notes / updates
April 25, 2025 | Feedback on today's quiz.
\( f(x) = \ln(1-x^2) \rightarrow f'(x) = \frac{-2x}{1-x^2}\)
\( 1 + \left(\frac{-2x}{1-x^2}\right)^2 = \frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2} =...= \frac{(1+x^2)^2}{(1-x^2)^2}\)
\( L = \int\limits_0^{1/2} \sqrt{1+(dy/dx)^2} dx = \int\limits_0^{1/2} \frac{1+x^2}{1-x^2} dx \)
Divide before setting up partial fractions, or rewrite as:
\( \int\limits_0^{1/2} \frac{-(1-x^2)+2}{1-x^2} dx = \int\limits_0^{1/2} -1 + \frac{2}{(1-x)(1+x)} dx = \int\limits_0^{1/2} -1 + \frac{A}{1-x} + \frac{B}{1+x} dx =... \)
The final answer is \( \ln(3) - \frac{1}{2} \).
April 21, 2025 | Feedback on today's quiz.
No points were awarded if in any of the parts below you set up a dx type of integral as a dy or vice-versa, even if the answer came out to be the same.
P1.
a) The sketch was fine. As a reminder, the curves were \(y = x^2, y = \sqrt{x} \).
b) \( V_{washer} = \pi \int\limits_0^1 \left(\sqrt{y} \right)^2 - \left(y^2 \right)^2 dy = ... = \frac{3\pi}{10} \)
c) \( V_{washer} = \pi \int\limits_0^1 \left(1-y^2 \right)^2 - \left(1-\sqrt{y} \right)^2 dy \)
d) \( V_{washer} = \pi \int\limits_0^1 \left(1-x^2 \right)^2 - \left(1-\sqrt{x} \right)^2 dx \)
P2.
a) the drawing was fine here, too. As a reminder, the curves were: \( x = 1 + y^2, x = 4-2y^2\).
b) \( A = \int\limits_{-1}^1 \left( 4-2y^2 \right) - \left( 1 + y^2 \right) dy = ... = 4 \)
Note: if using symmetry, be sure to explain why!
c) \( V_{washer} = \pi \int\limits_{-1}^1 \left( 6-(1+y^2) \right)^2 - \left( (6-(4-2y^2) \right)^2 dy \)
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 13, 2025 | Feedback on recent quiz.
P1.
\( x(t) = t-2\cos t , y(t)=2-2\sin t, \text{ for } 0\leq t \leq 10\)
a) For the particle to fly horizontally, we must set \( dx/dt \neq 0 \) AND \( dy/dt = 0 \).
Find three solutions: \( t = \frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2} \). (Note that \( 7\pi/2 > 10\).)
b) For the particle to fly vertically, we must set \( dx/dt = 0 \) AND \( dy/dt \neq 0 \).
Find three solutions: \( t = \frac{7\pi}{6},\frac{11\pi}{6}, \frac{19\pi}{6} \). (Note that \( 23\pi/6 > 10\).)
P2.
\( x(t) = \sqrt{t} - 2, y(t) = 2t^{3/4} \text{ for } 1\leq t \leq 16\).
\( (dx/dt)^2 = \frac{1}{4t}\)
\( (dy/dt)^2 = \frac{9}{4\sqrt{t}} = \frac{9\sqrt{t}}{4t} \)
\( L = \int\limits_1^{16} \sqrt{\frac{1+9\sqrt{t}}{4t}}dt = \int\limits_1^{16} \frac{1}{2\sqrt{t}} \sqrt{1+9\sqrt{t}}dt \)
Solve the integral with \( u = 1 + 9\sqrt{t} \). Be sure to update the bounds when switching variables.
April 3, 2025 | Common Maclaurin Series with Desmos
February 24, 2025 | Feedback on today's quiz.
P1.
Be sure to draw and shade the region. When drawing the graph, give a little more detail (for example, the vertical asymptotes).
\( A = \int_2^{\infty} \frac{4}{x^2-1} dx = \lim\limits_{t\to \infty} \int_2^t \left( \frac{A}{x-1} + \frac{B}{x+1} \right) dx = ... = 2\ln 3 \)
Show work when calculating \( A = 2, B=-2 \).
P2.
Recognize the integral as improper due to the discontinuity. This means we NEED to use limits. Use u-substitution.
