FRQs Set #1
Rates of Change, Estimating Instantaneous Change, Using
the IVT, and Introduction to Derivatives
Tips:
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Every FRQ is worth a total of 9 points
Correct units must be used
All answers must have either calculus and/or verbal justification.
All numeric answers must be correct up to 3 significant figures to the right of the
decimal point.
Most FRQs award majority of points for process/justification over single numeric
answers.
FRQ1 – Estimating Instantaneous Change
Calculator Active
The velocity of a particle moving along the x-axis is given by the following graph and
table, where t is the time in seconds and v(t) is the velocity in centimeters per second.
a. Estimate the rate of change in the velocity (i.e. the acceleration) of the particle
when t = 30. Indicate the units.
b. Use your answer from part (a) to approximate the velocity of the particle
when t = 31. Indicate the units.
c. Does the particle ever change direction? If so, at what time(s)? Justify your
answer.
d. Mike Boardman concludes that since the slope of the secant line joining the
point (0, 1) to the point (30, -1) is negative, the particle must have been
decelerating over the entire interval from t = 0 to t = 30. Is Mike correct? Give
reasons for your answer drawn from the graph or the table.
FRQ 2 –IVT and Linear Approximation
(Calculator Active)
A marble is released on an inclined table. The distance the marble rolls over a given
time interval is then measured. The table below gives various values of s(t), the
distance the marble is away from where it was released (measured in inches) as a
function of time t, measured in seconds from the moment the marble was released.
t
s(t)
0
0
2
1.5
4
6
6
13.5
8
24
10
37.5
a. Find the average rate at which the marble is rolling during the time interval t =
4 to t = 10. Show the computations that lead to your answer. Indicate units of
measure.
b. Estimate the instantaneous rate at which the marble is rolling at t = 6 seconds.
Show the computations that lead to your answer. Indicate units of measure.
c. Must there be a time when the marble has rolled exactly 10 inches? Justify your
answer.
d. The data in the table for s(t) is modeled by the function g(t) = 0.375t2. Use the
definition of the derivative to find the equation of the tangent line to g(t) at t =
6 seconds.
e. Use your answer from part (d) to estimate the distance the marble will have
rolled at t = 6.2 seconds. Indicate units of measure.
FRQ 3 – IVT and Continuity
Calculator Active
This is a calculator active question.
Given the function g(x) defined so that:
a. Verify that if k = 5, the Intermediate Value Theorem applies to g(x) on the
closed interval [2, 3], and predicts that g(x)has at least one root on that interval.
Find that root.
b. Find k so that g(x) is a continuous function.
c. For the value of k found in part (b), evaluate
. If it does not exist, so
state and explain why.
d. For the value of k found in part (b), evaluate g ' (1). If it does not exist, so state
and explain why.
FRQ 1 Standard:
FRQ 2 Standard:
a. (37.5 - 6)/(10 - 4) = 5.25 inches/second
1pt: answer based
on evidence
b. (24 - 6)/(8 - 4) = 4.5 inches/second OR (24 - 13.5)/(8 - 6) = 5.25
1pt: answer based
inches/second OR (13.5 - 6)/(6 - 4) = 3.75 inches/second
on evidence
c. Yes. Since marbles roll in a continuous fashion, s(t) must be a continuous 1pt: answer
function on [0, 10]. Therefore, the IVT applies. Since s(4) < 10 and s(6) > 10, 1pt: appeal to the
by the IVT there must be a value of c on [4, 6] where s(c) = 10.
IVT
d. g(6) = 13.5.
1pt: limit of
difference quotient
1pt: g'(6)
1pt: equation
Tangent line: y - 13.5 = 4.5(t - 6)
e. Distance ≈4.5 (6.2 - 6) +13.5 = 14.4 inches
1pt: answer
1pt: units in parts
a, b, and e.
FRQ 3 Standard:
a.If k = 5, g(x) = 5 - 2x on [2, 3]. Therefore,
1 pt. states g(x) is continous on the interval
g(x) is continuous on [2, 3], so the IVT applies.
Since g(2) = 5 - 2(2) = 1 > 0 and g(3) = 5 - 2(3)
1 pt. states g(a) > 0 and g(b) < 0 for some a, b
= -1 < 0, by the IVT there must be at least one
on [2, 3]
root of g(x) on [2, 3].
5 - 2x = 0, so x = 5/2.
b. 12 + 2k = k - 2(1)
k = -3
c. 12 + 2(-3) = -5
d. g ' (x) = 2x for x < 1 and g ' (x) = -2 for x ≥ 1.
1 pt. solves g(t) = 0
1 pt. sets two branches of g at x = 1 equal
1 pt. solves for k
1 pt. answer
1 pt: conclusion
Since
, g'(1) 1 pt: evaluates
does not exist because the left- and right-handed
limits of the difference quotient disagree.
1 pt: evaluates