CALCULUS  Check my answers :)
posted by Samantha on .
I'd appreciate corrections for the ones I get wrong. :) Thanks!
1. Determine which of the following is true for the function f(x) = x^3 +5x^2  8x +3.
I. f(x) has a relative minimum at x = 2/3.
II. f(x) has a relative maximum at x = 2/3.
III. f(x) has a zero at x = 2/3
I only
*II only
III only
I and III only
II and III only
2. Determine the equation of the normal line to y = x^2 + 5 at the point (2, 9)
y=1/4x + 17/4
*y=1/2x + 8
y=4x+1
y=4x34
y=1/4x + 19/2
3. If 12 ft^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
2 ft^3
4 ft^3
5 ft^3
*8.5 ft^3
9 ft^3
4. Find the tangent line approximation, L(x), of f(x)=x^2/3 at x = 8.
L(x) = 4/3x20/3
L(x) = 2/3x+8
L(x) = 4(x8)
L(x) = 1/3x + 4/3
*L(x) = 4/3x  8/3
5. A balloon is rising at a constant speed of 5 ft/sec. A boy is cycling along a straight road at a constant speed of 15 ft/sec. When he passes under the balloon, it is 5 feet above him. Approximately how fast is the distance between the boy and the balloon increasing 3 seconds after he has passed underneath it?
12 ft/sec
16 ft/sec
20 ft/sec
*25 ft/sec
30 ft/sec
6. A factory is manufacturing a rectangular storage container with an open top. The volume of the container is 10 ft^3, and the length of the base is twice the width. The material for the base costs $10 per square foot, and the material for the sides costs $6 per square foot. Find the cheapest cost to make the container, given the conditions.
$27.85
$46.19
*$87.24
$147.85
$163.54
7. The edge of a cube was found to have a length of 50 cm with a possible error in measurement of 0.1 cm. Based on the measurement, you determine that the volume is 125,000 cm^3. Use tangent line approximation to estimate the percentage error in volume.
0.6%
0.9%
*1.2%
1.5%
1.8%
8. An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft^3/min, find the rate of change of the depth of the water when the water is 10 feet deep.
0.007 ft/min
0.449 ft/min
0.018 ft/min
*0.051 ft/min
0.065 ft/min
9. For the function f(x)=Inx/x^2, find the approximate location of the critical point in the interval (0, 5).
(0.5, −2.773)
(1, 0)
(1.649, 0.184)
*(2, 0.173)
(0.778, −1.813)

1. Determine which of the following is true for the function f(x) = x^3 +5x^2  8x +3.
I. f(x) has a relative minimum at x = 2/3.
II. f(x) has a relative maximum at x = 2/3.
III. f(x) has a zero at x = 2/3
I only
*II only
III only
I and III only
II and III only
================================
f' = 3 x^2 + 10 x  8
zero at x = 2/3 and at a = 2
f" = 6 x + 10
curvature is POSITIVE at x = 2/3
the function is headed UP
That is a MINIMUM 
2. Determine the equation of the normal line to y = x^2 + 5 at the point (2, 9)
y=1/4x + 17/4
*y=1/2x + 8
y=4x+1
y=4x34
y=1/4x + 19/2
==================================
dy/dx = 2x
at x = 2, slope = 2*2 =4
so m of normal= 1/slope = 1/4
y = 1/4 x + b
9 = 1/4 *2 + b
9 = 1/2 + b
b = 19/2 so I say the LAST one 
3. If 12 ft^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
2 ft^3
4 ft^3
5 ft^3
*8.5 ft^3
9 ft^3
=================================
side length s and height h
A = 4 s h + s^2 = 12
4 s h = 12  s^2
h = (3/s)  s/(4)
V = s^2 h
V = 3 s  s^3/4
dV/ds = 3  (3/4)s^2
at max or min dV/ds = 0
so
3/4 s^2 = 3
s = 2
then find h
h = 3/2  2/4
h =1
volume = h s^2 = 1 (4) = 4 
I am getting bored

Well Damon, calculus isnt supposed to be.. interesting. Haha

Is number 4 the last one?

And 5 would be 25?

#4, the way you typed it, ...
y' = 2x/3
when x = 8, y' = 16/3, which is not found in any of your choices as the slope
if you meant , f(x) = x^(2/3)
then y' = (2/3)x^(1/3) = (2/3)(1/x^(1/3))
when x = 8
y' = (2/3)(1/2) = 1/3
the only one that has a slope of 1/2 is
L(x) = 1/3x + 4/3 
#5
At a time of t seconds after the boy passed under the balloon,
distance covered by boy = 15t
height of balloon = 5 + 5t
let the distance between them be d ft
d^2 = (5t+5)^2 + (15t)^2
2d dd/dt = 2(5t+5)(5) + 2(15t)(15)
dd/dt =( 5(5t+5) + 15(15t) )/d
when t = 3,
d^2 = 20^2 + 45^2
d = √2425
dd/dt = (5(20) + 15(45))/√2425
= 775/(5√97)
= appr 15.74
which is none of the answers unless it was rounded off to 16 ft/sec