AP Calculus AB: 2.1 Defining Average and Instantaneous Rates of Change at a Point – Exam Style questions with Answer- FRQ

Question

The figure above shows the polar curves r = f (θ) q = 1+ sin θ cos  (2θ) and r = g(θ)  = 2 cos θ for \(\(0 \leq \theta \leq \frac{\pi }{2}.\)\)  Let R be the region in the first quadrant bounded by the curve r = f (θ)  and the x-axis. Let S be the region in the first quadrant bounded by the curve r = f (θ) , the curve r = g (θ), and the x-axis.
(a) Find the area of R.
(b) The ray θ = k, where \(0 < k < \frac{\pi }{2},\) divides S into two regions of equal area. Write, but do not solve, an equation involving one or more integrals whose solution gives the value of k.
(c) For each θ, \(0 \leq \theta \leq \frac{\pi }{2},\) let w(θ) be the distance between the points with polar coordinates (f(θ), θ) and (g(θ), θ). Write an expression for w(θ). Find wA,  the average value of w(θ) over the interval \(0 \leq \theta \leq \frac{\pi }{2}.\) 
(d) Using the information from part (c), find the value of θ for which w(θ) = wA. Is the function w (θ) increasing or decreasing at that value of θ ? Give a reason for your answer.

▶️Answer/Explanation

Ans:

(a) \(\frac{1}{2}\int_{0}^{\pi /2}(f(\theta ))^{2}d\theta =0.648414\)
The area of R is 0.648.

(b) \(\int_{0}^{k}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta =\frac{1}{2}\int_{0}^{\pi /2}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta\)

-OR-

\(\int_{0}^{k}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta =\int_{0}^{\pi /2}(g(\theta ))^{2}-(f(\theta ))^{2}d\theta\)

(c) w(θ) = g(θ) – f(θ)

\(w_{A}=\frac{\int_{0}^{\pi /2}w(\theta )d\theta }{\frac{\pi }{2}-0}=0.485446\)

The average value of w(θ ) on the interval \(\left [ 0, \frac{\pi }{2} \right ]\) is 0.485.

(d) \(w(\theta )=w_{A}for 0\leq \theta \leq \frac{\pi }{2}\Rightarrow \theta =0.517688\)

w(θ) = wA at θ = 0.518 (or 0.517).

w'(0.518) < 0 ⇒ w(θ) is decreasing at θ = 0.518.

Question

Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A(t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A(t) over the interval 0 ≤ t ≤ 30. Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30 .
(d) For t >30, L (t), the linear approximation to A at t =30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

▶️Answer/Explanation

Ans:

(a) \(\frac{A(30)-A(0)}{30-0}=-0.197 (or-0.196) lbs/day)\)

(b) A'(15) = -0.164 (or – 0.163)

The amount of grass clippings in the bin is decreasing at a rate of 0.164 (or 0.163) lbs/day at time t = 15 days.

(c) \(A(t)=\frac{1}{30}\int_{0}^{30}A(t)dt\Rightarrow t=12.415 (or 12.414)\)

(d) L(t) = A(30) + A'(30) • (t-30)

A'(30) = – 0.055976

A(30) = 0.782928

L(t) = 0.5 ⇒ t = 35.054

 Question:

t
(minutes)
025910

H(t) 
(degrees Celsius)

6660524443

 

As a pot of tea cools, the temperature of the tea is modeled by a differentiable function H for 0 ≤ t ≤ 10, where time t is measured in minutes and temperature H(t) is measured in degrees Celsius. Values of H(t) at selected values of time t are shown in the table above.
(a) Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t = 3.5. Show the computations that lead to your answer.
(b) Using correct units, explain the meaning of \(\frac{1}{10}\int_{0}^{10}H(t)dt\) in the context of this problem. Use a trapezoidal  sum with the four subintervals indicated by the table to estimate \(\frac{1}{10}\int_{0}^{10}H(t)dt\).
(c) Evaluate \(\frac{1}{10}\int_{0}^{10}H'(t)dt\). Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 C∞ were removed from an oven. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that \(B'(t)=-13.84e^{-0.173t}.\) Using the given models, at time t = 10, how much cooler are the biscuits than the tea?

▶️Answer/Explanation

Ans:

(a) \(\frac{H(5)-H(2)}{5-2}=\frac{-8}{3}\frac{0_{C}}{min}\)

(b) \(\frac{1}{10}\int_{0}^{10}H(t)dt \approx \frac{\left [ 2\left ( \frac{66+60}{2} \right )+3\left ( \frac{52+60}{2} \right ) +4\left ( \frac{44+52}{2} \right )+1\left ( \frac{43+44}{2} \right )\right ]}{10}=52.95\)

This represents the average temperature in degrees celsius of the Tea over the interval 0 ≤ t ≤ 10

(c)

\(\int_{0}^{10}H'(t)dt = H(10)-H(0)=43-66 = -23^{0}C\)

This expression shows the total change in temperature in degree Celsius from t = 0 to t = 10.

