notes / updates | ap calculus ab

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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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