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Math 112 Exam I Solutions, Exams of Calculus

The solutions to exam i of math 112. It includes the steps to find the equation of the tangent line, the derivatives of arctan(x/3), the limits of certain functions, the setup of definite integrals representing volumes of solids, and the evaluation of integrals using integration by parts and other techniques.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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February 18, 2004
Math 112 Exam I
Show all work clearly; an answer with no justifying computations may not receive credit (except
in the “set up but do not evaluate” problems).
1. (15 points) (a) Find the equation of the tangent line to the graph of x2+ 2xy y2+ 4x= 5
at the point (1,2).
(b) Find d
dx (arctan(x/3)).
2. (21 points) Find the following limits:
(a) lim
xπ/2
(xπ
2) tan x(b) lim
x2
x22x+ 3
x2+ 3 (c) lim
x1x1/(x1)
3. (18 points) Set up but do not evaluate a definite integral that represents the volume of each
of the following solids:
(a) The base is the region bounded by the curves y=x3and x=y2, and the cross sections
perpendicular to the x-axis are right triangles with height 2.
(b) The solid of rotation generated by rotating about the x-axis the region above the x-axis
and under y= sin xbetween x= 0 and x=π.
(c) The solid of rotation generated by rotating the same region as in (c) about the line
x=1.
pf3
pf4

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February 18, 2004 Math 112 — Exam I

Show all work clearly; an answer with no justifying computations may not receive credit (except in the “set up but do not evaluate” problems).

  1. (15 points) (a) Find the equation of the tangent line to the graph of x^2 + 2xy − y^2 + 4x = 5 at the point (1, 2). (b) Find (^) dxd (arctan(x/3)).
  2. (21 points) Find the following limits:

(a) lim x→π/ 2 −

(x −

π 2 ) tan^ x^ (b)^ xlim→ 2

x^2 − 2 x + 3 x^2 + 3 (c)^ xlim→ 1 x

1 /(x−1)

  1. (18 points) Set up but do not evaluate a definite integral that represents the volume of each of the following solids: (a) The base is the region bounded by the curves y = x^3 and x = y^2 , and the cross sections perpendicular to the x-axis are right triangles with height 2. (b) The solid of rotation generated by rotating about the x-axis the region above the x-axis and under y = sin x between x = 0 and x = π. (c) The solid of rotation generated by rotating the same region as in (c) about the line x = −1.
  1. (10 points) A liquid weighing 50 lbs/ft^3 is lifted from a cistern, from its surface to a level 40 feet above at a rate of 2 ft/sec. The vessel, which weighs 5 lb and is drawn by a rope of negligible weight, starts with 3 ft^3 of liquid, but it is leaking at a rate of .2 ft^3 /sec. Set up but do not evaluate a definite integral that represents the work (in foot-pounds) done in lifting the liquid.
  2. (15 points) Find the antiderivative

∫ sin 4x cos 3x dx, not by a trigonometric identity for sin mx cos nx, but by integration by parts (twice).

  1. (21 points) Evaluate the following integrals:

(a)

∫ sec^4 x tan^2 x dx (b)

∫ sin^4 x cos^2 x dx (c)

∫ cos(cos x) sin x dx

Some possibly useful formulas:

sin A cos B = 1 2

(sin(A + B) + sin(A − B))

sin A sin B = 1 2

(cos(A − B) − cos(A + B))

cos A cos B = 1 2

(cos(A − B) + cos(A + B))

sin 2A = 2 sin A cos A cos 2A = cos^2 A − sin^2 A = 2 cos^2 A − 1 = 1 − 2 sin^2 A

∫ csc x dx = ln | csc x − cot x| + C

∫ cot x dx = ln | sin x| + C

y − y 0 = m(x − x 0 )

Subtracting the last term from both ends and dividing both ends by − 7 /9, we get ∫ sin 4x cos 3x dx = −

7 sin 4x^ sin 3x^ −^

7 cos 4x^ cos 3x^ +^ C.

  1. (a) Use sec^2 x = tan^2 x + 1 and the substitution u = tan x (so that du = sec^2 x dx): ∫ (tan^2 x+1) tan^2 x sec^2 x dx =

∫ (u^4 +u^2 )du =

5 u

5 +^1

3 u

3 +C =^1

5 tan

(^5) x+^1 3 tan

(^3) x+C.

(b) Use sin^2 x = 12 (1 − cos 2x) and cos^2 x = 12 (1 + cos 2x), then cos^2 2 x = 12 (1 + cos 4x), and in the last term the Pythagorean relation cos^2 2 x = 1 − sin^2 2 x and then the substitution u = sin 2x (so that du = 2 cos 2x dx): ∫ sin^4 x cos^2 x dx =

∫ (1 − cos 2x)^2 (1 + cos 2x)dx

= 1 8

∫ (1 − cos 2x − cos^2 2 x + cos^3 2 x)dx

= 1 8

( x − 1 2

sin 2x − 1 2

∫ (1 + cos 4x)dx +

∫ (1 − sin^2 2 x) cos 2x dx

)

( x − 1 2

sin 2x − 1 2

(x +^1 4

sin 4x) +^1 2

∫ (1 − u^2 )du

)

( x − 1 2

sin 2x − 1 2

(x +^1 4

sin 4x) +^1 2

(u − 1 3

u^3 )

)

  • C

= 1 8

( x − 1 2

sin 2x − 1 2

(x +^1 4

sin 4x) +^1 2

(sin 2x − 1 3

sin^3 2 x)

)

  • C

= 1 8

x − 1 8

sin 4x − 1 6

sin^3 2 x

)

  • C.

(c) Let u = cos x; then du = − sin x dx, so the integral becomes

∫ cos u du = − sin u + C = − sin(cos x) + C.