(d)

\( B'(t)= -13.84e^{-1.73t}\)

\(B(10)= \int_{0}^{10}-13.84e^{-1.73t}+100 = 100-65.817 = 34.1827\)

43 – 34.1827 = 8.817

= 8.817 0

Question:

The continuous function f is defined on the interval -4 ≤  x ≤ 3. The graph of f consists of two quarter circles and one line segment, as shown in the figure above. Let \(g(x)=2x+\int_0^x f(t)dt\).

(a) Find g(-3) . Find g'(x) and evaluate g'(-3) .
(b) Determine the x-coordinate of the point at which g has an absolute maximum on the interval -4 ≤  x ≤ 3. Justify your answer.
(c) Find all values of x on the interval -4 ≤  x ≤ 3 for which the graph of g has a point of inflection. Give a reason for your answer.
(d) Find the average rate of change of f on the interval-4 ≤  x ≤ 3.  There is no point c, -4 < c < 3,  for which f'(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.

▶️Answer/Explanation

Ans:

(a)

\(g(-3)=2(-3)+\int_{0}^{-3}f(t)dt = -6 -\frac{9\pi }{4}\)

\(g'(x)=\frac{d}{dx}\left ( 2x + \int_{0}^{x}f(t)dt \right )=2+f(x)\)

g'(-3) = 2 + f(-3) = 2 + 0 = 2

(b)

g'(x) = 0                        2 + f(x) = 0 

g'(x)                              f(x) = -2

                                       x = 5/2

                                    \(g(-4)=-8+\int_{0}^{-4}f(t)dt\)

                                    = -8 – 2π

\(g(5/2)= 5+\int^{\frac{5}{2}}_{0} f(t)dt = + \frac{5}{4}\)

\(g(3)= 6+\int_{0}^{3}f(t)dt = 6\)

x = 5/2 , because g’   going from t to – proves it as the only relative maximum and g(5/2) is greater than g at either endpoint.   

(c)

f'(x)                          g”(x) = d/dx    (g(x)) = f'(x)

The only point of inflection for g is at x = 0, since f'(x), which is equivalent to g”, only changes signs at x = 0 on the interval -4 ≤ x ≤ 3

(d)

Avg. Rate of change  = \(\frac{f(3)-f(-4)}{3-4}\)

\(\frac{-3-1}{3+4}=\frac{-2}{7}\)

Because mean value Theorem only applies when the function is continuous And differentiable on the interval, which doesn’t apply here since f(x) isn’t differentiable at x = 0.

Question:

Let f be a function defined by \(\left\{\begin{matrix}
1-2 sin x & for x \leq 0 & & \\e^{-4x}
&for x > 0. & &
\end{matrix}\right.\)

(a) Show that f is continuous at x = 0.
(b) For x ≠ 0, express f'(x) as a piecewise-defined function. Find the value of x for which f'(x) = -3.
(c) Find the average value of f on the interval [-1, 1] .

▶️Answer/Explanation

Ans:

(a) 

To be continuous 

i) f(0) = 1

ii) \(\lim_{x\rightarrow 0} 1 -2 sin x = \lim_{x\rightarrow 0}e^{-4x}\)

1  = 1    ∴ \(\lim_{x\rightarrow 0^{-}} f = \lim_{x\rightarrow 0^{+}}f\)

 ∴ \(\lim_{x\rightarrow 0} f = 1\)

iii) f(0) = \(\lim_{x\rightarrow 0} f = 1\)

∴ f is continuous for all values of x.

(b)

\(f'(x)=\left\{\begin{matrix}
-2 cos x &, &x<0 & \\-4e^{-4x}
&, &x>0 &
\end{matrix}\right.\)

\({f}'(x) = 3\)

Since -2 cos x oscillates between -2 and 2 there will be no such value in this function such that f'(x) = -3

but \({f}'(x) =- 3\)

-4e -4x = -3

e -4x = 3/4

-4x = In (3/4)  \(\therefore = \frac{-1}{4}In\left ( \frac{3}{4} \right )\therefore f’\left ( \frac{-1}{4}In\left ( \frac{3}{4} \right ) \right )= -3\)

(c)

\(f_{avg} = \frac{1}{1+1}\int_{-1}^{1}f(x)dx\)

