notes / updates
November 6 , 2025 | Feedback on recent quiz
P1.
Problem 1 went well overall. No drawing was expected or possible here.
Conclusion: The current is changing at a rate of \( \frac{3}{4} \) amperes per second
when V is 12 volts and R is 4 ohms.
P2.
Note that 'eastbound' means going east.
After making a drawing, let \( x \) be the distance from the car to the intersection, and let \( z \)
be the distance from the observer to the car.
The equation is: \( z^2 = 50^2 + x^2 \). We know that after 5 seconds, \(x = 225\) meters.
To find the value of \(z \), plug the value of \(x\) into the equation: \( z = \sqrt{50^2 + 225^2} = 25\sqrt{85}\) meters.
We also know that \( dx/dt = 45 m/s \). Fully differentiate with respect to time:
\( 2 z \frac{dz}{dt} = 2 x \frac{dx}{dt} \), to find: \( \frac{dz}{dt} = \frac{225 * 45}{25* \sqrt{85}} = \frac{405}{\sqrt{85}}\) meters per second.
Be sure to include a sentence as your conclusion.
P3.
Draw a big cone with an inner cone. The ratio of similar triangles is: \( \frac{r}{h} = \frac{5}{8}\).
The full volume is \( V =\frac{200\pi}{3} \), which means half of it is \( \frac{100\pi}{3} \).
The volume formula is: \( V = \frac{\pi}{3} r^2 h = \frac{25 \pi}{192} h^3 \).
Set this equal to the half volume to find \( h = 4 \sqrt[3]{4} \).
Note that \( h = 4 \) does not give you half the volume because this is a cone and not a cylinder.
After fully differentiating, find \( \frac{dh}{dt} = \frac{3 \sqrt[3]{4}}{125\pi} \) feet per hour.
Be sure to include a conclusion.
P4.
No issues with Problem 4. The unknown rate came out to 12 centimeters per minute.
November 4 , 2025 | Feedback on recent quiz
P1.
Make a neat drawing.
Let \( x \) be the distance from the telescope to the point on the ground closest to the drone,
and let \( \theta \) be the angle of elevation. The side opposite \( \theta \) is 1 km .
The equation is \( \tan\theta = \frac{1}{x} \) or \( \cot\theta = x \)
Differentiate with respect to time, to get:
\( -\csc^2\theta \frac{d\theta}{dt} = \frac{dx}{dt} \)
Plug in the details to find: \( \frac{dx}{dt} = \frac{2}{9\pi} \) km/min.
Be sure to include a full sentence as your conclusion: At the time when the angle of
elevation is \( \frac{\pi}{3} \) radian, the drone is traveling at a rate of \( \frac{2}{9\pi} \) km/min.
P2.
This is similar to the example we did in class on day 2 of related rates
Make a neat drawing.
Let \( x \) be the distance from the point where the beam hits the shore to the point on the
shore closest to the lighthouse, and let the angle that the beam forms with the closest path
from the lighthouse to the shore (1000 m) be \( \theta \).
The equation is: \( \tan\theta = \frac{x}{1000} \)
Differentiate with respect to time: \( \sec^2 \theta \frac{d\theta}{dt} = \frac{1}{1000} \frac{dx}{dt} \).
Next, plug in: \( \frac{d\theta}{dt} = 4 \pi \) and \( \sec^2(\theta^*) = \frac{17}{16} \).
Note that from precalculus you can easily compute secant from tangent: \( \sec^2 \theta = \tan^2 \theta + 1 = \left[\frac{250}{1000}\right]^2 + 1 = \frac{17}{16} \).
\( \frac{dx}{dt} = 4250 \pi\) meters per minute. Be sure to include a full sentence as your conclusion.
November 1 , 2025 | Feedback on recent quiz
P1. (Quiz 1)
\( 6 + x^3y = xy^2 \)
a) Differentiate implicitly to find \( \frac{dy}{dx} = \frac{y^2-3x^2y}{x^3-2xy} \text{ or } \frac{3x^2y-y^2}{2xy-x^3}\)
Note that when taking the derivative with respect to x, we needed the product rule (and chain rule) on both sides.
b) Plug in \( x= 1 \) into the curve's equation to find two solutions: \( y = -2, y=3 \)
Evaluate each slope at the points (1, -2) and (1,3) by plugging the values into your answer in part a).
