Non-Calculator Active Questions
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Revised 2008 |
| 1 We need to realize that question is referencing the definition of a
derivative at a point.
So we are looking for the derivative of the cosine function at the point
cos'(x)=sin(x)
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ANSWER A |
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2. We are looking for a portion of the graph where
 . This means
the slope must be negative and the concavity must be negative. This
would occur at point B. |
Answer B |
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| 3. Differentiating
implicitly and evaluate at (2,1):
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Answer C |
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4. Using the disk method: The
inside radius is 3-2x and the larger radius is 3-2x2.
Therefore the area of each cross section is
 and  .
Therefore the volume is found by completing the integral:
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Answer A |
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5.
 From the
first derivative we know that the sign of f' changes once at x = 3/2
From the second derivative we know that the sign of f" changes sign at x
= 0 and x = 1.
Therefore there is one relative extrema and two points of inflections.
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Answer C |
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| 6. 
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Answer B |
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| 7.
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Answer C |
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| 8.
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Answer C |
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9.
particle slow down from (0,3), momentarily stops at t=3, and speeds up
from (3,4). In the interval (4,5) the particle slows down and comes to
a momentary stop at t = 5. The particle then speeds up from
 .
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Answer E |
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| 10. 
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Answer D |
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| 11.
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Answer B |
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| 12.
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Answer B |
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| 13. Using the Chain Rule and the Second Fundamental
Theorem: 
In addition we know that
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Answer E |
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| 14. The correct slope field is E because when x=y
the slope is always 1. This is the only graph with this feature.
In addition when x and y are in the first and third quadrants, the slope
must be positive. When x and y are in the second and fourth
quadrants, the slope must be negative. There must be zero slope when x
= 0. |
Answer E |
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Calculator Active Questions
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| 15. Graph of the
velocity between [0,2]
V(1.5707)=0
Find the area under the absolute value of v(t) from (0,2) using numerical
integration:
Alternatively the distance traveled can be found on the graph:
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Answer D |
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16. Think of this problem as accumulating the area
around the lake. The area of the strip can be found by multiplying the
circumference ( ) by dr. This will
be the area which must be multiplied by the density f(r). We are only
considering the radius from r = 1 to r = 2.
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Answer D |
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| 17.
If the  exists
and f is not continuous at b. First look at x = 0 and x = 2. At
x = 0 there is a limit, but the function f is not continuous there. At
x = 2 the function has no limit and the function is not continuous there.
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Answer B |
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| 18. Based on the chart of information we can
see that the average rate of change of f on the intervals are
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(1.1,1.2) |
(1.2,1.3) |
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Therefore 1.8 < f '(1.2) < 2.0
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Answer D |
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| 19. Looking at the velocity graph for each
function yields the following graph. We can see there are 3 locations
where the velocity will be equal in the interval [0, 10]
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Answer D |
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20.
 so we are looking
for the area bounded by the f function between 0 and -1 and the t axis.
This is a triangular shape region. This area is positive, but since we
are integrating right to left the sign on the area is switched to a
negative.
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Answer B |
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| 21.Analyzing each picture it is clear that in A, B, and D
will have a zero average value because of the symmetry of the graphs about
the origin. Graph C will also have a zero average value because of the
area above and below the axes is equal. Therefore E is the only graph
whose average value will not be zero. It will be negative because
there is more negative area than positive area. |
Answer E |
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| 22. From the slope and the point the tangent line
would be y=4(x-5)+3 therefore y(4.5)=4(4.8-5)+3 or 2.2 |
Answer A |
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23.
To do numerical integration enter the Rate equation in Y1 and integrate
between 0 and 10:
Rounding to the nearest gallon is 8647 gallons. |
Answer D |
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| 24. Entering the function f '(x) in Y1 we can create a
function f using the Second Fundamental Theorem of Calculus:
Therefore using numerical integration we can find f(2)
Place f ' (x) in Y1 and then using numerical integration from 0 to 2.
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Answer E |
