Question 4
(a) Topic-5.9- Connecting a Function, Its First Derivative, and Its Second Derivative
(b) Topic- 1.16- Working with the Intermediate Value Theorem (IVT)
(c) Topic-6.3- Riemann Sums, Summation Notation, and Definite Integral Notation
(d) Topic- 4.5- Solving Related Rates Problems
4. An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function r, where r t( t ) is measured in centimeters and t is measured in days. The table above gives selected values of r'(t), the rate of change of the radius, over the time interval \(0\leq t\leq 12\).
(a) Approximate r “(8.5) using the average rate of change of r’ over the interval \(7\leq t\leq 10\). Show the computations that lead to your answer, and indicate units of measure.
(b) Is there a time t, \(0\leq t\leq 3\), for which r ‘(t ) = −6 ? Justify your answer.
(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of \(\int_{0}^{12}r'(t)dt\).
(d) The height of the cone decreases at a rate of 2 centimeters per day. At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time t = 3 days. (The volume V of a cone with radius r and height h is \(V=\frac{1}{3}\pi r^{2}h .\)
▶️Answer/Explanation
4(a) Approximate r′′(8.5) using the average rate of change of r′ over the interval 7 ≤ t≤ 10 . Show the computations that lead to your answer, and indicate units of measure.
\(r”(8.5)\approx \frac{r'(10)-r'(7)}{10-7}=\frac{-3.8-(-4.4)}{10-7}\)
\(=\frac{0.6}{3}=0.2\) centimeter per day per day
4(b) Is there a time t, 0 ≤ t≤ 3, for which r ′(t ) = −6? Justify your answer.
r (t ) is twice-differentiable. ⇒ r′( t) is differentiable.
⇒ r′( t) is continuous.
r′(0 ) = −6.1 < −6 < −5.0 = r′(3 )
Therefore, by the Intermediate Value Theorem, there is a time t, 0 < t< 3, such that r′( t) = −6.
4(c) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of
\(\int_{0}^{12} r′(t) dt\).
\(\int_{0}^{12} r′(t) dt ≈ 3r′(3) + 4r′(7) + 3r′(10) + 2r′(12) \)
= 3(−5.0) + 4(−4.4) + 3(−3.8) + 2(−3.5)
= −51
4(d) The height of the cone decreases at a rate of 2 centimeters per day. At time t = 3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with
respect to time, in cubic centimeters per day, at time t = 3 days. (The volume V of a cone with radius r and height h is \(V=\frac{1}{3}\pi r^{2}h\) .)
\(\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^{2}\frac{Dh}{dt}\)
\(\frac{dV}{dt}|_{t=3}=\frac{2}{3}\pi (100)(50)(-5)+\frac{1}{3}\pi (100)^{2}(-2)=-\frac{70,000\pi }{3}\)
The rate of change of the volume of the sculpture at t = 3 is approximately \(-\frac{70,000\pi }{3}\)cubic centimeters per day.