Set up the point-slope form of the tangent lines to find the equations: \(y=2x-4 \text{ and } y=3\)
c) In order to have a vertical tangent, the denominator of your answer in part a) should be zero,
which yields: \( y = \frac{x^2}{2} \). Plug this into the curve's equation and solve for x. Find \( x = \sqrt[5]{-24}\).
This is the equation of vertical tangent line.
In case you are curious, below is an embedded desmos plot of the curve and the three tangent lines.
P2. (Quiz 2)
Be sure to draw a neat cylinder with the inner cylinder of 'water'.
The radius radius \( r = 3 \) does not change, so you do not need a variable.
The height \( h \) should be labeled in the drawing.
The big equation is: \( V = \pi 3^2 h \text{ or } V = 9 \pi h \)
Take the derivative with respect to time:
\( \frac{dV}{dt} = 9\pi \frac{dh}{dt} \)
Plug in \( \frac{dV}{dt} = 5 \) to find \( \frac{dh}{dt} = \frac{5}{9\pi} \frac{m}{min} \).
Be sure to include a sentence as your conclusion.
P1. (Quiz 3)
\( x^2 y^2 + xy = 2 \)
If you prefer, you can rewrite the equation:
\( (xy)^2 + xy = 2 \)
Differentiate implicitly with respect to \( x\) :
\( 2 (xy) [x \frac{dy}{dx} + y] + [x \frac{dy}{dx} + y] = 0 \), then plug in \( dy/dx = -1\)
The resulting equation can be factored completely: \( (y-x)(2xy+1) = 0 \)
Case 1: \( y = x\)
Case 2: \( 2xy = -1 \)
Plug in each case into the original equation of the curve.
For case 1, find (-1,-1) and (1,1). For case 2, find that it is impossible.
October 28, 2025 | Feedback on recent quiz
P1.
\( f(x) = (2x+7)^6 (x-2)^2 \rightarrow f'(x) = 2(x-2) (8x-5) (2x+7)^5 = 0 \)
The x-values where the tangent line is horizontal are: \( 2, \frac{-7}{2}, \frac{5}{8} \)
P2.
\( f(x) = \frac{(x-3)^4} {x^2+2x} \rightarrow f'(x) = \frac{2 (x-3)^3 (x^2+6x+3)}{x^2 (x+2)^2} = 0 \)
The x-values where the tangent line is horizontal are: \( 3, -3-\sqrt{6}, -3 + \sqrt{6} \)
October 26, 2025 | Feedback on recent quiz
P1.
\( f(x) = (2x-1) \sqrt{4x+1}\)
Use the Product Rule and Chain Rule with care, before factoring completely:
a) \( f'(x) = 2 \sqrt{4x+1} + (2x-1) \frac{1}{2\sqrt{4x+1}}*4 = ... = \frac{12x}{\sqrt{4x+1}}\)
\( f'(x) = 0 \rightarrow x = 0 \)
Since it asks for the point(s), be sure to find the y-coordinate. The point where the tangent line is horizontal is \( (0, -1) \).
b)
At \( x = 2 \), find slope: \( f'(2) = 8, f(2) = 9 \)
Use the point-slope form to find and reduce the equation to \( y = 8x-7 \).
October 14, 2025 | Feedback on recent quiz
P1.
a)
\( f(x) = \frac{x-1}{1+x} \rightarrow f'(x) =\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} = \\
=\lim\limits_{h \to 0} \left( \frac{1}{h} \left[\frac{x+h-1}{1+x+h} - \frac{x-1}{1+x} \right] \right) = ... = \frac{2}{(1+x)^2} \)
b)
\( f'(x) = \frac{2}{(1+x)^2} = \frac{1}{2} \rightarrow (x+1)^2 = 4 \rightarrow x=-3, x=1 \)
The points are \( (1, 0), (-3, 2) \)
October 10, 2025 | Feedback on recent quiz
P1.
The function was: \( f(x) = -x^3+2x^2 \) at the point with x-coordinate \(-1\).
