notes / updates | ap calculus ab

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May 4, 2026 | Free Response Generator

For more Free Response practice, feel free to use the following generator. Each question should take approximately 15 minutes. After 2011, only Problems 1 and 2 are calculator active. Before 2011, problems 1, 2, and 3 were calculator active.
Free Response Generator

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April 10, 2026 | Quiz feedback.
P1.
The only issues here were:
1) show the step with integral signs.
2) pick the correct sign of the square root: \( f(x) =\pm \sqrt{x^2-2x+2} \) after declaring both.
3) C and 2C are not the same constant. Use a new letter like K to replace 2C if you must.


P2.
Separate variables:
\( \int \frac{1}{T-75} dT = \int k dt \)
Integrate both sides and add the integration constant:
\( \ln |T-75| = kt + C \) (Note the required use of the absolute value.)
Plug in \( t=0, T = 350 \) to find \( C = \ln 275 \approxeq 5.61677 \)
Plug in \( t=5, T = 340 \) to find \( k = \frac{\ln 265 - \ln 275}{5} \approxeq -0.0074 \)

The problem did not ask for the general solution to be stated, but if you must state the solution, be sure to write:
\( |T-75| = e^C e^{kt} \)
\( T - 75 = \pm e^C e^{kt}\)
\( T(t) = 75 \pm e^C e^{kt} \)
And then choose the positive option because the temperatures we are looking at are clearly more than 75 degrees.

To find the time it takes for the temperature to "cool" to 300 degrees, plug in T = 300:
\( \ln 225 = kt + \ln 275 \rightarrow t^{*} = \frac{\ln 225 - \ln 275}{k} \approxeq 27.087 \text{minutes} \). Below is a desmos plot for those interested..
https://www.desmos.com/calculator/saf1la8i6z

April 6, 2026 | Study Guide
In case you are interested, below I list the ISBN number for a recommended study guide for AP Calculus AB. Note that purchasing the study guide is not required.


ISBN-10 : 0593518217
ISBN-13 : 978-0593518212

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April 6, 2026 | Quiz notes
We went over problem 1 in class.
P2. \( y(t) = y_0 e^{kt} \rightarrow 0.75 y_0 = y_0 e^{5k} \rightarrow k = \frac{\ln(0.75)}{5} \approxeq -0.0575364 \)
a) The half-life of the substance is \( t_H = \frac{-\ln 2}{k} \approxeq 12.047 \) years.
b) Amount after t years: \( y(t) = y_0 e^{kt} \)

The percentage is: \( \frac{y(t)}{y0}*100 = e^{-0.0575364*t} *100 \) percent of the amount will remain t years later.
An interactive desmos demonstration is linked below. Use the slider at the top left to adjust the initial amount,
and observe the time values.
Exponential Decay Model

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April 3, 2026 | Slope Field Generator
Feel free to use these to confirm your slope field constructions(s).
Option 1: Desmos Slope field plotter Option 2: Geogebra Slope field plotter

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March 31, 2026 | Quiz Notes
a) \( f(x) = \ln(x-2), g(x) = x^2-6x \)
Sketch the region clearly indicating two intersection points. Use the TI-84 to find the
x-coordinates of the intersection points: \( m = 2.0003352, n = 6.2314943 \)
As was the case in the previous calculator-active quiz, refer to these coordinates by their name
in the remaining parts. The area of the region is:
\( A = \int\limits_m^n \left( f(x) -g(x) \right) dx \approxeq 28.375 \)
Note that we were asked to round to three decimal places.

b) Use the shell method: \( r=x-(-1) = x+1, h(x) = f(x)-g(x) \)
\( V_{shell} = 2\pi \int\limits_m^n (x+1) (f(x)-g(x)) dx \approxeq 851.650 \)

c) The typical cross section has area: \( A(x)= \frac{(f(x)-g(x))^2}{2} \)
The volume is: \( V = \int\limits_m^n A(x) dx \approxeq 108.853 \)
d) \( V_{washer} = \pi \int\limits_m^n (5-g(x))^2 - (5-f(x))^2 dx \)
e) Rewrite the integral as: \( V = 2\pi \int\limits_{-2}^2 (3-y) \left( (y^2+2 - \frac{1}{1+y^2}) \right) dy \)
State that the shell method is being used with \( y = 3 \) as a rotational axis. The region, which should be
shaded, is shown below with the aid of desmos:



f) Do the reading. Do the homework. Learn the calculator features by using the calculator.