\(= \frac{1}{2}\left [ \int_{-1}^{0} 1-2 sinx dx + \int_{0}^{1}e^{-4x}dx\right ]\)

\(= \frac{1}{2}\left [ [x+2cosx]_{-1}^{0}+[\frac{-1}{4}e^{-4x}]_{0}^{1} \right ]\)

\(= \frac{1}{2}\left [ 2+1-2cos(-1)\frac{-e^{-4}}{4} +\frac{1}{4}\right ]\)

\(=\frac{-\left [ 3-2cos(-1) \right ]+\left [ \frac{-1}{4e^{4}} +\frac{1}{4}\right ]}{2}\)

Question:

Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A (t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A (t) over the interval 0 ≤ t ≤ 30 . Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30.
(d) For t > 30,  L(t), the linear approximation to A at t = 30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.

▶️Answer/Explanation

Ans:

(a)

\(\frac{A(30)-A(0)}{30-0}=\frac{-5.904}{30}\approx -.197\frac{pounds}{day}\)

(b)

A'(t) = – .478(.931)t

A'(15) = -.164    \(\frac{pounds}{day}\)

The amount of grass clippings in the bin is decreasing (decomposing) at a rate of .164 pounds per day at time = 15 days

(c)

Average amount \(=\frac{1}{30}\int_{0}^{30}A(t)dt = 2.75263511\)

A(t)  = 2.75263511 = 6.687 (.931)t

this occurs at t = 12.414 days

(d)

L(t) is the tangent line to A(t) at t = 30

A(30) = .783 ⇒ (30, .783) = (t, A(t))

A'(30) = -.056               let  -.056  = M

(y-y1) = m (x, x1) so for this problem, 

(A(t) – .783) = -.056(t-30)

When there are .5 pounds of grass, A(t) = .5, 

(.5-.783) = -.056(t-30)

t = 35.054 days

Question:

t

(minutes)

025812

vA (t)

(meters/ minute)

010040-120-150

Train A runs back and forth on an east-west section of railroad track. Train A’s velocity, measured in meters per minute, is given by a differentiable function vA (t)  where time t is measured in minutes. Selected values for vA (t) are given in the table above.
(a) Find the average acceleration of train A over the interval 2 ≤ t ≤ 8.
(b) Do the data in the table support the conclusion that train A’s velocity is -100 meters per minute at some time t with 5 < t < 8 ? Give a reason for your answer.
(c) At time t = 2, train A’s position is 300 meters east of the Origin Station, and the train is moving to the east. Write an expression involving an integral that gives the position of train A, in meters from the Origin Station, at time t = 12. Use a trapezoidal sum with three subintervals indicated by the table to approximate the position of the train at time t = 12.
(d) A second train, train B, travels north from the Origin Station. At time t the velocity of train B is given by vB (t) = -5t2 + 60t + 25, and at time t = 2 the train is 400 meters north of the station. Find the rate, in meters per minute, at which the distance between train A and train B is changing at time t = 2.

▶️Answer/Explanation

Ans:

(a)

\(\frac{v(8)-v(2)}{8-2}\rightarrow \frac{-120-100}{6}\rightarrow \frac{-220}{6}\rightarrow \frac{-110}{3}m/min^{2}\)

(b)

Yes, because v(8) = -120 and v(5) = 40   and the function is differentiable and this continuous the train’s velocity must be -100 m/min at some point between 5 < t  < 8 accordingly to the intermediate value theorem.

(c)

\(x(12)= \int_{2}^{12}{v_{A}}^{(t)}dt + x(2)\rightarrow x(12)=\int_{2}^{12}{v_{A}}^{(t)}dt +300\)

\(x(12)\approx 3\cdot \frac{1}{2}\cdot (140)+3\cdot \frac{1}{2}\cdot (-80)+4\cdot \frac{1}{2}\cdot (-270)+300\)

210 – 120 – 540 + 300

210 – 240 – 120

-30-120

150 meaning it is 150 m W of origin station 

(d)

\(\frac{dB}{dt}= -5t^{2}+60t + 25 \rightarrow 125 m/min\)

-20 + 120 +25

120 + 5 → 135

\(\frac{dA}{dt}= 100 m/min\)

A2 + B2 + = D2

\(2A\frac{dA}{dt}+2B\frac{dB}{dt}=2D\frac{dD}{dt}\)

600(100) + 800 (125) = 1000 \(\frac{dD}{dt}\)

6000 + 100000 = 1000 \(\frac{dD}{dt}\)

\(\frac{160,000}{1000}=\frac{dD}{dt}\rightarrow \frac{dD}{dt}=160 m/min\)

