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Calculus and Analytic Geometry I: Exam II
Math 163 Section 51
November 7, 2002
Name: Signature:
- Questions 1–3 do not require any explanations. However, show all of your work. Partial credit will be very limited.
- Question 4 requires explanations written in complete sentences. Question 5 is an essay question. The marks which you earn on these questions will depend upon the quality of your writing. Answers which are unclear or which contain grammatical errors will not earn full credit.
- (15 pts.) Below is a table of values for the functions m, m′, n, and n′.
z − 1 0 1 2 3 4 5 6 m(z) 10. 1 9. 8 8. 1 6. 0 3. 3 0. 6 − 1. 5 − 1. 3 m′(z) − 0. 2 − 0. 9 − 1. 7 − 2. 0 − 2. 6 − 2. 6 − 2. 3 − 1. 1 n(z) 7. 7 3. 2 0. 2 − 1. 4 − 2. 0 − 1. 7 − 1. 0 0. 3 n′(z) − 5. 4 − 3. 7 − 2. 2 − 1. 0 0. 0 0. 2 0. 8 1. 4
(a) Let f (z) = 3m(z) −
n(z) 2
. Find f ′(1).
f ′(1) = 3m′(1) −
n′(1) 2
(b) Let g(z) = m(n(z)). Find g′(5).
g′(5) = m′(n(5))n′(5) = m′(− 1 .0)n′(5) = (− 0 .2)(0.8) = − 0. 16
(c) Let h(z) = m(z)/n(z). Find h′(0).
h′(0) =
m′(0)n(0) − m(0)n′(0) (n(0))^2
(3.2)^2
- (10 pts.) In celebration of the 150th anniversary of the founding of the town of Dandelion Creek, the city council of Dandelion Creek has constructed a large cylindrical tower made out of Jell-O in the Dandelion Creek’s central square. The base of the tower is a circle with a radius of 8 feet. The height of the tower is 50 feet. Over time, the Jell-O tower has gradually oozed, while retaining its cylindrical shape. The height is decreasing by 3 feet every day. Assuming that the volume of Jell-O remains constant, at what rate is the radius of the tower increasing?
Let V denote the volume of the Dandelion Creek tower (in cubic feet), r the radius of the base (in feet), and h the height of the tower (in feet). We can then write
V = πr^2 h.
Let t indicate time in days. Then differentiating everything in sight with respect to t, we get dV dt
= 2πrh
dr dt
dh dt
Since the volume of the tower stays constant, dV /dt = 0. We also know that dh/dt = − 3. Thus, we know the values of all of the quantities in the above equation except for dr/dt. We hence have that
0 = 2π(8)(50)
dr dt
− π(8)^2 (−3)
or 0 = 800π
dr dt
− 192 π.
Solving for dr/dt, we get
dr dt
= 0. 24 feet per day.
- (10 pts.) The motion of a particle at a time t is given by x = f (t) and y = g(t). To the right are graphs of f and g.
(a) Sketch the trajectory of the particle on the xy-plane.
x = f (t) y = g(t)
(b) Estimate the speed of the particle at t = 6.
The speed is (^) √ (f ′(6))^2 + (g′(6))^2 ≈
(−2)^2 + 1^2 ≈ 2. 24.
Note that we estimate f ′(6) and g′(6) graphically by examining the slopes of the appropriate tangent lines.
- (20 pt.) Let P be a function whose derivative P ′^ is a decreasing concave down function and which satisfies P ′(2) = 5. What can you say about P? What can you say about the critical points and inflection points of P? Analyze the properties of P in a well-organized essay. Include formulas, graphs, and tables where appropriate.
Solutions may vary. However, we can ascertain the following properties of P :
- Since P ′^ is decreasing, P must be concave down
- The function P has one critical point which is a global maximum. The global maximum must occur at some point greater than 2. Note that the function P ′^ must have a root because it is concave down. A possible graph of P ′^ is shown below.
Graph of P ′
- Since P ′^ has no local extrema, P has no inflection points.
Good solutions should...
... describe properties of P which follow from the given criteria ... clearly explain why the properties of P must follow from the above criteria. ... be well-organized, with few grammatical and spelling errors. ... utilize a combination of graphical, numerical, and algebraic methods in the explanation.