\( \text{slope} = m = f'(-1) = \lim\limits_{x \to -1} \frac{f(x)-f(-1)}{x+1} = \\
= \lim\limits_{x to -1} \frac{-x^3+2x^2-3}{x+1} = \lim\limits_{x \to -1} \frac{(x+1)(-x^2+3x-3)}{x+1}\\
= \lim\limits_{x \to -1} (-x^2+3x-3) = -7 \\
y - 3 = -7 (x+1) \text{ or } y = -7x -4 \) is the equation of the tangent line.
Note 1: In the process above you can factor the cubic by dividing by \( (x+1) \) using long or synthetic division.
Note 2: The "h" definition should give you the same result for slope.
October 8, 2025 | Feedback on recent quiz
General reminders: Show arrows to check where the expressions go. Use equal signs. Use limit notation.
P1.
State that the function \( f(x) = \sqrt{x} -\cos x - 1 \) is continuous in its domain and therefore continuous on \( [0, 2\pi ] \).
State that you are using the Intermediate Value Theorem on the given interval.
Evaluate \( f(0) = -2 < 0, f(2\pi) = \sqrt{2\pi} -2 > 0 \).
Conclude, using a full and grammatically correct sentence, that by IVT, the function must have an x-intercept at least once in the given interval.
This, of course, is the same as concluding that the original equation has a solution in the given interval.
P2.
a) Use the Squeeze Theorem. Start with the double inequality: \( -1 \leq \sin(\frac{2}{x^3}) \leq 1 \)
Multiply all three sides by \( |x| \).
Your conclusion should include a reference to the Squeeze Theorem with the setup and confirmation that the outside functions both go to zero.
The overall limit is zero.
b) Positive infinity with work / justification.
October 5, 2025 | Feedback on recent quiz
General reminders: Show arrows to check where the expressions go. Use equal signs. Use limit notation.
P1.
a) Determine the domain of \( f(x) =\frac{x^2-6x-16}{(x^2-7x-8)(x^2-4)} \) after factoring completely:
\( f(x) =\frac{(x-8)(x+2)}{(x-8)(x+1)(x-2)(x+2)} \)
The domain is all real numbers except for \( -2, -1, 2, 8 \)
b) The denominator's factors that are shared with the numerator's factors will yield removable discontinuities. The remaining factors of the denominator are vertical asymptotes.
Therefore, removable discontinuities at \( x=-2, x=8 \)
and infinite discontinuities at \( x=-1, x=2 \).
P2.
a) \( \tan t = \frac{\sin t}{\cos t} \) approaches \( -\infty \) as \( t\to \pi/2^+ \) because sine goes to 1 and cosine goes to zero from the left side.
Therefore, the overall limit is \( e^{-\infty} = \frac{1}{e^{\infty}} = 0 \)
b) Check that the fraction is a zero over zero form (use arrows), then break up it up into two limits:
\( \lim\limits_{x\to0} \frac{3x}{x} -\lim\limits_{x\to0} \frac{\sin \pi x}{x} = ... \)
Multiply the second limit by \( \pi \) on the outside and divide by \( \pi \) inside. The overall limit is \( 3-\pi \).
September 30, 2025 | Feedback on recent quiz
Show arrows to check where the expressions go. Use equal signs. Use limit notation.
P1.
Check where the top and bottom go, then use the conjugate.
\( \lim\limits_{x\to -3} \frac{\sqrt{x+4}-1}{x+3} = ... = \frac{1}{2} \)
P2.
Check where the top and bottom go, then factor completely:
\( \lim\limits_{x\to 1/2} \frac{2x^2+3x-2}{2x-1} = \lim\limits_{x\to 1/2} \frac{(2x-1)(x+2)}{2x-1} = ... = \frac{5}{2} \)
P3.
Check where the top and bottom go, then use the difference of cubes formula to factor completely:
\( \lim\limits_{x\to 2} \frac{x^3-8}{x^2-4} = \lim\limits_{x\to 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = ... =3 \)
P4.
Check where the top and bottom go, then rewrite the tangent expression:
\( \lim\limits_{\theta \to \pi} \frac{\sin\theta}{\tan\theta} = \lim\limits_{\theta\to \pi} \frac{\sin\theta \cos\theta}{\sin\theta} = ... =-1 \)
September 28, 2025 | Feedback on recent quiz
P1.
\( \lim\limits_{x\to 4^+} \frac{5-x}{x^2+x-20} =\lim\limits_{x\to 4^+} \frac{5-x}{(x-4)(x+5)} \)
Use arrows to show that the numerator goes to 1 and the denominator goes to 0 from the positive side. Therefore, the fraction approaches positive infinity.
