WAMAP Exemplar Course

Hello and welcome to the WAMAP Exemplar course.

Here you'll find videos explaining how each part of the various components of WAMAP work. If you can't find what you're looking for or have a suggestion for a video that you think should be added, email [email protected]. 

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The video content about how to do stuff is up there.

The items that were used/created as demonstrations in the above videos are down here.

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All grades (including the final exam) have now been posted to Wamap.

Note that, as you check your scores, the final exam replacing the lowest quiz and/or test (described in the syllabus) actually shows up in the gradebook as a "Quiz Replacement" and "Test Replacement" in their respective categories. Then, the lowest score in each category is dropped (indicated by the "d" next to the particular test/quiz). This is mathematically equivalent to what's described in the syllabus, but ...

• Is WAY easier for "Teacher-person" to enter into wamap.
• Allows you to see all of your old scores, even after the "replacement"

For your convenience, the "percentage to 4.0 scale" table from the syllabus is posted below.

Have a great Winter break! :) 

https://youtu.be/52ZlXsFJULI 

Free online grapher

The original question...

"There is a line through the origin that divides the region bounded by the parabola `y = 4 x - 6 x^2`  and the x-axis into two regions with equal area. What is the slope of that line?"

Oh my, this question is a great application of going beyond what we've done using what we know (meaning it's likely that most students hate it ;) )

Okay, so the line equation would be `y=mx`  . That line will cut the area into to sections of varying size that depend on the slope of that line (shown below).

So, when we choose the right m value, the area bound by the red line and the quadratic will be half of that total blue area. That total area is ` int_0^(2/3)(4x-6^2)dx=8/27`. So we need to come up with an integral for the area bound between the red line and the quadratic and make sure that that new integral comes out as half of the total, that is `4/27` . The integrand of this new integral is the quadratic minus the line `(4x-6x^2)-(mx)=(4-m)x-6x^2` . Now we have to recognize that the lower bound is 0 for any value of `m` , but the upper bound for the integral changes. We need to solve `4x-6x^2=mx` for `x` . Use the quadratic formula (recognizing that `a=-6,b=4-m,c=0` and we get `x=0` which we already knew, and `x=(4-m)/6` , Now we can set `int_0^((4-m)/6)((4-m)x-6x^2)dx=4/27`  The antiderivative with respect to x is `((4-m)x^2)/2-2x^3` . Replace x with`(4-m)/6` and this becomes `(4-m)^3/216` . At the lower bound, the antiderivative is 0. This means that `int_0^((4-m)/6)((4-m)x-6x^2)dx=(4-m)^3/216=4/27` .

We can (finally) focus on finding the m number, by solving `(4-m)^3/216=4/27` . Clear the fractions to get `(4-m)^3=32` , so `4-m=root(3)(32)` or `m=4-root(3)(32)`  (cue exhaustion collapse).