Class, Take Your Seats
Solution Key by David Pleacher
Published in the
Winter 1991 issue of
The Virginia Mathematics Teacher
#2. Find the minimum point by taking the derivative and setting it equal to zero.
y' = 2x - 2.
So, the minimum point is at x = 1, y = 3.
Therefore the normal line is x = 1.
And there are five points in our domain that satisfy that condition:
(1, -1), (1, 0), (1, 1), (1, 2), and (1, 3).
#3. To find the point of inflection, determine the first and second derivatives.
Check concavity on both sides of x = -2.
So, Newton sits at (-2, 0).
#4. First locate Boole in clue #9 (Boole is located at (2, 1)).
Let s = distance from Boole (2, 1) to Euler (x, y).
We want this distance s to be a minimum, so take the derivative and set = 0.
Therefore, Euler sits at (2, 0).
#5. To find the relative maximum, take the derivative and set = 0:
y' = 3x2 - 6x - 9
y'' = 6x - 6
x2 -2x - 3 = 0
(x - 3)(x + 1) = 0
So, x = -1, 3
At x = -1, there is a rel max
At x = 3, it is a rel min
So, MacLaurin sits at the point (-1, 1).
#6. Since f(x) = -x2 + 4x - 1 is a parabola opening down, the absolute maximum point
will be the relative maximum. So, take the derivative and set = 0.
y' = -2x + 4.
x = 2. Since y''= - 2, it is a relative maximum point.
Therefore, Saccheri sits at (2, 3).
#7. To find critical points, take the first derivative and set = 0.
y' = x3 - 3x2 + 2x
x (x - 2) (x - 1) = 0
So, x = 0, 1, 2.
So, Riemann must sit at either (0, -1) or (2, -1).
#8. To find the point of inflection, take the second derivative and set = 0,
and then check the concavity on either side to be sure that it changes.
y' = 2x - k x-2
y'' = 2 + 2k x-3
x3 = -k
k = -1
Therefore, y = x2 - 1/x
So, Zeno sits at (1, 0).
#9. The curve (x - 2)2 + y2 = 1 is a circle with center at (2, 0) and radius 1.
Therefore, the maximum point is located at (2, 1).
You could also do this taking the derivative and setting = 0.
#10. First draw a picture with the given information:
Since we are looking for the largest area of a rectangle,
this is a maximum / minimum problem.
Find an equation for the area, then take its derivative,
and set it equal to zero.
So, Archimedes sits at one of the four points:
(1, -1), (1, 1), (-1, 1), or (-1, -1).
#11. To find the point where the slope is 48,
Take the first derivative and set it equal to 48.
Then solve the equation for the x-coordinate.
y' = 6x2 - 12x
6x2 - 12x = 48
x2 - 2x - 8 = 0
(x + 2) (x - 4) = 0
x = -2, 4
The only point in our domain is (-2, 3), so Thales sits here.
#12. To determine the 99th derivative of cos(x), examine the following table of derivatives:
Since the derivatives repeat every 4th one, simply divide by 99
and look at the remainder (R = 3).
So, the 99th derivative of cos(x) will be the same as the 3rd derivative.
Set sin(x) = 0
Therefore, x = 0 and y = 1
#13. The tangent to a curve has the same slope as the curve at the point of tangency.
So take the first derivative: y' = 8x - 22
The slope at x = 3 is m = 24 - 22 = 2
Now use the slope and point to find the equation of the tangent:
y - 5 = 2 (x - 3)
So, Kronecker sits on the line y = 2x - 1, and the points
in our domain that fall on this line are (1, 1), (0, -1), and (2, 3).
#14. For the point of inflection, take the second derivative and set = 0.
y' = 3x2 - 12x + 33
y'' = 6x - 12
So, x = 2.
Check concavity to be certain that it changes:
At x = 1, y'' = -6 so it is concave down,
and at x = 3, y'' = +6, so it is concave up.
So, Fermat sits at the point (2, -1).
#15. To find critical points, take the first derivative and set = 0.
y' = -12x3 + 12x
Dividing by -12, we obtain x3 - x = 0
x (x - 1) (x + 1) = 0
So, x = 0, -1, +1
Descartes must sit at one of the points (0, 0), (-1, 3), or (1, 3).
#16. First determine the maximum point of the curve.
Note that the curve is a parabola opening down.
The derivative y' = -2x + 10.
So, the maximum point occurs at x = 5 and y = 0.
Therefore the tangent line is the x-axis or y = 0.
So, Cantor sits at (-1, 0), (-2, 0), (0, 0), (1, 0), or (2, 0).
#17. To check for the absolute maximum over an interval, first find any relative maximums,
and then check those y-coordinates with the y-coordinates of the endpoints of the interval.
So, Gauss sits at the point (-1, 3).
#18. Since Gauss sits at (-1, 3) and Kronecker at (1, 1), Viete must sit at either (0, 2) or (2, 0).
(The slope is -1)
#19. To find the point of inflection, take the second derivative and set = 0.
y' = 3x2 - 6x + 3
y'' = 6x - 6
So, x = 1 and y = 2.
Check the concavity on either side of x = 1.
Therefore, Heron is located at (1, 2).
#20. First, find the point of inflection by taking the second derivative.
12y' = -12x - 3x2
12y'' = -12 - 6x
0 = -12 - 6x
x = -2
After checking concavity, we see that (-2, 0) is the point of inflection.
The slope is given by the first derivative: y' = -x -x2/4
So, the slope of the tangent at (-2, 0) is m = 1
The equation of the tangent at this point is given by:
y - 0 = 1 (x + 2)
The points in our domain that fall on the line y = x + 2 are:
(-2, 0), (-1, 1), (0, 2), and (1, 3).
Many thanks to Vivian Robert of the Korea International School in South Korea for correcting my error in #12 above.
Many thanks to Philip Chen of the William Lyon Mackenzie Collegiate Institute in Toronto, Canada for correcting my error in #20 above.