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March 27, 2026 | Quiz Notes
P1.
The region was more or less given. All you needed to do is shade in the first quadrant and set up the Shell Method:
Use u-substitution with care when computing the integral. The steps are not included here.
\( V_{shell} = \int_{x=1}^{x=2} \left( 2\pi x e^{-x^2} \right) dx = ... = \frac{\pi (e^3-1)}{e^4} \)

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March 5, 2026 | Quiz Notes
P1.
a) Region:
The intersection point has coordinates \((m,n) = (3.4595647,1.2411428) \).
The area is: \( A = \int_{y=0}^{y=n} \left( (5-y^2) -(e^y) \right) dy \approxeq 3.109\)(ti-84)

b) \( V_{washer} = \pi \int_0^n \left( (5-y^2)^2 - (e^y)^2 \right) dy \approxeq 62.079 \)(ti-84)

c)\( V_{washer} = \pi \int_0^n \left( (6-e^y)^2- (6-(5-y^2))^2 \right) dy \approxeq 55.122 \) (ti-84)

d) The area of a typical cross section is: \( A(y) = \frac{\pi r^2}{2} = \frac{\pi}{8} \left( 5-y^2 - e^y \right)^2 \)
The volume is: \( V = \int_0^n A(y) dy \approxeq 3.713 \).

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March 2, 2026 | Quiz Notes
P1.
a) Region:

b) \(V_{washer} = \pi \int_0^{\pi/3} \left( 3-1 \right)^2 - \left((\sec x + 1) - 1\right)^2 dx = ... \)
c) \( V_{washer} = \pi \int_0^{\pi/3} \left( 4-(\sec x + 1) \right)^2 - \left((4 - 3\right)^2 dx \)
d) \( A=\int_0^{\pi/3} \left( 3 - (\sec x + 1) \right) dx =\int_0^{\pi/3} \left( 2- \sec x \right) dx \)

P2.
We went over the region in class. Use the washer method for the volume.
Since we are revolving about \(y=-1\), the outer and inner radii are:
\( R(x) = \cos x - (-1) = \cos x + 1 \)
\( r(x) = \sin x - (-1) = \sin x + 1 \)
For the evaluation itself, use the double angle formula for cosine.
\( V_{washer} = \pi \int_0^{\pi/4} \left( \cos x + 1 \right)^2 - \left( \sin x + 1 \right)^2 dx = ... \)

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February 23, 2026 | Snow Day - Office Hours
Office Hours - 12:30-1:30pm

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January 22, 2026 | Riemann Sums' Calculator
https://www.geogebra.org/m/Fv6t696j

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January 21, 2026 | Feedback on yesterday's quiz
\( a(t) = -32 \rightarrow \int (-32) dt = -32t + C = v(t)\)
Since the initial velocity is zero, it follows that \(C = 0 \).
The velocity when the stone hits the ground is -120. The time at which this happens can be determined by solving:
\( v(t) = -32t = -120 \rightarrow t=\frac{15}{4} \).
the position function is: \( s(t) = \int v(t) dt = \int (-32 t) dt = -16t^2 + K \).
The position is zero when \( t = \frac{15}{4} \), so find \( K = 225\) feet.
The constant is also the height of the cliff since it is the value of the position function \( s(t) = -16t^2 + 225 \) when \( t =0 \)

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January 15, 2026 | Feedback on today's quiz
Make a drawing of the paper (3 by 8) and the four equal squares that are cut off from the margins.
If the dimension of the squares is \( x\), then the box will have volume: \( V(x) = x (8-2x) (3-2x) \).
Expand the volume formula to get: \( V(x) = 4x^3 - 22x^2 +24 x \).
The domain of the cubic function is: \( 0 < x < 3/2\).
Differentiate and factor: \( V'(x) = 12x^2-44x + 24 = 4 (3x-2)(x-3) \)
The critical numbers are \(x = 2/3, x=3 \). Only the first critical number is in the domain.
Study the sign change of the derivative at the critical number to confirm that \( V(2/3) \) is a global maximum
by the First Derivative Test for Global Extrema.
Note that we could have used the Second Derivative Test for Global Extrema as well.
The maximum volume is \( V(2/3) = \frac{200}{27}\) cubic units.

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