APPM 1360 EXAM 1 SOLUTIONS FALL 2008
INSTRUCTIONS: Write your name, your instructor’s name, your recitation number, and a grading
table on the front of your bluebook. Start each problem on a new right-hand page. Justify your work
clearly and box your final answer. A correct answer with incorrect or no supporting work may receive no
credit, while an incorrect answer with relevant work may receive partial credit. Text books, class notes,
and crib sheets NOT permitted. Electronic devices may not be used during the exam.
1. (15 points) Compute dy
dx for the following functions and simplify your answers.
(a) y= tanh−1(sin x)
dy
dx =cos x
1−sin2x=cos x
cos2x= sec x
(b) y=xsinh−1(x)−p1 + x2
dy
dx = sinh−1(x) + x
√1 + x2−2x
2√1 + x2= sinh−1(x)
2. (15 points) Evaluate the following integrals and simplify your answers.
(a) Zdx
√5x2−3=1
√5Zdx
px2−(3/5) =1
√5cosh−1 x
p3/5!+C=1
√5cosh−1 r5
3x!+C
(b) Zdx
x2+ 8x+ 25 =Zdx
x2+ 2 ·4x+ 16 −16 + 25 =Zdx
(x+ 4)2+ 9 =1
3tan−1x+ 4
3+C
3. (25 points) Consider the solid generated by revolving about the y-axis, the region in the first quadrant
bounded by y= cos(x2) for 0 ≤x≤rπ
2.
(a) Calculate the volume of the object. Using shells, we get...
V=Z√π/2
x=0
2π x cos(x2)dx =πZπ/2
u=0
cos(u)du =π
(b) Set up, but do not evaluate, the integral calculations required to determine the surface area of
the top of the object.
SA =Z2π r ds =Z√π /2
x=0
2π x s1 + dy
dx2
dx where dy
dx =d
dx cos(x2) = −sin(x2)2x
4. (25 points) Consider a thin plate of uniform thickness t, density ρ=1
1 + x, with a shape defined by
the region in the fourth quadrant above the curve by y=x2−1 and below the x-axis.
(a) Clearly graph (not sketch) the region described. Be sure to find and label all intersection points.
-1.5 -1 -0.5 0 0.5 1 1.5
-1
-0.5
0.5
(b) Calculate the area of one side of the plate (the area of the region).
A=Z1
x=0
[0 −(x2−1)] dx =Z1
x=0
(1 −x2)dx = 2/3