\( \int_{-4}^2 \frac{x}{\sqrt{16-x^2}} dx = \lim\limits_{b\to -4^+} \frac{-1}{2} \int_b^2 \frac{-2x}{\sqrt{16-x^2}} dx = ... = -2 \sqrt{3} \)
P3.
Rewrite the integral by completing the square:
\( \int \frac{1}{\sqrt{(2y-1)^2-4}} dy\)
Use trigonometric substitution with: \( 2y-1 = 2\sec\theta \).
Be sure to draw a right triangle. The final answer comes out to:
\( \frac{1}{2} \ln \left| \frac{2y-1+\sqrt{4y^2-4y-3}}{2} \right| + C\)
P4.
\( \int_0^1 \ x \sqrt{2-\sqrt{1-x^2}}dx\)
Option 1: Use u-substitution with \( u = 1-x^2 \), then w-substitution with \( w = 2-\sqrt{u}\).
Option 2: Use trigonometric substitution with \( x = \sin\theta \).
No matter which option you choose, be sure to update the bounds carefully whenever you switch variables.
The final answer is: \( \frac{2(8\sqrt{2}-7)}{15}\)
February 17, 2025 | Feedback on today's quiz.
P1.
\( \int \frac{1}{\sqrt{x^2-a^2}} dx \)
Use trigonometric substitution with: \( x = a sec\theta \).
Be sure to draw a right triangle with all three sides labeled.
\( dx = a \sec\theta \tan\theta d\theta \)
After the substitution of the expressions for \( x \) and \( dx \), the integral reduces to:
\( \int \sec\theta d\theta = \ln |\sec\theta + \tan\theta | + C = \ln\left| \frac{x +\sqrt{x^2-a^2}}{a} \right| + C \)
P2.
One of your classmates pointed out that I was missing a \( dx \) at the end of the integral for this problem.
\( \int_0^{3\sqrt{3}/2} \left( \frac{x^3}{(4x^2+9)^{3/2}} \right) dx \)
Option 1: \( u = 4x^2 +9, du = 8 x dx, x^3 = x^2 x = \frac{u-9}{4} x \)
Remember to switch the bounds of integration to \( 9 \leq u \leq 36 \) when switching from \( dx \) to \( du\).
Option 2: Use trigonometric substitution with \( x = \frac{3}{2} \tan\theta \)
Be sure to draw a right triangle and update the bounds: \( 0 \leq \theta \leq \frac{\pi}{3}\)
The final answer with either approach comes out to 3/32.
January 15, 2025 | Problem 59 (Section 4.1)
The initial velocity is \( v(0) = 16\) ft/s (positive because the object is moving up).
The initial position is \( s(0) = 64 \) ft.
a)
Find the velocity function: \( v(t) = -32 t + 16 \) by taking the anti-derivative of acceleration and using the initial value.
Find the position function \( s(t) = -16t^2 + 16t + 64 \) by taking the anti-derivative of velocity and using the initial value.
To find the time when the object hits the ground, set the position expression equal to zero, then solve.
b) Plug the t-value into the velocity function.
January 13, 2025 | Feedback on Today's Quiz
P1.
We know for all four parts that \( f'(x) = v(x)\).
a) \( \int 2 v(x) dx = 2 \int v(x) dx = 2 f(x) + C \), so statement 1 is true.
b) The derivative of \( f(2x) \) is \( 2 f'(2x) = 2 v(2x) \), so the second statement is false.
\( f(x) = sin(x), v(x) = \cos(x) \) should work as an example.
c) \( \int (v(x) + 1) dx = f(x) + x + K \), so the third statement is false.
d) If we use the same example as the one in part b), we can see that \( (\sin x)^2 \) is not an anti-derivative of \( (\cos x)^2\). Therefore the fourth statement is false.
P2.
\( a(t) = -32 \rightarrow \int (-32) dt = -32t + C = v(t)\)
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 is also the height of the cliff since it is the value of the position function \( s(t) = -16t^2 + 225 \) when \( t =0 \)
P3.
a. \( \int \left[ \frac{1}{2\sqrt{1-x^2}} -\frac{3}{1+x^2} \right] dx = \frac{\arcsin x}{2} - 3\arctan x + C \)
b. \( \int \frac{\sec x + \cos x}{2 \cos x} dx = \int \left(\frac{1}{2}\sec^2x+\frac{1}{2} \right) dx = \frac{1}{2} \tan x + \frac{1}{2} x + C\)