Math 112 (Calculus I)
Final Exam Form A KEY
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The solutions to the math 112 calculus i final exam. It includes the calculations for finding limits, definite integrals, and derivatives, as well as the optimization problem for building a fence around an emu ranch. The document also covers topics such as logarithmic differentiation, implicit differentiation, and the relationship between the area and circumference of a circle.
Typology: Exams
2012/2013
1 / 4
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download Math 112 Calculus I Final Exam Solutions and more Exams Calculus in PDF only on Docsity!
Math 112 (Calculus I)
Final Exam Form A KEY
- (a) (4 points) Find lim x→∞
x^2 + 1 − x.
Solution:
lim x→∞
x^2 + 1 − x = lim x→∞
x^2 + 1 − x)(
x^2 + 1 + x) √ x^2 + 1 + x
= lim x→∞
x^2 + 1 + x
(b) (4 points) Find lim x→∞
x ln(x)
x^2
Solution:
lim x→∞
x ln(x)
x^2
= lim x→∞
ln x
x
= lim x→∞
1 x 1
using L’Hopital’s rule.
- (5 points) Compute the definite integral
1
x √ 9 + x^2
dx.
Solution:
Use u = 9 + x 2
. Then, du = 2x dx, and ∫ (^2)
1
x √ 9 + x^2
dx
10
u − 1 / 2 du = u 1 / 2 | 13 10 =^
- (5 points) You own an emu ranch, and want to build a rectangular fence around your herd.
One side of the fence must be built along the side of a river so the emu’s have access to fresh water. You want to enclose at least 1000 square meters. The fencing material costs 6 dollars per meter, but the fencing along the river costs 10 dollars per meter (to allow the emu to drink the water). What are the dimensions which minimize the cost of your fence? What is the cost?
Solution: Let x be the length along the river while y is the length perpendicular to it. The total cost is C = 16x + 12y.
Since the area is 1000, we have
xy = 1000, or y =
x
Hence,
C = 16x +
x
Notice
C ′ = 16 −
x^2
16 x^2 − 12000
x^2
A critical point is at 16 x 2 − 12000 = 0 or x 2 = 750.
Thus, x = 5
30 feet, and
y =
Notice this is a minimum because
C ′ =
x^3
which is positive for x > 0, and there is only one critical point in (0, ∞).
- (5 points) The area of a circle is increasing at a rate of 3 m/s
2
. When the area is π m 2 , how fast is the circumference growing?
Solution:
A = πr 2 ,
so dA
dt
= 2πr
dr
dt
Since the rate of change of the area is 3,
dr
dt
2 π
Since the circumference is C = 2πr,
dC
dt
= 2π
dr
dt
so dC
dt
- (5 points) The equation e x+y = y 2 implicitly defines y as a function of x. Find dy/dx.
Solution: (1 + y ′ )e x+y = 2yy ′
(e x+y − 2 y)y ′ = −e x+y
y ′ =
ex+y
2 y − ex+y^
- (5 points) Find the equation of the tangent line to the curve y = e x (x 2 + ln(x + 1) + 3) at the point (0, 3).
Solution:
y ′ = e x (x 2
- ln(x + 1) + 3) + e x (2x +
x + 1
y ′ (0) = 1 · 3 + 1 · 1 = 4.
y − 3 = 4x
y = 4x + 3