P2.
\( \lim\limits_{h\to 0} \frac{9-\sqrt{h^2+81}}{h} = \lim\limits_{h\to 0} \frac{9-\sqrt{h^2+81}}{h} \frac{9+\sqrt{h^2+81}}{9+\sqrt{h^2+81}} = ... \)
Check that both numerator and denominator go to 0, then use the conjugate, a technique we covered in class.
The limits turns out to be 0.
P3.
The statement is false. As an example, consider: \( f(x) =\frac{|x|}{x}, g(x) = \frac{-|x|}{x} \)
The individual limits of these two functions as \(x\) approaches zero do not exist. The limit of the sum is 0, however.
September 23, 2025 | Working with Lists - TI NSPIRE
For those of you who own an NSPIRE, feel free to use the following tutorial with text and graphics.
Stop after the section where it says "Selecting a Column (or Row)".
Link:
TI NSPIRE TUTORIAL
September 17, 2025 | Notes on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
a)Factor by grouping: \( f(x) = 3x^3-x^2 -9x+3 = ... = ((3x-1) (x-\sqrt{3}) (x+\sqrt{3}) \)
b) x-intercepts: \( (\sqrt{3},0), (-\sqrt{3},0), (1/3,0) \)
y-intercept: \( (0, 3) \)
c) Sketch is not shown here. Label your axes with arrows (not the cubic curve). Be sure to indicate intercepts clearly.
d) The domain of \( h(x) = \frac{1}{g(x)} \) consists of all x-values that make \( g(x) \) nonzero. This means the domain is all reals except for the x-intercept values in part a).
P2.
Factor completely: \( (\cos x - 2) (\cos x + 1) = 0 \) and set each factor equal to zero.
\( \cos x = 2 \) is impossible since the range for cosine is \( [-1, 1] \).
From the second factor, we get \( \cos x = -1 \rightarrow x=\pi \).
P3.
Use the laws of logarithms to solve:
\( \log_5 (x-2000)^3 = 6 \)
\( 3 \log_5 (x-2000) = 6 \)
\( \log_5 (x-2000) = 2 \)
\( x- 2000 = 5^2 = 25 \)
\(x=2025 \)
September 11, 2025 | Notes on recent quiz
If there are typos or mistakes, let me know so I can fix them here.
P1.
a) \( f(x) = \sqrt{(1-x)(x+3)} \rightarrow (1-x)(x+3) \geq 0 \)
Solve graphically or algebraically or on the number line with test points, to conclude that the domain is: \( [-3, 1] \)
b) \( g(x) = \sqrt{-x-2} \rightarrow -x-2 \geq 0 \). The domain is: \( (-\infty, -2] \)
c) The domain of \( h(x) = f(x) + g(x) \) is the intersection of the two domains above, which is \( [-3, -2] \).
d) Similar to the example in part a), we can form a function such as: \( f(x) = \sqrt{(1-x)(x-5)}\). Be sure to show some work. There are other examples that work, too.
September 9, 2025
Hi everyone. Below is an answer key to yesterday's homework. I typed it fast so please let me know if you find mistakes.
1. x = 1
2. x = 6
3. x = 2 +- sqrt(11)
4. x = 1/4 (-3 +- sqrt(73))
5. x < 3/5
6. x <= -23/4
7. 14 < x < 22
8. x <= -13 OR x >= -2
11. Quadrant IV
12. Quadrant III
13. Quadrant II
14. Quadrant I
15. Quadrants III or IV
16. Quadrants I or IV
17. Quadrant III
18. Quadrant III
19. Quadrants I or III
20. Quadrants II or IV
September 8, 2025
AP registration instructions at this [
link ].
If you think you'll be changing your schedule or dropping the class altogether, please hold off on joining the AP classroom.
For your course section, use join code: NW792Q
If you have any questions, contact Mr. Dan Monahan (AP Coordinator).
September 1, 2025 | Introductory note.
Welcome to AP Calculus AB. In this area of the course page we will post important course-related information such as links to resources, announcements, reminders, commentary or feedback on assessments etc. All entries will be dated and the most recent one will be at the top of the page. As far as daily course information goes, the Assignments and Bulletin tabs are the most important.