 Question:

t
(minutes)
012202440
v (t)
(meters per minute)
0200240-220150

Johanna jogs along a straight path. For 0 ≤ t ≤ 40, Johanna’s velocity is given by a differentiable function v.
Selected values of v (t), where t is measured in minutes and v (t) is measured in meters per minute, are given in the table above.
(a) Use the data in the table to estimate the value of v'(16).
(b) Using correct units, explain the meaning of the definite integral \(\int_{0}^{40}|v(t)|dt\) in the context of the problem.
Approximate the value of \(\int_{0}^{40}|v(t)|dt\) using a right Riemann sum with the four subintervals indicated in the table.
(c) Bob is riding his bicycle along the same path. For 0 ≤ t ≤ 10, Bob’s velocity is modeled by B(t) = t3 – 6t + 300, where t is measured in minutes and B (t) is measured in meters per minute.
Find Bob’s acceleration at time t = 5.
(d) Based on the model B from part (c), find Bob’s average velocity during the interval0 ≤ t ≤ 10.

▶️Answer/Explanation

Ans:

(a)

\(V'(14)\approx \frac{v(20)-v(12)}{20-12}=\frac{240-200}{8}=\frac{40}{8}=\frac{20}{4}=5\frac{m}{min^{2}}\)

(b)

\(\int_{0}^{40}|v(t)|dt\) represents the total distance in meters Johanna traveled between times. t = 0 and t = 40 minutes.

\(\int_{0}^{40}|v(t)|dt\approx \left [ 12(200)+8(240)+4(220)+16(150) \right ]\)

= 2400 + 1920 + 880 + 2400

= 7600 meters

(c)

acceleration  = B'(t) = 3t2 – 12t

B'(5) = 3(5)2 – 12(5)

= 75-60 = 15 \(\frac{m}{min^{2}}\)

(d)

Avg. Velocity = \(\frac{1}{10}\int_{0}^{10}(t^{3}-6t^{2}+300)dt\)

\(=\frac{1}{10}\cdot \left [ \frac{t^{4}}{4} -2t^{3}+300t\right ]_{0}^{10}\)

\(=\frac{1}{10}\cdot \left [ \frac{10000}{4} -2000+300t\right ]\)

\(=\frac{1}{10}\left [ 3500\right ]=350\frac{m}{min}\)

 Question:

t
(hours)
01368
R (t)
(liters / hour)
13401190950740700

Water is pumped into a tank at a rate modeled by \(W(t)= 2000e^{-t^{2}/20}\) liters per hour for 0 ≤ t ≤ 8, where t is measured in hours. Water is removed from the tank at a rate modeled by R (t) liters per hour, where R is differentiable and decreasing on 0 ≤ t ≤ 8. Selected values of R (t) are shown in the table above. At time t 0, there are 50,000 liters of water in the tank.
(a) Estimate R'(2). Show the work that leads to your answer. Indicate units of measure.
(b) Use a left Riemann sum with the four subintervals indicated by the table to estimate the total amount of water removed from the tank during the 8 hours. Is this an overestimate or an underestimate of the total amount of water removed? Give a reason for your answer.
(c) Use your answer from part (b) to find an estimate of the total amount of water in the tank, to the nearest liter, at the end of 8 hours.
(d) For 0 ≤ t ≤ 8, is there a time t when the rate at which water is pumped into the tank is the same as the rate at which water is removed from the tank? Explain why or why not.

▶️Answer/Explanation

Ans:

(a)

\(R'(2)\approx \frac{R(3)-R(1)}{3-1}=\frac{950-1190}{2}\)

R'(2) ≈ -120 L/hr2

(b)

\(\int_{0}^{8}R(t)dt\approx L_{4}=(1)(1340)+(2)(1190)+(3)(950)+(2)(740)\)

\(\int_{0}^{8}R(t)dt\approx 8,050 L\)

This is an underestimate because we are taking the left Riemann sum of a decreasing function. Therefore the left endpoints are greater than the right endpoints in each subinterval. 

(c)

t = + 50, 000 L

Total left ≈ 50,000 + \(\left ( \int_{0}^{8}W(t)dt – \int_{0}^{8}R(t)dt \right )\)

\(\int_{0}^{8}W(t)dt = 7836.19532455\)

Total left ≈ 50,000 + (7836.19332455-8050)

Total left ≈  49,786 L

(d)

T(t) = W(t) – R(t)

T(0) = 2000 – 1340 = 600                                                   -648.47 < 0 < 600

T(8) = 81.524-700 = -618.47

When T(t) = 0, the rate of water being pumped in to the tank is equal to the rate of water being pumped out. Because the function T(t) is continuous and differentiable by IVT these exists some t on the interval [0, 8] where T(t) = 0.

Scroll to Top