AP Calculus BC: 7.8 Exponential Models with Differential  Equations – Exam Style questions with Answer- FRQ

Question

t (minutes)0123456
C(t) (ounces)05.38.811.212.813.814.5

Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, 0 ≤ t ≤ 6, is given by a differentiable function C, where t is measured in minutes. Selected values of C(t), measured in ounces, are given in the table above.
(a) Use the data in the table to approximate C'(3.5). Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, 2 ≤ t ≤ 4 at which C'(t) = 2 ? Justify your answer.
(c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of \(\frac{1}{6}\int_{0}^{6}C(t)dt.\) Using correct units, explain the meaning of \(\frac{1}{6}\int_{0}^{6}C(t)dt\) in the context of the problem.
(d) The amount of coffee in the cup, in ounces, is modeled by B(t) = 16 – 16e-0.4t. Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5.

Answer/Explanation

Ans:

(a) \(C'(3.5)\approx \frac{C(4)-C(3)}{4-3}=\frac{12.8-11.2}{1}=1.6 ounces/min\)

(b) C is differentiable ⇒ C is continuous (on the closed interval) \(\frac{C(4)-C(2)}{4-2}=\frac{12.8-8.8}{2}=2\)

Therefore, by the Mean Value Theorem, there is at least
one time t, 2 < t < 4,  for which C'(t)=2.

(c) \(\frac{1}{6}\int_{0}^{6}C(t)dt\approx \frac{1}{6}\left [ 2\cdot C(1)+2\cdot C(3)+2\cdot C(5) \right ]\)

\(= \frac{1}{6}(2\cdot 5.3+2\cdot 11.2+2\cdot 13.8)\)

\(= \frac{1}{6}(60.6)= 10.1 ounces\)

\(\frac{1}{6}\int_{0}^{6}C(t)dt\) is the average amount of coffee in the cup, in ounces, over the time interval 0 ≤ t ≤ 6  minutes. 

(d) \(B'(t)=-16(-0.4)e^{-0.4t}=6.4e^{-0.4t}\)

\(B'(5)=6.4e^{-0.4(5)}=\frac{6.4}{e^{2}}ounces/min\)

Question

 This problem is solved without a calculator. The slope of a function f at any given point (x, y) is \(\frac{2y}{3x^{2}}\) . The point (3, 4) is on the graph of f.
(A) Write an equation of the tangent line to the graph of f at x = 3.
(B) Use the tangent line in part (a) to approximate f (5). Is this approximation an overestimate or underestimate? Justify your answer using the second derivative test.
(C) Solve the separable differential equation \( \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\) with initial condition f(3) = 4.
(D) Use the solution in part (c) to find f(5).

Answer/Explanation

(A) The slope at x=3 (i.e.,the point (3,4)) is
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(4)}{3(3)^{2}}=\frac{8}{27}\)
Now write the equation of the tangent line at (3, 4) in point-slope form.
\(y-y_{0}=m(x-x_{0})\Rightarrow y-4=\frac{8}{27}(x-3)\Rightarrow y=\frac{8}{27}(x-3)+4\Rightarrow y=\frac{8}{27}x+\frac{28}{9}\)
(B)\(f(5)=\frac{8}{27}(5-3)+4=\frac{124}{27}\) This is an overestimate because the graph is concavedown over the interval (3,5) with a negative second derivative of \(\left ( (3x^{2})y’-6x(2y) \right )/(3x^{2})^{2}\)
(C) Start by separating the variables and integrating both sides
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\Rightarrow \frac{1}{2y}dy=\frac{1}{3x^{2}}dx\Rightarrow \int \frac{1}{2y}dy=\int \frac{1}{3x^{2}}dx
\frac{1}{2}\int \frac{1}{y}dy=\frac{1}{3}\int x^{-2}dx\Rightarrow \frac{1}{2}\ln |y|=\left ( \frac{1}{3} \right )\frac{x^{-1}}{-1}+c_{1}\Rightarrow \frac{1}{2}\ln \left | y \right |=-\frac{1}{3x}+c_{1}
\ln \left | y \right |=-\frac{2}{3x}+c_{2}\Rightarrow e^{\ln \left | y \right |}=e^{-\frac{2}{3x}+c_{2}}=e^{-\frac{2}{3x}.e^{c_{2}}}\)
\(y=c_{3}e^{-\frac{2}{3x}}\)
Now use the initial condition f(3) = 4 to solve for \(c_{3}\).
\(4=c_{3}e^{-\frac{2}{3(3)}}\Rightarrow c_{3}=4.9954\)
\(y=4.9954e^{-\frac{2}{3x}}\)
(D) Evaluate the equation in C at f(5).
\(f(5)=4.9954e^{-\frac{2}{3(5)}}\approx 4.37184\)

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 \(\int_{0}^{10}H'(t)dt.\) Using correct units, explain the meaning of the expression in the context of this problem.
(d) At time t = 0, biscuits with temperature 100 0C 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)

\(\frac{1}{10}\int_{0}^{10}H(t)dt\approx \frac{\left [2(\frac{66+180}{2}) +3(\frac{52+60}{2})+4(\frac{44+52}{2})+1(\frac{43+44}{2}) \right ]}{10}=52.95\)

This represents the average temperature in degree 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)

\(B'(t)=-13.84e^{-.173t}\)

\(B(10)=\int_{0}^{10}-13.84e^{-.173t}+100 = 100-65.817 = 34.1827\)

43-341827 = 8.817

= 88170C Celsius 

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