 19 B 20 B 21 A 22 C 23 B 24 A 25 D 26 B 27 B 28 C.
April 18, 2025 | Feedback on today's quiz
a) Be sure to draw a curve that "fits" the slope field and passes through (1,1). The field below is generated with desmos.
b) Separate variables: \( \int y \text{ } dy = \int (1-x) dx \), then integrate:
\( \frac{y^2}{2} = x -\frac{x^2}{2} + C \).
Plug in (1,1) to find \( C = 0\).
The general solution is: \( y =\pm \sqrt{2x-x^2} \) but since (1,1) is the initial condition and \( y = 1 > 0 \),
we conclude that the specific solution is: \( y = \sqrt{2x-x^2} \).
P2.
\( \frac{dT}{dt} = k(T-75); T(0) = 350; T(5) = 340\)
Separate variables and integrate both sides to find the general solution:
\(\ln|T-75| =kt+C \).
\( T-75 = \pm e^{kt+C}\) (Note: the plus option makes sense here. Why?)
\( T(t) = 75 + e^C e^{kt} \)
To find the constants \(k\) and \( C \), plug in the two points listed above.
\( C = \ln 275 \approxeq 5.61677109767 \approxeq 5.617 \).
\( k = \frac{\ln(265/275)}{5} \approxeq -0.00740825433607 \approxeq -.007\)
To determine the time it takes for the calzone to reach 300 degrees, set \(T=300 \) and solve.
Using graphing on the TI-84, set \( y_1 = 300, y_2 = 75 + 275 e^{-0.007408254 x} \) (note the use of x rather than t)
The solution is \( t \approxeq 27.087 \) minutes.
Below find the desmos version with details:
P1.
After separating variables, set up the integrals as follows:
\( \int \frac{1}{1+y^2} dy = \int (1+x) dx \)
\( \arctan y = x + \frac{x^2}{2} + C \)
The general solution is \( y = \tan(x+\frac{x^2}{2}+C) \).
P2.
a) \( y = y_0 e^{kt}; y(5) = 0.7 y_0 \) (since 30% is gone)
\( 0.7 y_0 = y_0 e^{5k} \rightarrow k =\frac{\ln(0.7)}{5} \approxeq -0.071 \)
\( t_H = \frac{\ln(1/2)}{k}= \frac{-\ln 2}{k} \approxeq 9.763 \) years. (Does this make sense in the context of this problem?)
b) The percentage remaining is: \( 100 \times \frac{y(t)}{y_0} = 100\times \frac{y_0 e^{kt}}{y_0} = 100 \times e^{-.071 t} \% \) of the original.
April 18, 2025 | AP Classroom
The following link takes you to the
ap classroom. The link is also under the 'links' section on the bulletin board of your course page.
FYI - You registered for access to this resource at the start of school year.
April 14, 2025 | DESMOS FOR THE AP EXAM
Hey everyone. If you are a complete beginner with desmos, go over the following links in order. My sample file includes derivatives and integrals.
1. Getting Started
2. Graph Settings
3. Mr. Shubleka's Sample Desmos document
4. The Desmos version used in the AP Exam
April 14, 2025 | Feedback on today's quiz
The function is even, it has a horizontal asymptote of \( y = 0 \) at both ends of the picture, is increasing for \( x < 0 \) and decreasing for \( x > 0 \).
The graph is concave down on \( \left( \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \) and concave up everywhere else. See inflection points on graph.
Since we are revolving about the y-axis, the Shell Method is a reasonable choice here. The Washer Method would need to be broken into two integrals and one of them is especially difficult to solve.
\( V_{shell} = \int\limits_1^2 2\pi x e^{-x^2} dx \).
Solve this integral with \( u = -x^2\).
A desmos sketch is below.
April 3, 2025 | Feedback on today's quiz
a)
Draw the region neatly. Label the x and y-intercepts: (2,0) and (0,2).
b)
We are revolving about the x-axis. Use the Disk Method: \( V_{disk} = \pi \int\limits_0^2 (2-x)^2 dx = ... = \frac{8 \pi}{3} \) cubic units.
There is a gap, so we use the Washer method with respect to y. First find the outer and inner radii:
\( R = 2- 0 = 2. r = 2-x = 2 - (2-y) = y\)
\( V_{washer} = \pi \int\limits_0^2 (2^2-y^2) dy \)
Note: do NOT evaluate.
d)
Draw a typical cross-section and set up the area: \( A(x) = y*y = (2-x)^2 \)
The volume calculation is nearly identical to the calculation in part b).
\( V = \int\limits_0^2 (2-x)^2 dx = ... = \frac{8}{3} \) cubic units.
April 2, 2025 | Visualizing Solids of Known Cross-Sections
https://www.geogebra.org/m/XFgMaKTy
February 27, 2025 | Study Guide
As we get closer to the end of the curriculum, now is a good time to invest in a study guide for AP Calculus AB. The Princeton Review at the link below is recommended. Older editions of the same series should be fine, too. If you don't like Amazon, buy elsewhere after searching by title or ISBN. Purchasing the study guide is NOT required.
Publisher: Princeton Review; Premium edition
ISBN-10: 0593516737
ISBN-13: 978-0593516737
https://www.amazon.com/Princeton-Review-Calculus-Premium-Prep/dp/0593516737
January 13, 2025 | Thoughts on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
\( s(t) = t^3-9t^2+24t \)
\( v(t) = 3t^2-18t+24 = 3 (t-2)(t-4) \)
\( a(t) = 6t-18 = 6(t-3) \)
Study the sign behavior of velocity and acceleration, to conclude that that particle is slowing down on the intervals (0, 2) and (3,4) and speeding up on the intervals (2, 3) and \( (4, \infty) \).
Be sure to include a neat schematic drawing with domain \( t\geq 0 \) and s(0) = 0, s(2)=20, s(3)=18, s(4)=16.