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TEST 2
Math 105 11/9/12 Name: (^) ︸ ︷︷ ︸ by writing my name I swear this work is my own
Read all of the following information before starting the exam:
- Show all work, clearly and in order, if you want to get full credit. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
- Circle or otherwise indicate your final answers.
- Please keep your written answers brief; be clear and to the point. I will take points off for rambling and for incorrect or irrelevant statements.
- This test has 8 problems and is worth 100 points, It is your responsibility to make sure that you have all of the pages!
- Good luck!
1. (3 points) Solve the following differential equation y′^ = 5y with initial condition y(0) = R.
y = Re^5 x
2. (14 points)
x f (x) g(x) j(x) f ′(x) g′(x) j′(x) -2 0 1 -1 3 2 1 -1 1 3 2 -1 3 0 0 2 1 1 2 -2 2 1 3 1 -1 0 3 1 2 -2 2 1 3 0 3 3 -1 1 -1 1 -2 2
a. (7 pts) H(x) = f (j(x)) + g(x^2 − 1). Find H′(2).
H′(x) = f ′(j(x)) · j′(x) + g′(x^2 − 1) · 2 x
H′(2) = 0 ∗ 3 + (−2) ∗ 4 = − 8
b. (7 pts) F (x) = g(x)^2 (x + 1)j(x)
. Find F ′(0).
F ′(x) =
2 g(x)g′(x)(x + 1)j(x) − (j(x) + (x + 1)j′(x))g(x)^2 (x + 1)^2 j(x)^2
F ′(0) =
2 ∗ 1 ∗ (−2)(1)(1) − [(1 + 1 ∗ 2) ∗ 1
3. (20 points) Find y′.
a. (10 pts) y =
sin^4 (x)(3x^4 − 2 x + 5)^3 e^3 x (x + 1)^2 (2x − 3)^5
using logarithmic differentiation.
y′^ =
4 cos(x) sin(x)
36 x^3 − 6 3 x^4 − 2 x + 5
x + 1
2 x − 3
sin^4 (x)(3x^4 − 2 x + 5)^3 e^3 x (x + 1)^2 (2x − 3)^5
b. (10 pts) y = 4
tan^2 (2x + 1) + 2cos(3x)^ − arcsin(3x^2 )
1 4
(tan^2 (2x + 1))−^3 /^4 · 2 tan(2x + 1) · sec^2 (2x + 1) · 2 + ln(2)2cos(3x)^ · (−3 sin(3x)) −
6 x √ 1 − 9 x^4
Or see that the original function can be simplified: y = 2
tan(2x + 1) + 2cos(3x)^ − arcsin(3x^2 ).
1 2
(tan(2x + 1))−^1 /^2 · sec^2 (2x + 1) · 2 + ln(2)2cos(3x)^ · (−3 sin(3x)) − 6 x √ 1 − 9 x^4
These are equivalent.
6. (16 points) Evaluate the following limits. Only use L’Hˆopital’s rule when appropriate. Show your
work!!
a. (8 pts) lim x→∞
(ln x)^2 x Type:
lim x→∞
(ln x)^2 x
= lim x→∞
2 ln(x) · (x−^1 ) 1
= lim x→∞
2 ln(x) x
= lim x→∞
2 x−^1 1
= lim x→∞
x
b. (8 pts) lim x→∞
k x
)x , for k a constant.
y = lim x→∞
k x
)x
ln(y) = lim x→∞ x ln
k x
, Type ∞ · 0
ln(y) = lim x→∞
ln
1 + kx
1 x
, Type
ln(y) = lim x→∞
ln
1 + kx
1 x
= lim x→∞
−kx−^2 1+ kx −x−^2
= lim x→∞
k 1 + kx
= k
y = ek
7. (3 points) Does f (x) =
5 x^3 + 6x − 1 2 x^3 + 10
have a horizontal asymptote? If so, where is it?
lim x→∞
5 x^3 + 6x − 1 2 x^3 + 10
, Type
x^ lim→∞
5 x^3 + 6x − 1 2 x^3 + 10 = (^) xlim→∞
15 x^2 + 6 6 x^2 = (^) xlim→∞
30 x 12 x
Yes, the function has a horizontal asymptote at y = 52.
8. (15 points) A farmer wishes to build a corral for his cows in the shape below (rectangle with semi-
circles on the ends). She has 1000ft of fencing and she won’t fence-in the straight edge along the water (just bolded sides). What is the maximal area for the corral?
Label Water side x and the top of the rectangle y. The objective is the area of the corral. The constraint is the 1000ft of fencing for the perimeter.
A = xy +
πy^2 4
1000 = x + πy
Therefore, x = 1000 − πy and A = (1000 − πy)y + πy
2
A = 1000y −
3 π 4 y^2
A′^ = 1000 −
6 π 4
y
Set the derivative to 0.
6 π 4 y = 1000 → y =
6 π
Check to see if this is a max or min.
A′′^ = −
6 π 4
Since the second derivative is always negative, then y = (^40006) π is a local maximum. Check for global maximums on the interval [0,318.3]. A(0) = 0 sq. ft., A(1000/π) = 79,577.47 sq. ft (this is just a circle), A( (^40006) π ∼=212.2) = 106,103 sq. ft. The maximal area is 106,103 sq. ft.
